PDA

View Full Version : Deal or No Deal, probabilities not like Let's make A Deal

Pages : [1] 2 3

justice
12-29-2005, 03:36 PM
In Let's Make A Deal [Spanish version Trato Hecho], contestant picks one of three doors; two doors have a goat, one has a NEW CAR.

Monte hall then reveals one of the two unchosen doors to show a goat.

Should you switch. This created quite stir in actuarial circles. Are there any interesting prob questions with this new show "Deal or No Deal"?

12-29-2005, 03:46 PM
What kind of car? The only person who can't figure this problem out is the World's Smartest Woman, Marilyn Von Savant.

joe2
12-29-2005, 03:51 PM
I don't see how it would make a difference, switching or not. unless of course you wanted a goat.

DW Simpson
12-29-2005, 06:07 PM
I don't see how it would make a difference, switching or not. unless of course you wanted a goat.

http://en.wikipedia.org/wiki/Monty_Hall_problem

The Monty Hall problem is a puzzle in game theory involving probability that is loosely based on the American game show Let's Make a Deal. The name comes from the show's host, Monty Hall. In this puzzle a player is shown three closed doors; behind one is a car, and behind each of the other two is a goat. The player is allowed to open one door, and will win whatever is behind the door. However, after the player selects a door but before opening it, the game host (who knows what's behind the doors) must open another door, revealing a goat. The host then must offer the player an option to switch to the other closed door. Does switching improve the player's chance of winning the car? The answer is yes — switching results in the chances of winning the car improving from 1/3 to 2/3.

violaactuary
12-29-2005, 06:19 PM
I want to know how to get the job of the actuary who is making the offers by phone. How many exams must you have?

DW Simpson
12-29-2005, 06:47 PM
I want to know how to get the job of the actuary who is making the offers by phone. How many exams must you have?

What?

violaactuary
12-30-2005, 03:00 PM
In the new show... there is an "actuary" "upstairs" from the stage proper who calls the stage proper and offers a take-this-money-and-go-home deal to the contestant after the required cases are opened.

Of course I was kidding about the job and exams, that is unless you can get me the job :tup:

Actuary321
12-30-2005, 03:31 PM
In the new show... there is an "actuary" "upstairs" from the stage proper who calls the stage proper and offers a take-this-money-and-go-home deal to the contestant after the required cases are opened.

Of course I was kidding about the job and exams, that is unless you can get me the job :tup:"actuary", I thought he was called the "banker."

louisdj_KSU
12-30-2005, 03:32 PM
Clearly the Monty Hall Problem doesn't exists in this show. The reason is that the contestest chooses the suitcase (door) to open throughout the show, not the show host. As a result, there are = chances of opening good things and bad things, leaving you with a 50/50 chance. In 'Lets Make A Deal', you were gauranteed to see a goat, and this was key to the 2/3 change, 1/3 stay probabilities. My only concern is that this assumes that all money values are indeed assigned to a suitcase before the show starts (that little speech about how some security personnel (?) filled the suitcases and no one at NBC even knows what's in them, at the beginning).

As for this show, I think the big 'puzzle' to look at would be the banker's deals. I only saw the show one night, so there's still lots to pick up for me. However, one peculiar thing stuck out, and that was that the offers rarely exceeded the expected values of the remaining possibilities. The most peculair of these would be when the remaining values are at both ends of the spectrum, with one end having way more than the other (1, 2000, 5000, 10,000, 100,000, 500,000 and 1 million (Have a feeling these aren't proper numbers; like I said, I've only seen it once.); and 0.01, 1, 2, 5, 10, 100, and 1 million. To me 2 important factors play a role with the deal. The first is the relationship with the expected value, mentioned above. The other is the probability that my suitcase is bigger than the deal. This should give y'all something to think about for the new year.

Actuary321
12-30-2005, 03:33 PM

louisdj_KSU
12-30-2005, 03:34 PM
"actuary", I thought he was called the "banker."

I wonder what his background in options is.

violaactuary
12-30-2005, 04:04 PM
"actuary", I thought he was called the "banker."
Maybe he is... I didn't get to see the show until the last two nights it was on, so I missed the introduction. My uncle told me they said it was an actuary which is why he mentioned it to me.

Wag, the Dog
12-30-2005, 05:30 PM
It is highly probable that the banker and Howie Mandel will soon be unemployed. This show is about as interesting as watching a stranger scratch an instant win lottery ticket. Good for about two, maybe three, watches.

Krieger
12-30-2005, 06:57 PM
It is highly probable that the banker and Howie Mandel will soon be unemployed. This show is about as interesting as watching a stranger scratch an instant win lottery ticket. Good for about two, maybe three, watches.

The show is coming back in March. It's already a hit in other countries. Very low probability the banker and Howie will be unemployed.

Krieger
12-30-2005, 06:59 PM
Maybe he is... I didn't get to see the show until the last two nights it was on, so I missed the introduction. My uncle told me they said it was an actuary which is why he mentioned it to me.

He's definitely called "The Banker."

renaissoxx
02-06-2006, 02:13 PM
That result is incorrect. Part of the information that is given when the host opens a goat door is that you did not choose that goat.

Here is their argument:

When you start you have a 2/3 chance of choosing a goat.

So
(1/3) * You chose goat 2, and the car is in the space you would switch to.
(1/3) * You chose goat 1, and the car is in the space you would switch to.
(1/3) * You chose the car, and switching would make you lose.

OH LOOK SWITCHING = GOOD.

But when he opens a door he eliminates the possibility that you chose goat 1 or 2 (whichever you didn't pick) and you are left with a 50/50 chance.

Also you might reason intuitively that there is nothing to differentiate this situation from one where you had 2 doors, 1 goat, 1 car. Nothing about starting with 3 doors actually gives you any information above and beyond this one. All you did was choose without knowing.

Gandalf
02-06-2006, 02:22 PM
:shake:

Westley
02-06-2006, 02:29 PM
renaissoxx, I think you should stick to your efforts at trying to increase math learning efficiency.

Alto Reed on a Tenor Sax
02-06-2006, 02:32 PM
That result is incorrect. Part of the information that is given when the host opens a goat door is that you did not choose that goat.

Here is their argument:

When you start you have a 2/3 chance of choosing a goat.

So
(1/3) * You chose goat 2, and the car is in the space you would switch to.
(1/3) * You chose goat 1, and the car is in the space you would switch to.
(1/3) * You chose the car, and switching would make you lose.

OH LOOK SWITCHING = GOOD.

But when he opens a door he eliminates the possibility that you chose goat 1 or 2 (whichever you didn't pick) and you are left with a 50/50 chance.

Also you might reason intuitively that there is nothing to differentiate this situation from one where you had 2 doors, 1 goat, 1 car. Nothing about starting with 3 doors actually gives you any information above and beyond this one. All you did was choose without knowing.

Another way of putting this reasoning: There is one goat that the host is going to show. You have a 50% chance of choosing the other goat, and a 50% chance of choosing the car. The goats are interchangeable, so this does not alter the problem.

It's like my wife says, when I buy a Powerball ticket, there are only two options, I win, or I don't. So my odds are 50-50, right?

Right?

JMO
02-06-2006, 03:22 PM
That result is incorrect. Part of the information that is given when the host opens a goat door is that you did not choose that goat.

Your analysis assumes that Monty Hall is following a particular strategy, namely that he will choose a door with a goat. How do you know that? Think what you would do if sometimes the door he opens shows a goat. :oyh:

Gandalf
02-06-2006, 04:56 PM
Your analysis assumes that Monty Hall is following a particular strategy, namely that he will choose a door with a goat. How do you know that? Think what you would do if sometimes the door he opens shows a goat. :oyh:
I believe you mean Think what you would do if sometimes the door he opens does not show a goat.

renaissoxx
02-06-2006, 05:54 PM
It's like my wife says, when I buy a Powerball ticket, there are only two options, I win, or I don't. So my odds are 50-50, right?

Right?

Are you being serious? What I meant was, hes going to show a goat. Given that its not the one hes shown, (prob 2/3) whats the probability that its in either door? (1/3)/(2/3) which is .5

I believe you mean

Didn't he say Monty always chose a door with a goat?

Im really not interested in any typical sheepish responses to someone diagreeing with the commonly accepted answer to a problem. Agreement with the commonly accepted is not an indication of intelligence. It only serves to demonstrate that you don't have the courage to challenge what other people think and do regardless of weather or not it is correct, or at best that you have leadership skills (if your the person who came up with a commonly accepted solution)

This is because the only knowledge a human being is capable of having is the absence of dissenting information. This means instead of trying to block out or demotivate dissenting opinions you pursue them as a source of information. Furthermore if someone had an understanding that encompassed any counter claims like mine, then they would be able to deal with them. If you don't have the guts to challenge a claim like the 2/3 one until you understand it well enough to objectively deal with any disagreements, then you dont have the guts to come to me with an attitude when I do.

I am very familiar with problems of this nature. It is the logic behind how they solved the problem that I have a problem with. If you know you didn't choose goat 1 or 2 because you are looking at it through the door Monty opened, then how can you still look at it like they say on Wiki. You now KNOW you did not choose that goat.

I see a general fallacy in the explanation given by wiki anyways. It does not attempt to specify and deal with the intuition that drives the belief it is 50/50%. Instead it simply gives a superficially appealing argument, and then takes a "prove me wrong" approach. When there is 2 intuitively appealing explanations like that, one of them has to explain why the other one is appealing and yet is still wrong to the point where it is no longer confusing. Instead this approach simply tries to make the person pay more attention to their explanation. On the other hand I can see why this explanation is appealing and yet still wrong.

Choose any of the explanations and I will explain what is wrong with it. Besides the one i already explained, the pictoral one can be debunked for one reason like this. If the 2 non chosen doors have a 2/3 probability because there are 2 of them, then the 2 doors that are not revealed also have 2/3 probability because there are 2 of them with 1/3 probability each. But how can the same door have a 2/3 probability when its grouped one way and a 1/3 probability when its grouped another? Its a self defeating argument.

Westley
02-06-2006, 06:51 PM
Choose any of the explanations and I will explain what is wrong with it.
I choose this one, since it's correct.

Here is their argument:

When you start you have a 2/3 chance of choosing a goat.

So
(1/3) * You chose goat 2, and the car is in the space you would switch to.
(1/3) * You chose goat 1, and the car is in the space you would switch to.
(1/3) * You chose the car, and switching would make you lose.

OH LOOK SWITCHING = GOOD.
Good luck, kid. You'll need it.

Actuary321
02-06-2006, 07:11 PM
What is wrong with their explaination of the 100 doors? Seems like they are trying to prove this by induction. Start with N=3 then the probablility of winning by swithing is (N-1)/N = 2/3 Now let N=4 on to N=100 and the problem with the 100 doors. The probablility of winning with 100 doors is 99/100. That seems clear.

The only problem would be that you have to open more doors in the one with 100. But the crux is that all but one door you did not pick is opened to show a goat.

Gandalf
02-06-2006, 08:22 PM
Didn't he say Monty always chose a door with a goat?

Agreement with the commonly accepted is not an indication of intelligence. It only serves to demonstrate that you don't have the courage to challenge what other people think and do regardless of weather or not it is correct, or at best that you have leadership skills (if your the person who came up with a commonly accepted solution).
:shake: Agreement with the correct answer is more an indication of intelligence.

You seem to agree that before Monty shows you a goat, there is a 100% chance there is a goat behind at least one of the other doors and there is only a 1/3 chance you have the right door. 100% of the time Monty can show you a goat. 100% of the time Monty will show you a goat (as JMO said, this is a key assumption). Yet when Monty confirms that there is a goat behind at least one of the doors you didn't choose - a fact you knew when you concluded you had a 1/3 chance of winning, all of a sudden your odds on having the right door increase. :shake:

Dr T Non-Fan
02-06-2006, 09:06 PM
renaissoxx, I think you should stick to your efforts at trying to increase math learning efficiency.
Those who can't do.....

Dr T Non-Fan
02-06-2006, 09:29 PM
I think this explanation is apt, similar to what Actuary321 is stating:

Consider a 52-card deck. One of the cards is the Ace of Spades.
1. You are to guess which one is the Ace of Spades. You choose one, but are not shown the card.
2. I take away 50 of the unchosen cards, leaving you now with a choice between two cards, the one you originally chose, and one unchosen. I say, "One of these is the Ace of Spades, and you are allowed to switch." Should you switch?

What has happened is that your original choice has a 1/52 probability of being the Ace of Spades and still does, while the other 51 cards have a combined 51/52 probability of being the Ace of Spades. When the 50 cards are removed, the probability that the unchosen card remaining is the Ace of Spades is still 51/52, since the probability that you've chosen the Ace of Spades has not changed. You have been given a lot of information about the unchosen card, but information about your chosen card has not changed.

Think about it in practice: you choose one card, a low probability that it's the Ace of Spades. The dealer looks through a pack of cards for the Ace of Spades, an average of 51 out of 52 times -- and the 1 time in 52 when you actually have it, he picks a random card-- drops it on the table and says, "One of these cards is the Ace of Spades."

Same explanation, using the three door situation: your first choice has 1/3 probability of being correct. The rest of the doors combine for a probability of 2/3 being correct. The two unchosen doors are collapsed into one choice, but the 2/3 probability remains the same.

OK, one last look at this, using the deck of cards again:
1. The goal is NOT to choose the Ace of Spades.
2. Pick one card from the deck but don't look at it.
3. I discard 50 unchosen cards, take the remaining one card, put it on the table and declare, "One of these cards is NOT the Ace of Spades and you're allowed to switch." Should you switch?
I leave it to the reader to explain.

OK, really, THE last attempt to explain this:

Consider a 52-card deck. One of the cards is the Ace of Spades.
1. You are to guess which one is the Ace of Spades. You choose one, but are not shown the card.
2. I don't take away any of the unchosen cards, leaving you now with a choice between your already chosen card, and the rest of the deck. "Choose the set of cards in which the Ace of Spade is an element." Should you switch?

The real problem with seeing how the three-choice works is that 3 is so close to 2, while 52 (in my example) is very far away from 2.

Dr T Non-Fan
02-06-2006, 09:50 PM
But how can the same door have a 2/3 probability when it's grouped one way and a 1/3 probability when it's grouped another? It's a self defeating argument.
How can your chosen door have a 1/3 probability when it's grouped one way and then 1/2 probability when it's grouped another? Have you learned something about the contents of your chosen door when Monte Hall shows you a goat behind one of the unchosen doors? You know he was going to show you a goat. It's not as if there wasn't a goat to show you.

Here's what this contest isn't:
"Three doors, two with goats, and one with a car. Now, before you choose, I'm going to eliminate one of the doors with a goat behind it. Now, choose between the two remaining doors."

Westley
02-06-2006, 10:05 PM
Here's what this contest isn't:
"Three doors, two with goats, and one with a car. Now, before you choose, I'm going to eliminate one of the doors with a goat behind it. Now, choose between the two remaining doors."

You actually believe this; this is because the only knowledge a human being is capable of having is the absence of dissenting information.

Dr T Non-Fan
02-06-2006, 10:31 PM
You actually believe this; this is because the only knowledge a human being is capable of having is the absence of dissenting information.
r.n.

Alto Reed on a Tenor Sax
02-07-2006, 01:46 AM
Didn't he say Monty always chose a door with a goat?

Im really not interested in any typical sheepish responses to someone diagreeing with the commonly accepted answer to a problem. Agreement with the commonly accepted is not an indication of intelligence. It only serves to demonstrate that you don't have the courage to challenge what other people think and do regardless of weather or not it is correct, or at best that you have leadership skills (if your the person who came up with a commonly accepted solution)

This is because the only knowledge a human being is capable of having is the absence of dissenting information. This means instead of trying to block out or demotivate dissenting opinions you pursue them as a source of information. Furthermore if someone had an understanding that encompassed any counter claims like mine, then they would be able to deal with them. If you don't have the guts to challenge a claim like the 2/3 one until you understand it well enough to objectively deal with any disagreements, then you dont have the guts to come to me with an attitude when I do.

Gentlemen, I present you with someone who has never sat at home with a friend and actually EXPERIMENTED with the Monty Hall problem.

I don't care what kind of sophistry you have at your disposal, no one who has actually condicted a few dozon or so trials of this problem, and seen their actual result approach more and more closely to a perfect 2/3, can still believe that 1/2 is the correct answer.

JMO
02-07-2006, 07:23 AM
(as JMO said, this is a key assumption)

Actually, what I meant to say was:
Think what you would do if sometimes the door he opens shows the car.

Swiper
02-07-2006, 09:25 AM
I'll try too. Two doors have a goat and one has a car. You pick a door and have a 1/3 chance of being correct. Monty then opens a door that does not have the car (this is the key assumption).

Assume the car is behind door 1. If you pick door 1 Monty can open EITHER doors 2 or 3. If you pick door 2, Monty can ONLY open door 3 (since he will not open the door with the car). Likewise, if you pick door 3 he can ONLY open door 2.

Therefore, There are three steps with the following probabilities

1. You pick a door (each having 1/3 proability of being correct)

2. Monty opens a door
50% chance of opening 2 and 50% chance of opening 3 if you pick 1
100% chance of opening 3 if you pick 2
100% chance of opening 2 if you pick 3

3. You keep your door or you switch

Here are the probabilities of the different scenarios when you switch doors:

You pick door 1, Monty opens door 2, you switch to 3 = LOSE
1/3 * 1/2 = 1/6

You pick door 1, Monty opens door 3, you switch to 3= LOSE
1/3 * 1/2 = 1/6

You pick door 2, Monty opens door 3, you switch to 1 = WIN
1/3 * 1 = 1/3

You pick door 3, Monty opens door 2, you switch to 1 = WIN
1/3 * 1 = 1/3

As you can see, the probabilities of the WINs = 2/3 when you switch.

Gandalf
02-07-2006, 09:38 AM
I know one trial won't prove anything, but let's try an experiment: someone choose 2, 4 or 5.

Westley
02-07-2006, 09:44 AM
I know one trial won't prove anything, but let's try an experiment: someone choose 2, 4 or 5.4

Gandalf
02-07-2006, 09:50 AM
Good guess?

Your goal is to find the "R". There were two "L"s, worth 0. "R" has no higher monetary value, but will give you bragging rights if you can find it.

1 2 3 4 5
T ? O ? L

To keep the game comparable to the original, I hid the R in position 2, 4 or 5. (and revealed 1 & 3 now; you had to be faced with a 1 in 3 chance at inception). You guessed 4. I revealed an L in 5.

Do you want to keep position 4 as the "R", or switch to 2?

Westley
02-07-2006, 10:14 AM
Do you want to keep position 4 as the "R", or switch to 2?I will switch to 2. :lol:

ShakeNBakes
02-07-2006, 10:17 AM
Good guess?

Your goal is to find the "R". There were two "L"s, worth 0. "R" has no higher monetary value, but will give you bragging rights if you can find it.

1 2 3 4 5
T ? O ? L

To keep the game comparable to the original, I hid the R in position 2, 4 or 5. (and revealed 1 & 3 now; you had to be faced with a 1 in 3 chance at inception). You guessed 4. I revealed an L in 5.

Do you want to keep position 4 as the "R", or switch to 2?:lol: I'd like to solve the puzzle, Pat.

Gandalf
02-07-2006, 10:17 AM
Right you are. And to prove that you would have lost to keep your door, I'll open it too.

T R O L L

Thanks for playing.

Gandalf
02-07-2006, 01:04 PM
Gentlemen, I present you with someone who has never sat at home with a friend and actually EXPERIMENTED with the Monty Hall problem.

I don't care what kind of sophistry you have at your disposal, no one who has actually condicted a few dozon or so trials of this problem, and seen their actual result approach more and more closely to a perfect 2/3, can still believe that 1/2 is the correct answer.
Excellent point, Alto Reed. So here, renaissoxx, is your own data set to experiment with:

"Correct" is the door where the car is. "Door 1" is the door Monty will open if you choose Door 1; similarly for Door 2. You make your initial guesses for each of the 99 trials BEFORE looking at the "Correct" column. Then Monty opens the indicated door. What a surprise: it's a goat. You believe you'll win 1/2 the time by not changing, so don't change. Report back how many of the 99 trials you originally had the right door, and how many after not changing you had the right door.

I think we all agree that the initial guess should be right about 1/3 of the time, so I'll count the experiment as typical if you are 27 to 39 right (of 99) before Monty opens the door.

If your theory is right, you should win at least 50%, or at least 49, after seeing a goat and not changing. Of course, on any given try you might not win that often. So I'll concede your theory is right if you win at least 40 of 99 in a typical experiment (i.e., 27 to 39 right before the goat is shown).

Play as often as you like. Let me know if you ever win at least 40 with 27-39 right initially. I'll trust you; just report what your initial guess was on each trial.

Trial Correct Door 1 Door 2 Door 3

1 1 3 3 2
2 1 3 3 2
3 3 2 1 2
4 1 3 3 2
5 3 2 1 2
6 1 3 3 2
7 2 3 1 1
8 2 3 1 1
9 3 2 1 1
10 2 3 1 1
11 1 3 3 2
12 1 3 3 2
13 3 2 1 2
14 2 3 1 1
15 3 2 1 1
16 1 2 3 2
17 2 3 1 1
18 2 3 1 1
19 1 3 3 2
20 2 3 1 1
21 1 3 3 2
22 1 3 3 2
23 3 2 1 2
24 3 2 1 1
25 3 2 1 1
26 2 3 3 1
27 1 2 3 2
28 1 2 3 2
29 1 2 3 2
30 2 3 1 1
31 1 3 3 2
32 2 3 3 1
33 3 2 1 1
34 2 3 1 1
35 2 3 3 1
36 3 2 1 2
37 3 2 1 1
38 2 3 3 1
39 1 3 3 2
40 1 2 3 2
41 3 2 1 1
42 2 3 3 1
43 1 3 3 2
44 2 3 1 1
45 2 3 1 1
46 2 3 1 1
47 3 2 1 1
48 2 3 1 1
49 2 3 1 1
50 2 3 3 1
51 1 2 3 2
52 2 3 3 1
53 1 2 3 2
54 3 2 1 2
55 1 2 3 2
56 3 2 1 1
57 1 2 3 2
58 2 3 1 1
59 1 2 3 2
60 3 2 1 1
61 2 3 3 1
62 2 3 1 1
63 2 3 3 1
64 3 2 1 2
65 3 2 1 2
66 3 2 1 2
67 3 2 1 2
68 2 3 3 1
69 1 2 3 2
70 1 3 3 2
71 1 3 3 2
72 3 2 1 2
73 2 3 3 1
74 1 2 3 2
75 3 2 1 2
76 2 3 1 1
77 3 2 1 1
78 3 2 1 1
79 3 2 1 1
80 1 2 3 2
81 3 2 1 1
82 1 3 3 2
83 3 2 1 1
84 3 2 1 1
85 3 2 1 2
86 1 2 3 2
87 1 2 3 2
88 2 3 1 1
89 3 2 1 1
90 2 3 3 1
91 1 2 3 2
92 2 3 3 1
93 3 2 1 1
94 2 3 3 1
95 2 3 1 1
96 1 3 3 2
97 3 2 1 2
98 1 2 3 2
99 3 2 1 1

thtevie
02-07-2006, 01:28 PM
Excellent point, Alto Reed. So here, renaissoxx, is your own data set to experiment with:

"Correct" is the door where the car is. "Door 1" is the door Monty will open if you choose Door 1; similarly for Door 2. You make your initial guesses for each of the 99 trials BEFORE looking at the "Correct" column. Then Monty opens the indicated door. What a surprise: it's a goat. You believe you'll win 1/2 the time by not changing, so don't change. Report back how many of the 99 trials you originally had the right door, and how many after not changing you had the right door.

I think we all agree that the initial guess should be right about 1/3 of the time, so I'll count the experiment as typical if you are 27 to 39 right (of 99) before Monty opens the door.

If your theory is right, you should win at least 50%, or at least 49, after seeing a goat and not changing. Of course, on any given try you might not win that often. So I'll concede your theory is right if you win at least 40 of 99 in a typical experiment (i.e., 27 to 39 right before the goat is shown).

Play as often as you like. Let me know if you ever win at least 40 with 27-39 right initially. I'll trust you; just report what your initial guess was on each trial.

WTMTOYH

JMO
02-07-2006, 01:40 PM
WTMTOYH
Huh?

Alto Reed on a Tenor Sax
02-07-2006, 02:42 PM
Huh?

Without goolgling it, I'll take a stab in the dark and say that stands for "Way Too Much Time On Your Hands."

Which, incidentally, is also the answer to a question from another thread, "Why is actuary rated the number one job?"

Ailing Factuary
02-07-2006, 05:06 PM
Anyone that cannot understand the correct answer to that stupid let's make a deal goat problem, after it is explained to them, should not be an actuary.

Dr T Non-Fan
02-07-2006, 05:10 PM
Anyone that cannot understand the correct answer to that stupid let's make a deal goat problem, after it is explained to them, should not be an actuary.
[as renny] Anyone who cannot explain the correct answer should not be an actuary.[/as renny]

JMO
02-08-2006, 07:19 AM
Anyone that cannot understand the correct answer to that stupid let's make a deal goat problem, after it is explained to them, should not be an actuary.
But what if Monty only opens a door if you have already chosen the car? That would be HIS best (game theory) strategy, if he wanted you to get his goat.:horse:

renaissoxx
02-08-2006, 08:09 AM
I choose this one, since it's correct.

Good luck, kid. You'll need it.

1/3 You choose goat 1
1/3 You choose goat 2
1/3 You choose car

Monty chooses goat 1

The top one is now eliminated, and you are left with 1/3 goat 2, 1/3 car, which means 1/2 chance each given you didnt choose goat 1.
Didn't I already say that? If your going to bring that up again attack my counter argument.

What is wrong with their explaination of the 100 doors? Seems like they are trying to prove this by induction. Start with N=3 then the probablility of winning by swithing is (N-1)/N = 2/3 Now let N=4 on to N=100 and the problem with the 100 doors. The probablility of winning with 100 doors is 99/100. That seems clear.

The only problem would be that you have to open more doors in the one with 100. But the crux is that all but one door you did not pick is opened to show a goat.

That problem with that explanation is the same as the problem with the original. It just also happens to contain a straw man version of the argument against their result.

When you choose 1 out of 100 doors, you probably chose a goat. But when he reveals all the doors he shows you didn't choose any of the other 98 goats, see the above response. The illusion is that your choosing a door changes the probability that the car is behind it from when after all but 2 doors are eliminated and there would be a 50/50 chance.

:shake: Agreement with the correct answer is more an indication of intelligence.

You seem to agree that before Monty shows you a goat, there is a 100% chance there is a goat behind at least one of the other doors and there is only a 1/3 chance you have the right door. 100% of the time Monty can show you a goat. 100% of the time Monty will show you a goat (as JMO said, this is a key assumption). Yet when Monty confirms that there is a goat behind at least one of the doors you didn't choose - a fact you knew when you concluded you had a 1/3 chance of winning, all of a sudden your odds on having the right door increase. :shake:

If humans had some god like connection to the correct answer all the time, then perhaps your statement might have more meaning to me. But is it not the case that the only knowledge a human is capable of having is the abscence of information counter to what you believe? And if that is the case, is it not true that you must FACE any information that might be counter to what you believe rather than being in denial of it, attempting to squelch it, etc? Although people often assume someone who disagrees with them is wrong for some reason, how can you know you are not thinking of a straw man argument of what they are saying? You cannot until they admit otherwise. Something which insulting them makes it harder for them to do. Many people disagree on many things and think everyone who disagrees with them are stupid and these disagreements cause many problems. This type of behavior is at the root of that problem. Not "everyone being stupid" which of course cannot be true if everyone with different belief about ethics etc claims it about everyone who disagrees with him.

Yes they increase. You knew monty would show you a goat, but you didn't know which goat. Before Monty opens his door, you know your door can contain Goat 1, Goat 2, or the car. After you know your door can contain only Goat 2 or the car. Goat 2 and the car both had an equal chance to be at your door at the beginning, so did Goat 1 (which you now know isn't)

I can just as easily ask how your choosing a door changes the probability that the car is not behind it from a situation where you come on the scene after a goat has been revealed. You didn't actually do anything or gain any different information from this situation? Try directly dealing with what I am saying like I try dealing with what you are saying. That way even if I ultimately turn out to be wrong, a better explanation of the problem has been created.

SharksFan08
02-08-2006, 08:23 AM
Just do us all (including yourself) a favor: get two black aces and a red queen, sit down with a coworker or friend, and do some trials.

Although maybe don't do it with a coworker, because he/she probably understands this problem already and it will just make you look bad at work...

Westley
02-08-2006, 08:39 AM
1/3 You choose goat 1
1/3 You choose goat 2
1/3 You choose car

Monty chooses goat 1
You're assuming that I didn't choose goat 1 - I get to choose before Monty does. Which is a pretty stupid assumption for you to make, since I will, in the long run, choose it 1/3 of the time.

Westley
02-08-2006, 08:43 AM
The illusion is that your choosing a door changes the probability that the car is behind it from when after all but 2 doors are eliminated and there would be a 50/50 chance.The choosing the door is NOT what changes it - that's what everybody here but you understands. The change is because Monty knows which door has the car, and he is intentionally showing you a door that he knows don't have the car. That's what changes the probability.

Try this: there are two doors, and there's a 2/3 chance that there's a car behind one of them. Do you want to guess at which one it is, or would you rather I open one - which I promise you will not have the car - and then you guess?

Gandalf
02-08-2006, 08:45 AM
But what if Monty only opens a door if you have already chosen the car? That would be HIS best (game theory) strategy, if he wanted you to get his goat.:horse:
Unless this is a secret strategy by Monty, then it is no better for him than the original. In response, the player's action becomes:
Never switch if Monty opens a door.
Always switch if Monty doesn't open a door.
Player wins the 1/3 where Monty opens the door and you keep the car.
Player wins 1/2 of the other 2/3: player knows he must switch, and guesses right half of those.

If Monty is always going to open a door if the player has started with a car, the best Monty can do is also open a door half of the times the player started with a goat (and in those cases always show the other goat.) Player wins 50%.

Best Monty can do is never open any doors, but that defeats the purpose of the game. Another good strategy for Monty is to find contestants like renaissox.

Westley
02-08-2006, 08:45 AM
If humans had some god like connection to the correct answer all the time, then perhaps your statement might have more meaning to me. But is it not the case that the only knowledge a human is capable of having is the abscence of information counter to what you believe? And if that is the case, is it not true that you must FACE any information that might be counter to what you believe rather than being in denial of it, attempting to squelch it, etc? Although people often assume someone who disagrees with them is wrong for some reason, how can you know you are not thinking of a straw man argument of what they are saying? You cannot until they admit otherwise. Something which insulting them makes it harder for them to do. Many people disagree on many things and think everyone who disagrees with them are stupid and these disagreements cause many problems. This type of behavior is at the root of that problem. Not "everyone being stupid" which of course cannot be true if everyone with different belief about ethics etc claims it about everyone who disagrees with him.And, posting a bunch of gibberish that you read in a philosophy book somewhere doesn't change anything either.

Gandalf
02-08-2006, 08:52 AM
Also don't forget what I said about the "prove me wrong" tactic (assuming this doesn't convince you).
How about responding to my post 40 in the thread? It's all laid out for you. You can even manipulate the results greatly. My only restriction is that you don't start by picking the correct door more than 39 times out of 99 (or less than 27, but any with less than 27 would also work for me). I'll let you choose which 39 you start with the car. Even letting you choose your 39, I don't require you to win 1/2 after never changing, all you need to do is win 40 of 99 to convince me.

Or are you going to claim those 99 games were carefully constructed? :shake:

renaissoxx
02-08-2006, 11:03 AM
You're assuming that I didn't choose goat 1 - I get to choose before Monty does. Which is a pretty stupid assumption for you to make, since I will, in the long run, choose it 1/3 of the time.

When are you making the decision? After Monty chooses. There is no assumption being made here. You have a 100% gaurantee that you didn't choose the one Monty chose at the time you make your final decision.

Westley
02-08-2006, 11:10 AM
When are you making the decision? After Monty chooses. There is no assumption being made here. You have a 100% gaurantee that you didn't choose the one Monty chose at the time you make your final decision.But you assumed that Monty chose goat 1 - that's your stated assumption. I choose first. After I choose, then Monty gets to make a choice (which is why the odds change, because he provides me new information), and then I make my final choice. I choose first. Monty cannot choose goat 1 if that's the one I chose, which I will 1/3 of the time. Which means that your stated assumption is wrong 1/3 of the time.

Gandalf
02-08-2006, 11:14 AM
How long does it take you to play those 99 games I laid out for you?

Don't like my 99? This should give you even better odds. You create the 99 games. You decide where the car is on each. You decide which of the other doors Monty will open for each possible choice by the player (always showing a goat). You decide all the initial choices for the player. Then the player sees a goat, and doesn't switch because there is no advantage (under your theory) to switching.

The only restriction is that your player's initial guesses can't be too good: no more than 39 correct guesses. Show me under these conditions, completely under your control, how the player can win even 40 of 99.

Are you skurd? :tfh:

Actuary321
02-08-2006, 11:15 AM
I think I finally figured out what ren believes. He believes that the game begins AFTER Monte opens the door. That the act of opening a door with a goat provides no information at all.

That in reality the game is that there are two closed doors and one open one. The open one has a goat and the other 2 have a goat and a prize. Now pick from the two closed doors. This is the 50-50 that ren is looking at.

That problem would imply that Monte always has the choice to open either door with the goat. But he doesn't. If you first chose a door with a goat then Monte is restricted in his choice of doors. This restriction (loss of a degree of freedom) would imply that the result does not have the probability that would be assumed from looking at the problem without that additional information.

Gandalf
02-08-2006, 11:20 AM
ren: Since it's completely under your control, you can arrange it so that in all 99 Monty always displays Goat 1. (In some cases, he wouldn't display Goat 1 with a different player initial door choice, but you can force the player never to choose the Goat 1 door.) What more could you ask?

Westley
02-08-2006, 11:21 AM
I think I finally figured out what ren believes. He believes that the game begins AFTER Monte opens the door. That the act of opening a door with a goat provides no information at all.

That in reality the game is that there are two closed doors and one open one. The open one has a goat and the other 2 have a goat and a prize. Now pick from the two closed doors. This is the 50-50 that ren is looking at.

That problem would imply that Monte always has the choice to open either door with the goat. But he doesn't. If you first chose a door with a goat then Monte is restricted in his choice of doors. This restriction (loss of a degree of freedom) would imply that the result does not have the probability that would be assumed from looking at the problem without that additional information.
Yes, that's exactly it. I figured this out a while ago, but didn't explain it nearly as well as you. DTNF explained it like 30 posts ago, but ren ignored him and then wrote about how DTNF is a sheep for believing what everybody else believes.

Gandalf
02-08-2006, 11:23 AM
I think I finally figured out what ren believes. He believes that the game begins AFTER Monte opens the door. That the act of opening a door with a goat provides no information at all.

That in reality the game is that there are two closed doors and one open one. The open one has a goat and the other 2 have a goat and a prize. Now pick from the two closed doors. This is the 50-50 that ren is looking at.

That problem would imply that Monte always has the choice to open either door with the goat. But he doesn't. If you first chose a door with a goat then Monte is restricted in his choice of doors. This restriction (loss of a degree of freedom) would imply that the result does not have the probability that would be assumed from looking at the problem without that additional information.
I disagree. If he is reading the thread at all, he knows that isn't how the game works. He thinks the game is equivalent to that; he doesn't think that is the game.

renaissoxx
02-08-2006, 11:40 AM
The choosing the door is NOT what changes it - that's what everybody here but you understands. The change is because Monty knows which door has the car, and he is intentionally showing you a door that he knows don't have the car. That's what changes the probability.

Try this: there are two doors, and there's a 2/3 chance that there's a car behind one of them. Do you want to guess at which one it is, or would you rather I open one - which I promise you will not have the car - and then you guess?

Anyways about your claim... Heres a sample space of your and monty's choice that I think shows up (or something similar) on the wiki site.

1/3 You choose Goat 1, Monty Goat 2
1/3 You choose Goat 2, Monty Goat 1
1/6 You choose Car, Monty Goat 1
1/6 You choose Car, Monty Goat 2

I think it is obvious that this is a sample space created at the beginning of the problem. If you were making a decision about what to do at the beginning of the problem then this would be the one to use right. But here that is a logical contradiction. The problem forces you to make a decision after seeing what Monty reveals. How can that sample space at that point contain a 1/3 chance to see the goat Monty reveals? That makes no sense. You need a new sample space at that point. The new sample space is considered given that you did not choose Monty's goat.

(1/3)/(2/3) = .5 Goat 1
(1/3)/(2/3) = .5 Car

What other information are you claiming Monty has revealed?

Actuary321
02-08-2006, 11:43 AM
I disagree. If he is reading the thread at all, he knows that isn't how the game works. He thinks the game is equivalent to that; he doesn't think that is the game.Sorry, I think you are correct about ren.

Westley
02-08-2006, 11:53 AM
The problem forces you to make a decision after seeing what Monty reveals. How can that sample space at that point contain a 1/3 chance to see the goat Monty reveals? It doesn't. Because Monty has now given you more information regarding the two doors that you didn't initial choose. But, he has given you no additional information about the one door you did choose. The fact that you don't consider the additional information he chose is why you keep getting te problem wrong.

You need a new sample space at that point.Yes.

The new sample space is considered given that you did not choose Monty's goat.No, because that's not what happened. It's not that I did not choose Monty's goat. I never had a chance to choose Monty's goat, because he chose after I did. I made my choice, and then he chose. So, your statement "given that you did not choose Monty's goat" makes no sense - I never could have chosen Monty's goat, it's not possible, and for you to state a conditional probability based on that is indisputably wrong. The fact that I "did not choose Monty's goat" provides no additional information, because that has a 100% certainty of happening because of the timing of the decisions. Therefore, in the sense of probabilities, there is no additional information given by the fact that I did not choosee Monty's goat.

The conditional is on Monty providing me information about the two doors I didn't pick.

But, I'm still waiting for you to explain what's wrong with this one:

Originally Posted by renaissoxx
Here is their argument:

When you start you have a 2/3 chance of choosing a goat.

So
(1/3) * You chose goat 2, and the car is in the space you would switch to.
(1/3) * You chose goat 1, and the car is in the space you would switch to.
(1/3) * You chose the car, and switching would make you lose.

OH LOOK SWITCHING = GOOD.

You tried oncec, and used an assumption that is clearly wrong. I'll let you try again, if you want.

Gandalf
02-08-2006, 11:53 AM
Anyways about your claim... Heres a sample space of your and monty's choice that I think shows up (or something similar) on the wiki site.

1/3 You choose Goat 1, Monty Goat 2
1/3 You choose Goat 2, Monty Goat 1
1/6 You choose Car, Monty Goat 1
1/6 You choose Car, Monty Goat 2

I think it is obvious that this is a sample space created at the beginning of the problem. If you were making a decision about what to do at the beginning of the problem then this would be the one to use right. But here that is a logical contradiction. The problem forces you to make a decision after seeing what Monty reveals. How can that sample space at that point contain a 1/3 chance to see the goat Monty reveals? That makes no sense. You need a new sample space at that point. The new sample space is considered given that you did not choose Monty's goat.

(1/3)/(2/3) = .5 Goat 1
(1/3)/(2/3) = .5 Car

What other information are you claiming Monty has revealed?
You are finally getting close. Try creating the new sample space by throwing out the parts of the original sample space that haven't occurred: You didn't choose Goat 1, and Monty didn't show Goat 2.

So 1/2 the original cases are left. Originally, the car was in your door in 1/3 of the cases where you would see Goat 1, and in 1/3 of the cases where you would see Goat 2. All Goat 2 cases are gone; the car is still in your door in 1/3 of the cases where you would/do see Goat 1.

Gandalf
02-08-2006, 11:54 AM
Play the 99 games. You'll see you are wrong. Then at least you'll know that you need to be figuring out why you are wrong.

whisper
02-08-2006, 11:56 AM
But what if Monty only opens a door if you have already chosen the car? That would be HIS best (game theory) strategy, if he wanted you to get his goat.:horse:

Consider it. If he only opens up a door if you got car, then if he opens a door - you have to stay on your selection. If he doesn't open up a door, then you should change your selection to one of the others.

Westley
02-08-2006, 11:57 AM
What other information are you claiming Monty has revealed?Of the two doors that we all know have a 2/3 chance of holding the car, he has shown me one of them that doesn't. So, if the car was there (2/3 chance), he has told me which door it is.

whisper
02-08-2006, 12:03 PM
Anyways about your claim... Heres a sample space of your and monty's choice that I think shows up (or something similar) on the wiki site.

1/3 You choose Goat 1, Monty Goat 2
1/3 You choose Goat 2, Monty Goat 1
1/6 You choose Car, Monty Goat 1
1/6 You choose Car, Monty Goat 2

I think it is obvious that this is a sample space created at the beginning of the problem. If you were making a decision about what to do at the beginning of the problem then this would be the one to use right. But here that is a logical contradiction. The problem forces you to make a decision after seeing what Monty reveals. How can that sample space at that point contain a 1/3 chance to see the goat Monty reveals? That makes no sense. You need a new sample space at that point. The new sample space is considered given that you did not choose Monty's goat.

(1/3)/(2/3) = .5 Goat 1
(1/3)/(2/3) = .5 Car

What other information are you claiming Monty has revealed?

You're missing an obvious part of the problem. The game cannot begin until you make the first selection. Since the game cannot begin until the 1st selection is made, it has an impact upon the second stage of the problem. The 1st selection limits Monty's next action - he can never reveal what's under the door you selected.

campbell
02-08-2006, 12:05 PM
Jumping in a bit late, did anybody do the extreme version of this game to argue as to what's going on.

Let me propose a new game:
1000 doors, 999 goats, 1 car.
Monty knows where the car is.

You pick a door. Monty will open 998 doors with goats behind them (as he knows where the car is, he'll never open that door.)
Now you're left with the door you originally picked, and another door.
Do you switch?

Last, and most extreme version:
Pick a real number between 0 and 10.
The number I'm thinking of is either that number or e + sqrt(pi).
What number do you think I'm thinking of?

campbell
02-08-2006, 12:06 PM
Oh, and to be clear on my first example, Monty does not open the door you picked, either.

tymesup
02-08-2006, 12:15 PM
Your real number between 0 and 10 game is similar to looking under the lamppost for the missing keys. It's not that the keys are more likely to be under the lamppost, it's that at least you can find them when that's where they are.

renaissoxx
02-08-2006, 12:21 PM
You are finally getting close. Try creating the new sample space by throwing out the parts of the original sample space that haven't occurred: You didn't choose Goat 1, and Monty didn't show Goat 2.

So 1/2 the original cases are left. Originally, the car was in your door in 1/3 of the cases where you would see Goat 1, and in 1/3 of the cases where you would see Goat 2. All Goat 2 cases are gone; the car is still in your door in 1/3 of the cases where you would/do see Goat 1.

Thats obvious if you were using that sample space still at the time you were making the final decision. The argument was that sample space no longer applies at the time of the decision cause you know you didn't choose the goat that monty did.

You're missing an obvious part of the problem. The game cannot begin until you make the first selection. Since the game cannot begin until the 1st selection is made, it has an impact upon the second stage of the problem. The 1st selection limits Monty's next action - he can never reveal what's under the door you selected.

So what? I repeat the question: What information has monty revealed above and beyond that you did not choose the goat he shows? Please specify if you think there is something else.

It doesn't. Because Monty has now given you more information regarding the two doors that you didn't initial choose. But, he has given you no additional information about the one door you did choose. The fact that you don't consider the additional information he chose is why you keep getting te problem wrong.

Yes.

No, because that's not what happened. It's not that I did not choose Monty's goat. I never had a chance to choose Monty's goat, because he chose after I did. I made my choice, and then he chose. So, your statement "given that you did not choose Monty's goat" makes no sense - I never could have chosen Monty's goat, it's not possible, and for you to state a conditional probability based on that is indisputably wrong. The fact that I "did not choose Monty's goat" provides no additional information, because that has a 100% certainty of happening because of the timing of the decisions. Therefore, in the sense of probabilities, there is no additional information given by the fact that I did not choosee Monty's goat.

The conditional is on Monty providing me information about the two doors I didn't pick.

But, I'm still waiting for you to explain what's wrong with this one:

You tried oncec, and used an assumption that is clearly wrong. I'll let you try again, if you want.

By the way just because you misinterpret or don't understand someones reasoning doesn't mean they are wrong. And since you do not know who is right or wrong unless or until the other person admits they are wrong, it is silly to state that over and over.

You had a choice of 3 doors. You could have chosen any of them including either goat. Monty gave you the information that you did not choose either goat 1 or goat 2. What do you mean you never had a chance to choose Monty's goat? Are you referring to some abstract goat thats only identity is whatever Monty ends up choosing? Why are we talking about obscure abstractions here? I am talking about the actual 2 goats. Theres 2 goats and a car. The door you chose has an equal chance of containing each of the two goats and a car. Monty reveals that one of the goats was behind another door- Eliminating the possibility that said goat is behind your door. If you had chosen that door its the same thing - he eliminates the possibility that you chose the other goat. That leaves the 1/3 chance you chose the other goat and the 1/3 chance you chose the car, which means 1/2 conditional probability.

Of the two doors that we all know have a 2/3 chance of holding the car, he has shown me one of them that doesn't. So, if the car was there (2/3 chance), he has told me which door it is.

This reasoning is also self contradictory. If there is a 2/3 chance of Car being behind (Door 1 OR Door 2) there is also a 2/3 chance of car being behind (Door 2 OR Door 3) If door 2 is revealed, the same reasoning can be used to claim that BOTH door 1 and door 3 have a 2/3 chance of having the car. That makes no sense.

Gandalf
02-08-2006, 12:25 PM
Try the 99 games. Decide by experiment if you are right or wrong. Then worry about why.

Gandalf
02-08-2006, 12:38 PM
Suppose I have 2 bags of marbles. One has 1 black marble and 999 white marbles. The other has 1000 black marbles. I choose a bag and draw a marble at random. It is black.

I draw another marble from the same bag, without replacement. What is the probability it is black?

Saying that not seeing a white marble here is irrelevant is equivalent to saying that not seeing Goat 2 in the Monty Hall problem is irrelevant.

Both are relevant: if a white marble or Goat 2 was possible, then the lack of a white marble and lack of Goat 2 is relevant.

Alto Reed on a Tenor Sax
02-08-2006, 12:39 PM
Play the 99 games. You'll see you are wrong. Then at least you'll know that you need to be figuring out why you are wrong.

Since he has obviously (and unwisely) placed Gandalf on his ignore list, I am repeating this sage advice.

And I agree with the previous poster who implied that this person should NOT be an actuary. Managers, I ask you, would you want an actuary working for you who chose to argue the validity of his theory, without ever checking his theory against the actual data he is giving you?

Dr T Non-Fan
02-08-2006, 12:44 PM
Jumping in a bit late, did anybody do the extreme version of this game to argue as to what's going on.

Post #27 uses a 52-card deck, which is more readily available than 1000 doors.

Some people will learn when physically shown. Or if their money is taken from them enough times.

Dr T Non-Fan
02-08-2006, 12:45 PM
Yes, that's exactly it. I figured this out a while ago, but didn't explain it nearly as well as you. DTNF explained it like 30 posts ago, but ren ignored him and then wrote about how DTNF is a sheep for believing what everybody else believes.

Westley, would you consider using this in an interview, to weed out poor thinkers?

renaissoxx
02-08-2006, 01:04 PM
Yes, that's exactly it. I figured this out a while ago, but didn't explain it nearly as well as you. DTNF explained it like 30 posts ago, but ren ignored him and then wrote about how DTNF is a sheep for believing what everybody else believes.

Of course I never said anything like that. By sheepish response I was referring to the fact that people often break down emotionally when you disagree with them and start using aggressive and passive aggressive tactics like insulting someone or trying to alter what someone said to make it look like they were calling someone a sheep. Don't bring me down to your level. If I do that on my own thats a different story.

1695814
02-08-2006, 01:06 PM
Some people will learn when physically shown. Or if their money is taken from them enough times....getting off topic now...I once devised an Excel spreadsheet to simulate flipping a coin 20 (or some large number of) times...I wrote a macro to recalculate the spreadsheet until there were 20 (or whatever) heads in a row & then calculate a 21st "flip".

Given some people's understanding of probability, "tails" should be "due", so I'm sure that they'd be willing to give me favorable odds on "heads". If I could get them to play enough games I could clean up!!! :burn:

Westley
02-08-2006, 01:08 PM
And since you do not know who is right or wrong unless or until the other person admits they are wrong, it is silly to state that over and over.I agree that you should not have done this.

renaissoxx
02-08-2006, 01:11 PM

Westley, would you consider using this in an interview, to weed out poor thinkers?

How do you expect someone to take a statement like this seriously when your reasoning is supposedly founded on things like:

If (Door 1 or Door 2) have 2/3 prob and Door 2 is shown not to have car then Door 1 has 2/3...

When it is obvious that this same reasoning can be used to show that door 3 SIMULTANEOUSLY has 2/3 probability?

Westley
02-08-2006, 01:13 PM
What do you mean you never had a chance to choose Monty's goat? Are you referring to some abstract goat thats only identity is whatever Monty ends up choosing?

I am referring to this statement:
Originally Posted by renaissoxx
The new sample space is considered given that you did not choose Monty's goat.

Which was incorrect, which is presumably why you edited it. I can never ever ever choose Monty's goat, it is impossible, because of the order that it takes place. So, your conditional probability is meaningless and your statement is wrong.

Dr T Non-Fan
02-08-2006, 01:14 PM
Just so everyone knows that you are sometimes correct:
You had a choice of 3 doors.
Correct.
You could have chosen any of them including either goat.
Correct.
Monty gave you the information that you did not choose either goat 1 or goat 2.
Yes, but even before you choose, you know he will do this. That's part of the set-up.
What do you mean you never had a chance to choose Monty's goat?
Because he waits until you've chosen before he chooses his goat. He cannot choose a door with a goat until you choose a door. He cannot choose your door.
Are you referring to some abstract goat thats only identity is whatever Monty ends up choosing?
It's not abstract. It's right there among the choices. His chosen goat depends on what you choose.
Why are we talking about obscure abstractions here?
We're not.
I am talking about the actual 2 goats.
Yes, I guess you are.
Theres 2 goats and a car.
Correct.
The door you chose has an equal chance of containing each of the two goats and a car.
Correct. 1/3 probability at the beginning of the game of your choosing each.
Monty reveals that one of the goats was behind another door- Eliminating the possibility that said goat is behind your door.
Yes, but you knew this was going to happen, as I've written above. In other words, it is not new information about your chosen door.
If you had chosen that door its the same thing - he eliminates the possibility that you chose the other goat.
Correct, but you knew that before you choose, so no new information about your door has been given.
That leaves the 1/3 chance you chose the other goat and the 1/3 chance you chose the car, which means 1/2 conditional probability.No, it doesn't. You have chosen a door. You don't know what's behind the door at the beginning of the game, but Monte does.
After Monte eliminates one door, something you knew he would do, he has not changed the probability of your door containing a goat or a car. He has collapsed the probabilities of the other two doors into the one remaining door.

The interesting thing about this problem and your approach to it, is that your reasoning is the supposedly intuitive one, the one that most people think. The actual solution requires thinking outside the box of intuition, and seeing that the probability of your original choice does not change after Monte provides no additional information about your door.

renaissoxx
02-08-2006, 01:16 PM
I am referring to this statement:
[/I]

Which was incorrect, which is presumably why you edited it. I can never ever ever choose Monty's goat, it is impossible, because of the order that it takes place. So, your conditional probability is meaningless and your statement is wrong.

Edited? No I didn't edit out anything. You could have chosen the actual physical goat that monty chose. Not the abstract goat that is whatever monty chooses. The physical goat that monty actually chose. You could have chosen it. But him revealing it shows that you didnt. Try making the goats wear a number in your head. If goat 2 is revealed by monty then it is given that you did not pick goat 2. This is what I am claiming.

Westley
02-08-2006, 01:19 PM
This reasoning is also self contradictory. If there is a 2/3 chance of Car being behind (Door 1 OR Door 2) there is also a 2/3 chance of car being behind (Door 2 OR Door 3) If door 2 is revealed, the same reasoning can be used to claim that BOTH door 1 and door 3 have a 2/3 chance of having the car. That makes no sense.You are correct that applying that logic to any two doors makes no sense.

The only thing that makes sense is applying that logic to the two doors that it applies to, which is the two doors that you did not select with your first choice.

renaissoxx
02-08-2006, 01:21 PM
Just so everyone knows that you are sometimes correct:
Correct.
Correct.
Yes, but even before you choose, you know he will do this. That's part of the set-up.
Because he waits until you've chosen before he chooses his goat. He cannot choose a door with a goat until you choose a door. He cannot choose your door.
It's not abstract. It's right there among the choices. His chosen goat depends on what you choose.
We're not.
Yes, I guess you are.
Correct.
Correct. 1/3 probability at the beginning of the game of your choosing each.
Yes, but you knew this was going to happen, as I've written above. In other words, it is not new information about your chosen door.
Correct, but you knew that before you choose, so no new information about your door has been given.
No, it doesn't. You have chosen a door. You don't know what's behind the door at the beginning of the game, but Monte does.
After Monte eliminates one door, something you knew he would do, he has not changed the probability of your door containing a goat or a car. He has collapsed the probabilities of the other two doors into the one remaining door.

The interesting thing about this problem and your approach to it, is that your reasoning is the supposedly intuitive one, the one that most people think. The actual solution requires thinking outside the box of intuition, and seeing that the probability of your original choice does not change after Monte provides no additional information about your door.

Again:

All of your reasoning (in this post) is based on this premise that the probability of (Door 1 OR Door 2) having the car which is 2/3 collapses into Door 2 when Door 1 is revealed to have a goat. But what about the 2/3 probability of (Door 1 OR Door 3) Why would this not collapse into Door 3?

Also, I did not know to begin with which goat I was not going to choose. Knowing which goat I did not choose allows me to cross off the chance that I chose that goat.

You are correct that applying that logic to any two doors makes no sense.

The only thing that makes sense is applying that logic to the two doors that it applies to, which is the two doors that you did not select with your first choice.

Suuurrrrreee... And why is that? Your reasoning was that (Door 1 OR Door 2), containing 2 1/3 probability slots contains 2/3 probability altogether correct? Does (Door 1 OR Door 3) not ALSO contain 2 1/3 probability slots?

SharksFan08
02-08-2006, 01:23 PM
Try making the goats wear a number in your head. If goat 2 is revealed by monty then it is given that you did not pick goat 2. This is what I am claiming.

OK, with you so far. Now lead us through the rest of it...? How does which goat Monty reveals change anything about the door you started with? When you picked the door, you knew it had a 1/3 chance of being a car. You also knew Monty would then show you a goat. You don't learn anything new.

If you don't switch your door, it's like you pick a door and then walk off the set, not caring about what Monty shows you. Would you agree that in this case you'd pick the car 1/3 of the time?

Seriously, would you PLEASE just try this 20 times with three playing cards?

Dr T Non-Fan
02-08-2006, 01:24 PM
How do you expect someone to take a statement like this seriously when your reasoning is supposedly founded on things like:

If (Door 1 or Door 2) have 2/3 prob and Door 2 is shown not to have car then Door 1 has 2/3...

When it is obvious that this same reasoning can be used to show that door 3 SIMULTANEOUSLY has 2/3 probability?
No, the same reasoning cannot be used, since no additional information has been revealed about Door 3 (your originally chosen door).

At least one of Door 1 and Door 2 does NOT have a car, agreed?
Monte will show you one that isn't the car, agreed?

1. Prob (Door 1 or 2 has a car) = ?
2. Prob (at least one of Door 1 or 2 does NOT have a car) = ?
3. Prob (Door 1 or 2 has a car | at least one of Door 1 or 2 does not have a car) = ?
4. Prob (Door 3 has a car) = ?
5. Prob (Door 3 has a car | at least one of Door 1 or 2 does not have a car) = ?

Westley
02-08-2006, 01:24 PM
Edited? No I didn't edit out anything. OK, troll

Last edited by renaissoxx : 02-06-2006 at 11:42 PM.

Last edited by renaissoxx : Yesterday at 12:01 AM.

Last edited by renaissoxx : Today at 01:58 PM.

Last edited by renaissoxx : Today at 04:44 PM.

Last edited by renaissoxx : Today at 05:57 PM.

Actuary321
02-08-2006, 01:24 PM
OK, one more try. There are 3 doors. 2 doors conceal goats and one conceals a car. You get to choose one door. Now I am going to change the next part slightly but it has the same result.

You have chosen a door. I now have 2 doors, at least one of which conceals a goat 100% of the time. Now you can keep the door you chose or you can opt for everything behind my doors.

What is the probability that you get a car out of this deal if you stay with your door? What is the probability you get a car if you switch?

No door has been opened but I did give you the ability to pick between 2 choices. Are you still going to argue that the chance of winning a car with the option of the second choise is 50-50?

Westley
02-08-2006, 01:26 PM
Seriously, he's obviously a troll, quit wasting your time.

SharksFan08
02-08-2006, 01:27 PM
OK, troll

Yeah, I will make one concession for Ren: he's the most successful troll I've seen in a while.

Dr T Non-Fan
02-08-2006, 01:30 PM

All of your reasoning (in this post) is based on this premise that the probability of (Door 1 OR Door 2) having the car which is 2/3 collapses into Door 2 when Door 1 is revealed to have a goat. But what about the 2/3 probability of (Door 1 OR Door 3) Why would this not collapse into Door 3?

Because you have chosen Door 3, and it is not a door that Monte can choose to reveal to you.
If you want to discuss Prob (Door 1 or Door 3), you'll have to first choose Door 2.
And, for completeness, Choose Door 1, and then collapse Doors 2 and 3.
We can discuss the solution without loss of generality by choosing which door you will choose.

Last post for me in this thread, too.

Swiper
02-08-2006, 01:32 PM
Maybe it is clearer if you only look at the case where you switch doors (ignore the case where you don't switch since we are only stating that when you switch you will win 2/3 of the time)

If you initially pick a goat, monty only has one other door he can open (the door with the other goat since he won't open the car and he won't open the door you picked). Therefore, in this case the door he didn't open (that is not also yours) is the car. If you switch you win the car.

Since you will initially pick a door with a goat 2/3 of the time, the scenario above (you switch and win the car) will happen 2/3 of the time.

The remaining 1/3 of the time (where you do pick the car initially) you will lose if you switch.

renaissoxx
02-08-2006, 01:33 PM
OK, troll

Last edited by renaissoxx : 02-06-2006 at 11:42 PM.

Last edited by renaissoxx : Yesterday at 12:01 AM.

Last edited by renaissoxx : Today at 01:58 PM.

Last edited by renaissoxx : Today at 04:44 PM.

Last edited by renaissoxx : Today at 05:57 PM.

edit does not = edit out...

Alto Reed on a Tenor Sax
02-08-2006, 01:34 PM
Damn, he is a troll, isn't he? I can't believe I fell for that.

Of course, most everybody else fell for it even harder. But personally, in cases like this, I refuse to waste my time trying to explain something like this until the person has TRIED THE SIMULATION. To paraphrase Gandalf, once you realize that you are very, very wrong, you are at least open to trying to figure out why.

I declare this thread over until he posts that he has actually tried the simulation. Until he does that, my assumption stands that he is a troll.

Westley
02-08-2006, 01:37 PM
edit does not = edit out...Again PLEASE READ THIS TIME: Ok, Troll

renaissoxx
02-08-2006, 01:38 PM
No, the same reasoning cannot be used, since no additional information has been revealed about Door 3 (your originally chosen door).

At least one of Door 1 and Door 2 does NOT have a car, agreed?
Monte will show you one that isn't the car, agreed?

1. Prob (Door 1 or 2 has a car) = ?
2. Prob (at least one of Door 1 or 2 does NOT have a car) = ?
3. Prob (Door 1 or 2 has a car | at least one of Door 1 or 2 does not have a car) = ?
4. Prob (Door 3 has a car) = ?
5. Prob (Door 3 has a car | at least one of Door 1 or 2 does not have a car) = ?

What information was revealed about the other door that you did not choose that was not revealed about the door you did choose that allows you to use the reasoning that the 2/3 prob of the 2 1/3 probability slots including the door you didnt choose collapses to that door, but doesn't allow you to use it regarding the 2 slots including the door you chose and the one revealed?

1. 2/3
2. Prob (Not 1 And 2) = 1
3. 2/3 Same question as 1...
4. 1/3
5. 1/3 same as 4..

Are you sure you meant these this way? Whats the point?

renaissoxx
02-08-2006, 01:45 PM

Usually when I edit something its to be more objective and less antagonistic.

DOH!... I double posted :(

Because you have chosen Door 3, and it is not a door that Monte can choose to reveal to you.
If you want to discuss Prob (Door 1 or Door 3), you'll have to first choose Door 2.
And, for completeness, Choose Door 1, and then collapse Doors 2 and 3.
We can discuss the solution without loss of generality by choosing which door you will choose.

Last post for me in this thread, too.

Why do I have to choose Door 2 in order to discuss Prob (Door 1 or Door 3)? What reasoning went in to the idea that Prob (Door 1 OR Door 2) = 2/3 other than that there is 2 doors included out of the original 3 1/3 probability doors (which would not isolate (Door 1 OR Door 2) from (Door 1 or Door 3)?
What does it matter that Monty cannot reveal it to you? Does that mean the probability of (Door 1 or Door 3) is incomprehensible? Gee I thought it was just 2/3...

Damn, he is a troll, isn't he? I can't believe I fell for that.

Of course, most everybody else fell for it even harder. But personally, in cases like this, I refuse to waste my time trying to explain something like this until the person has TRIED THE SIMULATION. To paraphrase Gandalf, once you realize that you are very, very wrong, you are at least open to trying to figure out why.

I declare this thread over until he posts that he has actually tried the simulation. Until he does that, my assumption stands that he is a troll.

Dont be rediculous. Look at the level of objectivity from me here, and then look at the level of objectivity displayed by eveyrone else... Almost everything I have posted here has been taking something someone else said, considering it, and showing a problem with it.

Alto Reed on a Tenor Sax
02-08-2006, 01:57 PM
renaissoxx

I have a simple question, which I do not think you will respond to, because you have not responded to any posts along these lines. I think you will ignore this post like the others because, as I have stated, I think you are a troll,but maybe you will surprise me:

HAVE YOU ACTUALLY TRIED THE SIMULATION YET?

And, if not,

WHY ON EARTH NOT??

Again, I don't anticipate an answer, but maybe your refusal to respond will effectively kill this thread, and keep people from feeding you.

Dr T Non-Fan
02-08-2006, 02:07 PM
Going back on my word....
What information was revealed about the other door that you did not choose that was not revealed about the door you did choose that allows you to use the reasoning that the 2/3 prob of the 2 1/3 probability slots including the door you didnt choose collapses to that door, but doesn't allow you to use it regarding the 2 slots including the door you chose and the one revealed?

Monte chose among the two doors that you didn't choose. If the game had the condition that he could choose to reveal your door as having a goat and then asked you if you want to switch, then that is a different game

1. 2/3
2. Prob (Not 1 And 2) = 1
3. 2/3 Same question as 1...
4. 1/3
5. 1/3 same as 4..

Are you sure you meant these this way? Whats the point?
Point is that you are not given any information about your door, so the probability that your door has a car does not change.

What Monte is asking you to do is not to switch from your door to one of the other doors. He's asking you to switch from your door to the contents of ALL the other doors, minus one goat. The probability that the car is behind one of the unchosen doors is 2/3. Keep what you have, or switch to ALL the other doors. What do you do?

whisper
02-08-2006, 02:15 PM
So what? I repeat the question: What information has monty revealed above and beyond that you did not choose the goat he shows? Please specify if you think there is something else.

You're missing the point. Information is flowing BOTH ways.

Lets look at two situations of the game.
The first situation is this:

There are three doors. Two doors have a goat, one door has a car. Monty selects one of the doors and reveals a goat. Now you have to decide which door to select. Monty knows where the car is, and he can't open that door. Monty doesn't know what door you're initially inclined to open. So, he has 2 doors to choose from and he randomly selects one of the doors to open - obviously with a goat behind it. Now, the contestant has a 1/2 chance of getting the car.

The second situahtion is the standard version of the game.
There are three doors. Two doors have a goat, one door has a car. The contestant has to select a door. Monty now knows two pieces of information: where the car is, and which door is you're initially inclined to select. Furthermore, your selection LIMITS Monty's next action. He cannot randomly select a door to open. The only scenario where he gets to do this is if you initially select a car. Otherwise, there is only one door he can open. He reveals a goat. Now you get to reselect.

Now, you're faced with a slightly different choice than the prior example. You still have two doors. But, those two doors are different from the above example. You have you're initial selection - which transmitted information and constraints to Monty which affected his actions, and the other door.

In both cases, the amount of information revealed to the contestant was the same. But, in the standard case, the contestants initial selection also transmitted information. That initial information transmissal is not trivial, and must also be included in the problem.

Gandalf
02-08-2006, 02:16 PM
Lots of people are missing the point, which is
renaissoxx

I have a simple question, which I do not think you will respond to, because you have not responded to any posts along these lines. I think you will ignore this post like the others because, as I have stated, I think you are a troll,but maybe you will surprise me:

HAVE YOU ACTUALLY TRIED THE SIMULATION YET?

And, if not,

WHY ON EARTH NOT??

Again, I don't anticipate an answer, but maybe your refusal to respond will effectively kill this thread, and keep people from feeding you.

Dr T Non-Fan
02-08-2006, 02:21 PM
Meh. It can be shown or proven without simulation. Whether the observer understands is not our responsibility.

renaissoxx
02-08-2006, 02:21 PM
OK, one more try. There are 3 doors. 2 doors conceal goats and one conceals a car. You get to choose one door. Now I am going to change the next part slightly but it has the same result.

You have chosen a door. I now have 2 doors, at least one of which conceals a goat 100% of the time. Now you can keep the door you chose or you can opt for everything behind my doors.

What is the probability that you get a car out of this deal if you stay with your door? What is the probability you get a car if you switch?

No door has been opened but I did give you the ability to pick between 2 choices. Are you still going to argue that the chance of winning a car with the option of the second choise is 50-50?

No. But what does this have to do with the original problem?

Gandalf
02-08-2006, 02:25 PM
Meh. It can be shown or proven without simulation. Whether the observer understands is not our responsibility.
Of course it isn't. So why do you persist, when he is probably a troll who understands? At least make him commit to what the simulation shows before wasting more time.

SharksFan08
02-08-2006, 02:26 PM
Meh. It can be shown or proven without simulation. Whether the observer understands is not our responsibility.

But it can't be shown or proven any more effectively than has been done 75 times in this thread already. A simulation might help. Of course, that's assuming that the misunderstanding on Ren's part is sincere, which at this point I think we can all agree, it isn't.

renaissoxx
02-08-2006, 02:26 PM
renaissoxx

I have a simple question, which I do not think you will respond to, because you have not responded to any posts along these lines. I think you will ignore this post like the others because, as I have stated, I think you are a troll,but maybe you will surprise me:

HAVE YOU ACTUALLY TRIED THE SIMULATION YET?

And, if not,

WHY ON EARTH NOT??

Again, I don't anticipate an answer, but maybe your refusal to respond will effectively kill this thread, and keep people from feeding you.

I will try the simulation when I have the chance. Why am I still debating without it? Because I believe the reality can be shown without simulation it just takes careful reasoning. Why do I think you guys are confident in the results I will obtain? Because in these situations it is often the case that you create an experiment that gives the results you want to see, whereas an experiment created to test my beliefs would give the results I thought I would see. In this case the only answer is that one of us doesn't understand the others belief.

JMO
02-08-2006, 02:27 PM
Meh. It can be shown or proven without simulation. Whether the observer understands is not our responsibility.
Go for it DTNF. If your way of explaining it finally gets through to him, everybody will be better off.

Dr T Non-Fan
02-08-2006, 02:33 PM
Go for it DTNF. If your way of explaining it finally gets through to him, everybody will be better off.
I'm glad to hear you have more confidence in me than I in him.

SharksFan08
02-08-2006, 02:35 PM
I can't believe I'm going to bite on this...

...whereas an experiment created to test my beliefs would give the results I thought I would see.

...but why don't you go ahead and construct such an experiment? Do you not "have time" to do our simulation because you're so busy posting on AO all day?

Gandalf
02-08-2006, 02:39 PM
I'm glad to hear you have more confidence in me than I in him.That's the point. Trolls can play dumb forever. What change do you expect from your fine efforts?

If he acknowledges what is right, there's a chance he'll acknowledge why it is right.

renaissoxx
02-08-2006, 02:42 PM
Going back on my word....

Monte chose among the two doors that you didn't choose. If the game had the condition that he could choose to reveal your door as having a goat and then asked you if you want to switch, then that is a different game

Point is that you are not given any information about your door, so the probability that your door has a car does not change.

What Monte is asking you to do is not to switch from your door to one of the other doors. He's asking you to switch from your door to the contents of ALL the other doors, minus one goat. The probability that the car is behind one of the unchosen doors is 2/3. Keep what you have, or switch to ALL the other doors. What do you do?

I am asking you why it matters which Door Monty chooses. I can read the wiki for myself. The only reason 2 doors grouped together have 2/3 probability is because there are 2 of them. This same reasoning can be applied to the combination of the other 2 unchosen doors to show that they have a combined 2/3 probability. Is this true or false? It has nothing to do with which one Monty chooses. How do the laws of statistics suddenly break down just for this game? (1 or 3) = 2/3 just like (1 or 2) = 2/3 in the beginning.

EDIT: Im going to read the Wiki discussion and reply next to an argument regarding what you are saying from there cause there is alot more.

MountainHawk
02-08-2006, 02:45 PM
Let's try this.

You pick one of three doors. Monty is going to show you a door with a goat, then show you your door. You are not allowed to switch, though. What's your probability of winning this game?

JMO
02-08-2006, 02:46 PM
I am asking you why it matters which Door Monty chooses. I can read the wiki for myself. The only reason 2 doors grouped together have 2/3 probability is because there are 2 of them. This same reasoning can be applied to the combination of the other 2 unchosen doors to show that they have a combined 2/3 probability. Is this true or false? It has nothing to do with which one Monty chooses. How do the laws of statistics suddenly break down just for this game? (1 or 3) = 2/3 just like (1 or 2) = 2/3 in the beginning.
Try this now.
1. If you know that the probability that the car is behind (door 1 or door 2) is 2/3 and you also know that it is not behind door 2, what is the probability the car is behind door 1?

2. If you know that the probability that the car is behind (door 1 or door 2) is 2/3 and you also know that it is not behind door 1, what is the probability the car is behind door 2?

Gandalf
02-08-2006, 02:49 PM
Let's try this.

You pick one of three doors. Monty is going to show you a door with a goat, then show you your door. You are not allowed to switch, though. What's your probability of winning this game?
Best explanation yet. Game, set, match, in all the judges' eyes.

whisper
02-08-2006, 02:50 PM
I am asking you why it matters which Door Monty chooses. I can read the wiki for myself. The only reason 2 doors grouped together have 2/3 probability is because there are 2 of them. This same reasoning can be applied to the combination of the other 2 unchosen doors to show that they have a combined 2/3 probability. Is this true or false? It has nothing to do with which one Monty chooses. How do the laws of statistics suddenly break down just for this game? (1 or 3) = 2/3 just like (1 or 2) = 2/3 in the beginning.

The laws of statistics aren't breaking down.

You're not taking into account Bayesian Statistics.

Alto Reed on a Tenor Sax
02-08-2006, 02:59 PM
Again, I posit to all of you that your continued explanations will do no good, because he is a troll who is exercising his bulls***-pseudo-nihilistic-debating skills. He understands the problem fine, he just wants to see how convoluted an argument he can make against it, while still sounding rational.

As actuaries, how would you deal with this person in real life? If anyone came to me with this "it's so mysteeerrrrrrrious how can either of us ever be right or wrooonnnggg when none of us truly understands the T-T-T-TRRUUTHHH that the other believes...?" the first thing I would do is say, go conduct 20 trials on your own, on your own terms, and tell me if your results match your expectation.

Until he posts and says that he has done some trials, you are wasting your time. He won't actually do this, of course, because as I said, he understands the problem perfectly well, and admitting he has experimented would lead him to one of two outcomes:

1) He claims that his result approaches 2/3, in which case he was "wrong," or,
2) He claims that his result approaches 1/2, in which case he exposes himself as a troll.

Which will it be, soxx?

JMO
02-08-2006, 03:00 PM
You're not taking into account Bayesian Statistics.

Isn't there a version of Bayesian logic where a statistical question is posed to a bunch of people, and the answers they give reproduces the distribution of the actual parameter they are estimating?

E.g., the question I recall was how much longer a cake (recipe not given) still needs to bake given it has been baking for n minutes. And the guesses from the group actually reproduced the peculiar distibution of cake-baking times less n.

I think maybe that's what is happening here.

whisper
02-08-2006, 03:07 PM
Isn't there a version of Bayesian logic where a statistical question is posed to a bunch of people, and the answers they give reproduces the distribution of the actual parameter they are estimating?

Yeah, here is the article:

SharksFan08
02-08-2006, 03:07 PM
Isn't there a version of Bayesian logic where a statistical question is posed to a bunch of people, and the answers they give reproduces the distribution of the actual parameter they are estimating?

E.g., the question I recall was how much longer a cake (recipe not given) still needs to bake given it has been baking for n minutes. And the guesses from the group actually reproduced the peculiar distibution of cake-baking times less n.

I think maybe that's what is happening here.

that's almost word-for-word what I was about to say... beat me to it.

renaissoxx
02-08-2006, 03:14 PM
Let's try this.

You pick one of three doors. Monty is going to show you a door with a goat, then show you your door. You are not allowed to switch, though. What's your probability of winning this game?

What I am debating here is that the 2/3 probability of the two doors other than the one you have chosen collapses into the one door you still didn't choose. My argument that any 2 slot grouping carries a 2/3 probability and therefore so does the (Door you chose OR Door monty revealed) was only meant to show that this explanation is lacking.

So to answer your question, the probability is 1/3. What is the usefulness of this?

MountainHawk
02-08-2006, 03:17 PM
What I am debating here is that the 2/3 probability of the two doors other than the one you have chosen collapses into the one door you still didn't choose. My argument that any 2 slot grouping carries a 2/3 probability and therefore so does the (Door you chose OR Door monty revealed) was only meant to show that this explanation is lacking.

So to answer your question, the probability is 1/3. What is the usefulness of this?

Fine. You have 1/3 probability of winning. That is correct. Now Monty shows you a door with a goat. You knew he was going to do that, so you still have 1/3 probability of winning. So what is the probability of the door he's NOT going to show you winning?

Alto Reed on a Tenor Sax
02-08-2006, 03:21 PM
Hey, soxx, have you gotten a chance to try that simulation yet? Or any simulation? Even 10 or 20 trials?

I think every time soxx posts another logical fallacy (thus proving that he has at least a little time on his hands), someone should ask him that.

Old Timer
02-08-2006, 03:24 PM
I couldn't stand the tension waiting for soxx to run the simulation, so I had to do it myself.

1,000,000 trials,

switching produced a car in 66.6925% of the trials.

Can we end this now.

Chubbs
02-08-2006, 03:27 PM
I can't belive noone has said this yet. Put your money where you mouth is.

Lets play a game. Behind one door is +\$1.50 for you. Behind the other two doors is -\$1.00 to you, (or +\$1.00 to me if you prefer).

You will pick your door, I will show you a door with -1 behind it, and then you must always keep your original choice.

Since the odds are 50/50, your expected value per trial is positive 25 cents.

(I don't expect you to know how to come up with that so here goes an explanation. If you win, you get 1.50, if you lose you get -1. Since you can expect to win and lose at about the same rate, each play is worth .25 to you)

Lets play this about 100 times and then settle up on party poker with a fund transfer.

Are you in, or just a troll?

Westley
02-08-2006, 03:34 PM
switching produced a car in 66.6925% of the trials.That's really not that far from 50%.

Gandalf
02-08-2006, 03:35 PM
Fine. You have 1/3 probability of winning. That is correct. Now Monty shows you a door with a goat. You knew he was going to do that, so you still have 1/3 probability of winning. So what is the probability of the door he's NOT going to show you winning?
Your prior post was checkmate, not "mate in x". Game over. You don't need to actually capture the king.

SharksFan08
02-08-2006, 03:39 PM

We forgive you. Especially if you up the ante, and actually get Ren to play you.

tymesup
02-08-2006, 03:42 PM
This problem is similar to the Restricted Choice problem in bridge.

The classic example is an opponent could have one of three "equally" likely possibilities - a J, a Q or both a J and a Q. When he has both the J and the Q, he will play either of them 50% of the time.

When the opponent actually plays the J, the percentage play is to assume that he has the J by itself. The logic is that if the J is by itself, he is forced to play it, while if he has the J and the Q, he might have played the Q instead.

Chubbs
02-08-2006, 03:43 PM
Yeah, I think that he will avoid the post, because he is just having fun with people. We should do a poll about how he will respond to it.

1. I don't gamble
2. You will create the game so that it favors you
3. Ignore completely
4. Play and don't pay.

I am not very creative, you others could come up with some funny options.

Alto Reed on a Tenor Sax
02-08-2006, 03:43 PM
That's really not that far from 50%.

Touche. Gentlemen, this Westley is good. Damn good.

Dr T Non-Fan
02-08-2006, 03:45 PM
My argument that any 2 slot grouping carries a 2/3 probability and therefore so does the (Door you chose OR Door monty revealed) was only meant to show that this explanation is lacking.

The P(Door you chose OR Door monty revealed) is not relevant to the game, since it's not part of the game. Monte chooses a goat from one of the doors you didn't choose. You don't know what's behind each of the unchosen doors, but you know there's a probability of 2/3 that the car is behind one of them.

Another game:
1. 2 goats, 1 car.
2. You choose a door.
3. Monte then allows you switch from your chosen door to ALL of the other doors.
4. Should you switch?

This is what the original game breaks down to, since Monte tells you nothing about your chosen door, but tells you something you already know about one of the remaining doors. It's as if he didn't reveal the location of the goat, and you can switch from your chosen door to all the remaining doors, and you can leave the goat(s) behind if you want.

Alto Reed on a Tenor Sax
02-08-2006, 03:46 PM
Yeah, I think that he will avoid the post, because he is just having fun with people. We should do a poll about how he will respond to it.

1. I don't gamble
2. You will create the game so that it favors you
3. Ignore completely
4. Play and don't pay.

I am not very creative, you others could come up with some funny options.

I choose 3), until he starts getting grief, at which time we will see a large helping of 2), perhaps with a shade of 1) for flavor, padded throughout by

5)incomprehensible bulls*** sophistry.

mr.c
02-08-2006, 03:53 PM
Poor DTNF is in a codependent relationship. We need an intervention.

Dr T Non-Fan
02-08-2006, 04:05 PM
Poor DTNF is in a codependent relationship.
With my keyboard.
Interesting thing is that I took some time out of here for lunch and to read the wiki explanation. For the first time, mind you. Amazing how many of my explanations and examples were there.

I don't see how any reasonable person can go through wiki's various explanations and not be convinced.
I can draw only one conclusion from this observation.

Thank you, mr.c. Aaaaaayyyy!

This thread should be moved from this forum and into reef, as it involves some unprofessional behavior.

Actuary321
02-08-2006, 04:13 PM
The P(Door you chose OR Door monty revealed) is not relevant to the game, since it's not part of the game. Monte chooses a goat from one of the doors you didn't choose. You don't know what's behind each of the unchosen doors, but you know there's a probability of 2/3 that the car is behind one of them.

Another game:
1. 2 goats, 1 car.
2. You choose a door.
3. Monte then allows you switch from your chosen door to ALL of the other doors.
4. Should you switch?

This is what the original game breaks down to, since Monte tells you nothing about your chosen door, but tells you something you already know about one of the remaining doors. It's as if he didn't reveal the location of the goat, and you can switch from your chosen door to all the remaining doors, and you can leave the goat(s) behind if you want.Wasn't that what I said?

1695814
02-08-2006, 04:14 PM
Yeah, here is the article:
http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?t=71812 (http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?t=71812[/quote] :lol):lol: check out the 2nd post in that thread. what a moran! no wonder I had no idea what jmo was referring to. :duh:

I'm going to go read it now.

tymesup
02-08-2006, 04:15 PM
Let's suppose you pick door 1. At this point, there are four possible outcomes which have different probabilities:

You picked goat number 1, Monty shows you goat number 2, 1/3 * 1 = 1/3

You picked goat number 2, Monty shows you goat number 1, 1/3 * 1 = 1/3

You picked the car, Monty shows you goat number 1, 1/3 * 1/2 = 1/6

You picked the car, Monty shows you goat number 2, 1/3 * 1/2 = 1/6

By switching, you win in two of the four outcomes, with total probability 2/3.

Westley
02-08-2006, 04:23 PM
By switching, you win in two of the four outcomesHow can this not be 50%? Are you really this clueless?

Dr T Non-Fan
02-08-2006, 04:25 PM
Wasn't that what I said?
Probably. I guess we've run out of ways to explain it in simple English.

You'd think that a web site could be created with the game...time to Google me some Truth!

Wow, that was easy:
http://montyhallgame.shawnolson.net/

http://www.shodor.org/interactivate/activities/monty3/

DoctorNo
02-08-2006, 04:32 PM
Here's another good one:

http://people.hofstra.edu/staff/steven_r_costenoble/MontyHall/MontyHallSim.html

I used to give the Monty Hall Problem talk to undergraduates when I was doing the professor job search.

I'm pretty sure that soxx was always that one guy in the back row that always seemed to show up. :lol:

Alto Reed on a Tenor Sax
02-08-2006, 04:35 PM
How can this not be 50%? Are you really this clueless?

Sweet Lord, you really ARE my wife! Let's work this through, with her logic:

When you play the Powerball lottery, with one ticket, there are two options,

In one of these two outcomes, you are a multi-millionaire.

HOW CAN THIS NOT BE 50%!?!?

Gandalf
02-08-2006, 04:38 PM
The problem with these simulations is that you can't be sure they are really randomizing things behind the scenes. If I had the expertise to create one of those, I could create one where the odds would be 50% that as to whether you should switch, by choosing the location of the car after you chose your door.

Or, from the alternative viewpoint, those sites could be getting the apparent 2/3 odds for switching by manipulating the locations to get 2/3.

1695814
02-08-2006, 04:42 PM
I tried programming this little problem in Excel, except that as I was trying to figure out how to program it, it became so obvious to me that 2/3 is the correct answer that I didn't bother finishing. qed.

Roomba
02-08-2006, 04:43 PM
I think you people spend too much time behind closed doors with goats.

Actuary321
02-08-2006, 04:44 PM
Probably. I guess we've run out of ways to explain it in simple English.

You'd think that a web site could be created with the game...time to Google me some Truth!

Wow, that was easy:
http://montyhallgame.shawnolson.net/

http://www.shodor.org/interactivate/activities/monty3/That was great DTFN, I played on the first link 3 times, I won 2 times lost once. See 2/3 probablity of winning.

At 100 trials it was 61 wins 39 losses.

At 200 trials it was 122 wins 78 losses.

Does it matter if my initial pick is always the same one? That one was for ren.

Dr T Non-Fan
02-08-2006, 04:45 PM
The problem with these simulations is that you can't be sure they are really randomizing things behind the scenes. If I had the expertise to create one of those, I could create one where the odds would be 50% that as to whether you should switch, by choosing the location of the car after you chose your door.

Or, from the alternative viewpoint, those sites could be getting the apparent 2/3 odds for switching by manipulating the locations to get 2/3.
(Insert tin-foil hat emoticon.)
And whatever YOU created would likely do the same thing!

And, really, what have they got to gain by it? Unless, of course, the intuition is correct and the critical reasoning is wrong, and all these web sites banded together to make sure it's always 2/3 when you switch, because this problem holds the key to omnipotence.... no, not "impotence."

Gandalf
02-08-2006, 04:47 PM
That was great DTFN, I played on the first link 3 times, I won 2 times lost once. See 2/3 probablity of winning.
You have just proved renaissoxx's point. You didn't specify whether you switched or not. As he has repeated to DTNF ad nauseum, by DTNF's logic the odds that the car is behind the potential switch door is 2/3, and by symmetry the odds behind the original door are also 2/3.

Gandalf
02-08-2006, 04:57 PM
(Insert tin-foil hat emoticon.)
And whatever YOU created would likely do the same thing!

And, really, what have they got to gain by it? Unless, of course, the intuition is correct and the critical reasoning is wrong, and all these web sites banded together to make sure it's always 2/3 when you switch, because this problem holds the key to omnipotence.... no, not "impotence."
I believe they are legit, because their results match my perception of reality. If they didn't, I would reexamine my position, but wouldn't just conclude they were right because they had nothing to gain. Or even because 4 apparently independent sites led to the same result.

Assuming (purely hypothetically) that renaissoxx believed the odds were 50-50, he would probably conclude that they were maliciously or incompetently programmed.

That's why I laid out my 99 trials, or invited renaissoxx to compute his own. Likewise the people who suggested playing with a co-worker with 3 cards have a convincing position. Chubbs's "challenge" would not be. I could make Chubbs'x challenge letting renaissoxx switch every time, so long as I was the one who told him, after he guessed, whether he was right or wrong.

Alto Reed on a Tenor Sax
02-08-2006, 05:11 PM
I could make Chubbs'x challenge letting renaissoxx switch every time, so long as I was the one who told him, after he guessed, whether he was right or wrong.

This is exactly what my roommate did with me 5 years ago. Before he got out the cards, I was quite adamant that 50% was the correct answer. After 10 or 20 trials, with me winning on the switch twice as often as losing, I started looking at him like he was doing black magic or something. Then he explained some of the mechanics to me, I thought through the rest, and a lifelong(so far, at least) interest in probability theory was born.

Dr T Non-Fan
02-08-2006, 05:26 PM
Then he explained some of the mechanics to me, I thought through the rest, and a lifelong(so far, at least) interest in probability theory was born.
And this step the difference between you and the steadfastly unconvinced.

whisper
02-08-2006, 05:27 PM
Its not surprising that people's initial belief of the problem is wrong.

You could see it in soxx's question about what additional information does Monty reveal when he opens one of the doors. They're looking at the problem one sided. What additional information does the host reveal to the player. They fail to account that information flows both ways. The initial selection gives information as well.

Thats why the game where the participant selects a door before the reveal and then can reselect is different from where the door is revealed before the participant selects anything.

Old Actuary
02-08-2006, 06:17 PM
This thread is deja vu all over again. After the original publication by Marilyn Vos Savant, which I believe was around 1989, many college math professors wrote to say that she was completely wrong to say that one should switch.

Suprisingly, Monty Hall was interviewed by the NY Times and stated that the solution depended on the exact definition of the problem and he didn't think that Marilyn's stated problem agreed with how he did the show.

In my view, the many comments about doing a Monte Carlo simulation to solve the problem are not correct; what's needed is a Monty Hall simulation.

wat?
02-08-2006, 06:18 PM
I haven't read the whole thread, but I think everyone's missing one big assumption.

Depending on what kind of car it is, I'd switch to the goat.

whisper
02-08-2006, 06:23 PM
In my view, the many comments about doing a Monte Carlo simulation to solve the problem are not correct; what's needed is a Monty Hall simulation.

:shake2:

Actuary321
02-08-2006, 08:10 PM
So how do you think ren would define the height of the right triangle? (http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?t=55811)

campbell
02-09-2006, 09:53 AM
I think you people spend too much time behind closed doors with goats.

Best quote of thread. I think it's time to shut things down.

Of course, because of this thread, I ended up explaining this problem to my non-math friends elsewhere, and they got it. One even got it before I finished my explanation.

1695814
02-09-2006, 09:57 AM
So how do you think ren would define the height of the right triangle? (http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?t=55811)shoot...you beat me to it...I was going to search for & post that link today...oh well...you are the master...I am but an apprentice.

Actuary321
02-09-2006, 11:20 AM
shoot...you beat me to it...I was going to search for & post that link today...oh well...you are the master...I am but an apprentice.I almost didn't find it. When I searched for triangle I came up with 8 pages of threads but not that one or even the poll that it inspired. Then I searched for hypotenuse to finally find it. I was ready to just leave it as an excersize for Claude.

renaissoxx
02-10-2006, 03:31 PM
Ok. Here is my question.

You have 3 possible outcomes equally likely.

What is the probability of the first two outcomes given the third outcome has not occured?

1/2 each Correct?

A: Outcome 1 occurs
B: Outcome 3 does not occur

(Prob(B|A)*Prob(A))/Prob(B) = (1*1/3)/(2/3) = 1/2

Also if you did an experiment with 100 trials, you should have close to 33 of each outcome right? What happens when you find out you didn't choose outcome 3? You cross all the outcome 3 results off your list and are left with about 33 1, 33 2.

So the next question is, is that dependent on how you got the information? If it is, I was never taught so in school or determined it was.

So in the monty problem, one piece of information given is that you did not choose the goat he shows you.

Now your saying there is a 2/3 chance you chose a goat, he had to show you the goat he did, and switching would cause you to win.

And a 1/3 chance you chose the car, he didn't have to show you the goat he did, and switching would make you lose.

But what I don't understand is, how can you still look at it like you had a 2/3 chance to choose a goat when its given that you didn't choose one of the goats you are now supposed to look at it like you had a 50% chance to have chosen the goat?

Isn't that what it means to say 50% chance given that you didn't choose outcome 3? Doesn't it mean from that point you are supposed to look at it like theres a 50% chance you chose either one? Im not saying that the fact that Monty reacted to your choice hasn't told you anything. Im saying that basing whatever information he gave you on your original 66% chance to choose a goat doesn't make sense to me when you are supposed to say that you had a 50% chance to have chosen a goat after you know you didn't choose one of them.

Now if its true that switching gets you the car 66% of the time, then the conditional probability must be dependent on how you got the information that you are given.

So... What is the probability of Outcome 1, given that outcome 3 did not occur with the conditional that you are only given the information that outcome 3 did not occur if Outcome 2 occured or 50% of the time outcome 1 occured?

Is there a more general form of Bayes' Theorem that takes into account how the given information was aquired, or the probability that the information being given means a certain outcome?
I never used this formula before. What does it look like?

JMO
02-10-2006, 03:35 PM
Try this now.
1. If you know that the probability that the car is behind (door 1 or door 2) is 2/3 and you also know that it is not behind door 2, what is the probability the car is behind door 1?

2. If you know that the probability that the car is behind (door 1 or door 2) is 2/3 and you also know that it is not behind door 1, what is the probability the car is behind door 2?

whisper
02-10-2006, 03:42 PM
Is there a more general form of Bayes' Theorem that takes into account how the given information was aquired, or the probability that the information being given means a certain outcome?
I never used this formula before. What does it look like?

Monty Hall Problem with Bayes Theorem:
http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html

whisper
02-10-2006, 03:51 PM
Since I can't copy and paste, let me paraphrase the site:

Given 3 doors, A - B - C.

A priori probability that car is behind any door p(x) = 1/3

Contestant chooses door A (doesn't matter).

Probability(reveal B| prize is behind A) = 1/2
Probability(reveal B | prize is behind B) = 0
Probability(reveal B | prize is behind C) = 1

Pr(reveal B) = pr(prize behind A) * pr(B | A) + pr(prize B) * pr(B| B) + pr(prize C) * pr(B | C) = 1/3 * 1/2 + 1/3 * 0 + 1/3 * 1 = 1/2.

By Bayes theorem
Pr(prize is behind A | B revealed) = pr(prize behind A) * pr(B reveal | A) / pr(reveal B) = 1/3 * 1/2 / 1/2 = 1/3

pr(rize is behind C | B revealed) = pr(prize behind C) * pr(B reveal | C) / pr(reveal B) = 1/3 * 1 / (1/2) = 2/3.

Actuary321
02-10-2006, 03:52 PM
Let me propose the reverse Monte Hall problem. I will define it this way:

There are 3 doors. Behind one door is a car. Behind the other 2 doors are goats.
Monte lets you choose any two doors.
He then opens one of the doors you picked to reveal a goat.
He then gives you the option of stiking with the unopened door you originally picked or you can switch to the door you did not pick.
What would you do?
Which stratagy will give you the best chances to win?
What are those chances?

Thanks for playing.

renaissoxx
02-10-2006, 03:53 PM
Monty Hall Problem with Bayes Theorem:
http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html

This isn't what I was asking.

This definition has the same problem of being dependent on probabilities that should be revised once Monty's door is opened.

Im looking for a formula that can answer this question:

What is the probability of Outcome 1, given that outcome 3 did not occur with the conditional that you are only given the information that outcome 3 did not occur if Outcome 2 occured or 50% of the time outcome 1 occured?

Actuary321
02-10-2006, 04:04 PM
This actually is kind of like the problem of the 3 guys who rent a room for \$30 but the room really only costs \$25, so the clerk gives 5 \$1 bills to the bell boy to return to the men. The bell boy realizes that he can't split the 5 one's equally amount the 3 men so he gives each guy 1 \$1 bill and keeps 2 for himself.

Now each man has paid \$9 * 3 = 27 + the 2 the bell boy has equals \$29. What happened to the other dollar?

If you don't look at the problem correctly you won't understand.

The thing you need to think about is that the information that Monte gives you in the end tells you absolutely nothing. You are given the choise of picking 1 door or picking 2 doors. Which do you want to pick?

renaissoxx
02-10-2006, 04:40 PM
Why would I look at the problem the way you want me to?
How can I know that the explanations given are not wrong and only by coincidence match experimental results?

Every explanation given at one point in the explanation requires you to assume a 2/3 chance of having chosen a goat when the information given to you by monty indicates that there is a 1/2 chance you chose the goat according to bayes formula which does not discriminate regarding how you got the information given.

What is the connection between choosing 2 doors and this game?

MountainHawk
02-10-2006, 04:43 PM
Fine. You have 1/3 probability of winning. That is correct. Now Monty shows you a door with a goat. You knew he was going to do that, so you still have 1/3 probability of winning. So what is the probability of the door he's NOT going to show you winning?

Gandalf
02-10-2006, 04:52 PM
Has renaissoxx at least agreed that the correct answer to the Monty Hall problem is 2/3 in favor of switching? Or is he still trolling on that issue?

SharksFan08
02-10-2006, 04:57 PM
I was so optimistic after nearly two days of inactivity on this thread that we were done feeding the troll...

Seriously, there is NO way that Ren is this dense/stupid.

whisper
02-10-2006, 05:00 PM
This isn't what I was asking.

This definition has the same problem of being dependent on probabilities that should be revised once Monty's door is opened.

Im looking for a formula that can answer this question:

What is the probability of Outcome 1, given that outcome 3 did not occur with the conditional that you are only given the information that outcome 3 did not occur if Outcome 2 occured or 50% of the time outcome 1 occured?

We need to establish the starting point. There isn't enough information provided.

What are all the possibilities?
Let 0 be not happened, let 1 be happened.

Can you have: [A B C]
[1 0 0]
[1 1 0]
[1 0 1]
[0 1 0]
[0 0 1]?

So, pr(A) = 3/5, pr(b) = 2/5 and pr(c) = 2/5?

Alto Reed on a Tenor Sax
02-10-2006, 05:03 PM
How can I know that the explanations given are not wrong and only by coincidence match experimental results?

Honey, once you've admitted that we are all right about the experimental results (and I take this phrase quoted above as a tacit admission thereof), it is up to YOU to figure out why. You have been led to the water, but you have to get your own damn drink.

Of course, I still think you are a troll, but now that you have at least conceded that your "theories" are not backed experimentally, maybe this thread can die for once and for all?

Alto Reed on a Tenor Sax
02-10-2006, 05:04 PM
How can I know that the explanations given are not wrong and only by coincidence match experimental results?

Honey, once you've admitted that we are all right about the experimental results (and I take this phrase quoted above as a tacit admission thereof), it is up to YOU to figure out why. You have been led to the water, but you have to get your own damn drink.

Of course, I still think you are a troll, but now that you have at least conceded that your "theories" are not backed experimentally, maybe this thread can die for once and for all?

whisper
02-10-2006, 05:11 PM
This isn't what I was asking.

This definition has the same problem of being dependent on probabilities that should be revised once Monty's door is opened.

Im looking for a formula that can answer this question:

What is the probability of Outcome 1, given that outcome 3 did not occur with the conditional that you are only given the information that outcome 3 did not occur if Outcome 2 occured or 50% of the time outcome 1 occured?

You also realize, that you're question is a straight conditional probability question, and not a Baysian situation?

Old Actuary
02-11-2006, 10:11 PM
Actually, I believe the Bayesian solution misstates the problem, even though it gets the right answer.

Contestant chooses door A (doesn't matter).

Probability(reveal B| prize is behind A) = 1/2
The problem doesn't require that the doors be selected at random. The only requirement is that Monty can't open the door with the car. The probability that Monty opens door B is undefined.

renaissoxx
02-13-2006, 04:07 PM
Has renaissoxx at least agreed that the correct answer to the Monty Hall problem is 2/3 in favor of switching? Or is he still trolling on that issue?

I am not trolling on anything. I don't agree with some of you guys' approach to intellectual tasks. Lets me state a few things and see if you guys can see where I am coming from.

1) None of you are responsible for pioneering the vast majority of this knowledge. Just because you are capable of following someone else's path of reasoning and no alarms go off in your mind telling you that there is a problem with it doesn't mean that

A) You did the same thing the person who invented it did. Did you consider all alternatives, did you accept only in absence of any dissenting information, were you open to dissenting information or did you sweep it under the carpet? Did you find that the results stood well in the face of all opposition in the view of others (ie not you)? If not, I don't believe you really understand it or know that

B) It is true. I remember being in a downtown crowd once at a dmv whent he subject of OJ simpson came up. And you know darn well what I did. Countless times did I hear things like "If the glove don't fit you must aquit OMG what part of that don't you understand?" People who are surrounded with or who have been impressed upon by certain groups and ideas develop belief in a power structure related to those ideas. Thats how they define when something makes sense or not.. by things like "does everyone agree with it". I don't think there is any difference between you guys and people like that OJ crowd. The power structure driven criteria for making sense some of you seem to follow is like "Will this be accepted by math community" or "Does this follow some convention of thinking about/ approaching problems". Im sorry if the power structure I follow being "Is this actually true?" makes me a troll to you. While I point out bits of information that don't agree with your explanations, you side step them and point a path of reasoning that your mind is capable of accepting. As part of the criteria for making sense that I follow, my mind forces me to look for and explain away all dissenting information before accepting something. Besides whatever initial experiences drove me down this path in the first place, I found it constantly being reinforced as I reinvented understanding of each subject (purposely isolating myself from information in order to better learn how new understanding of a subject is created) in my education.

2) If you are trying to figure something out with out other people giving you the "answer" it is insanely difficult. Its like trying to get the big picture with tunnel vision, a person can process so many relations (4 at most?) at once. It looks good when you consider these but this doesn't add up, someone who doesn't want to agree could claim this etc. Remembering something once you figure it out is easy. You guys aren't even subjecting the ideas to the level of scrutiny really needed to guarantee its accuracy (Someone told you X and you connected it to some other things you know and said ok without taking the time to look for or consider all dissenting Y's which weren't glaring) and then remembering it and then criticisizing me for not "understanding". Good job! I really value your criticisms alot. (cough cough) I could be wrong but I seriously doubt anyone who reguarly goes through the most rigorous process of determining truth ever does this type of thing, because they know better...

Anyways all that being said... No I don't agree that after seeing a goat you have a 1/3 chance of winning. I think after seeing a goat you have a 1/2 chance of winning (in a situation without the monty reaction type thing). If you had 1/3 chance to get each and one is eliminated, then do you have 1/3 chance to get the remaining? No, its 1/2 each. Bayes rule: (1/3)/(2/3), trials: 33 1, 33 2, 33 3, cross out all 3s left with 33 of each, 66 historical outcomes 33/66 = 1/2. Now I'm not saying it doesn't makes sense that monty is giving you information by reacting to your choice. But I wan't a precise explanation of this and one that depends on a 1/3 chance to get the car at a point where you should be looking at it as a 1/2 chance doesn't cut it for me.

Someone show me a version or explanation of Bayes formula where it talks about how you got the information given being signifigant. I can see how this idea might make sense as if you take

(Prob (Outcome 1|Not outcome 3) = .5) * (50% chance you were given Not Outcome 3, given Outcome 1, given not outcome 3 (irrel. given outcome 1)) = .25

(Prob (Outcome 2|Not outcome 3) = .5) * (100% chance you were told Not Outcome 3, given outcome 2, given not outcome 3 (again irrel.)) = .5

This is a totally different explanation because you are not using a 2/3 probability to choose a goat after you should be using a 1/2 probability to have chosen a goat. The .75 total represents the probability (Outcome 1 or Outcome 2) given (not outcome 3) given (not Outcome 3 Was given at a 50% chance if Outcome 1 or 100% chance Outcome 2). The .25 Compliment represents the meaningless probability (Outcome 1) given (not outcome 3) given (not outcome 2 was given at a .5 probability) ... which is meaningless in this case because not outcome 2 was not given.

The more general form of Bayes' theorem I used here for use when how often the information given is correlated with the outcome:

Prob(A|B) = (Prob(B|A)*Prob(A)*Prob(B was given|A))/(Prob(B)) / ((Prob(B|A)*Prob(A)*Prob(B was given|A))/(Prob(B)) + (Prob(B|A)*Prob(A)*Prob(B was given|A compliment))/(Prob(B))

Now I can figure out any similar game with differing amounts of money behind doors or no matter what monty does to decide what door to open. It is wrong to say "You had a 2/3 chance of choosing the goat in which case monty has to choose the other one" when the conditional probability of your having chosen the goat given you didnt choose the one he shows you is 1/2.

What happened was in the 1/3 occasion you chose goat 1 and the 1/3 occasion you chose goat 2, he shows you the other goat for sure whereas in the 1/3 occasion you got the car, he shows you the goat you see by 50% chance. That may have happened because there were 2 goats and 1 car, and you had a 2/3 chance to choose a goat until you see otherwise because there were 2 goats and 1 car, but it did not happen because you had a 2/3 chance to choose a goat at the time of the final decision.. you did not.

SharksFan08
02-13-2006, 04:24 PM
Good job! I really value your criticisms alot. (cough cough)

I liked this line the best.

Gandalf
02-13-2006, 04:28 PM
Has renaissoxx at least agreed that the correct answer to the Monty Hall problem is 2/3 in favor of switching, even if he does not yet agree why it is the correct answer?

SharksFan08
02-13-2006, 04:35 PM
Beats me. I'm confused what he means with "I think after seeing a goat you have a 1/2 chance of winning (in a situation without the monty reaction type thing)."

Doesn't the solution to the problem hinge on the "monty reaction type thing"?

Regardlesss, I'm (sadly) impressed by his persistence on trolling this thing.

Gandalf
02-13-2006, 04:36 PM
Doesn't the solution to the problem hinge on the "monty reaction type thing"?
Most of us think so. We're probably wrong. :roll:

Actuary321
02-13-2006, 04:50 PM
Personally, I think Bayes simply complicates this problem unnescesarily.

There are 3 doors, 2 goats and a car.
You pick a door, Monty gets the other 2 doors.
Would you now agree that you have a 1/3 chance of getting the car and Monty has a 2/3 chance of getting the car?
Now whether Monty shows you anything or not you are given the choice of sticking with your result (the door you picked) or switching to Monty's result (the 2 doors you didn't pick).

Monty doesn't need to show you anything because one of his doors has a goat by definition. He can't not have a goat.

The choice is not between the original door you chose and the door Monty didn't open. It is between success with the door you chose or failure with the door you chose. The odds of the car being behind the door you chose didn't change. By switching you get the odds of you having failed to choose the correct door in the original problem.

Dr T Non-Fan
02-13-2006, 04:51 PM
Has renaissoxx at least agreed that the correct answer to the Monty Hall problem is 2/3 in favor of switching, even if he does not yet agree why it is the correct answer?
I don't think so. He refuses to believe what anyone tells him. He refuses to understand and agree with the five or six explanations on wiki, ones with Venn diagrams and logical path charts. He refuses to play the games on the internet 100 times or so to see empirical evidence that would help him to understand which path is correct. He refuses to recognize equivalent-outcome games that allow the answer to be easier to understand.

He has to understand it himself from his own thoughts. Since any line of reasoning that leads him to the correct answer must have already been traveled, that path must not be taken. That leaves him only with paths to the wrong answer.

As for the trolldom, I find that impossible cases help my mind create better reasoning and teaching skills. If it were all easy, then we wouldn't learn how to explain things better. I asked my wife this question. She immediately answered 1/2, but that is the simple, intuitive solution. After I explained it to her (I used the 52-card deck exaggeration of the original problem), I think she understood.

whisper
02-13-2006, 04:52 PM
Anyways all that being said... No I don't agree that after seeing a goat you have a 1/3 chance of winning. I think after seeing a goat you have a 1/2 chance of winning (in a situation without the monty reaction type thing). If you had 1/3 chance to get each and one is eliminated, then do you have 1/3 chance to get the remaining? No, its 1/2 each. Bayes rule: (1/3)/(2/3),

:roll:
That isn't Bayes Theorem, not even close to it.

Now, lets look at your little example.
What is the pr(A=1 | C = 0) given
pr(C = 0 | B = 1) = 100% and pr(C = 0 | A = 1) = 1/2.

Well, lets look at all the possibilities.
[1 0 0]
[1 0 1]
[0 1 0]
[0 0 1]

each equally likely.

So, pr(A = 1 | C = 0) = pr(A = 1) / pr(C = 0) * pr(C = 0 | A = 1) = (1/2) / (1/2) * 1/2 = 1/2.

Now, compare the scenario above (bolded), to the Monty Hall situation:
[1 0 0]
[0 1 0]
[0 0 1]

Look at the 2 sets. They are fundamentally different. Yet, you're arguing the Bayes Theorem will result in the same set of probabilities. Can you see how that cannot be true?

Dr T Non-Fan
02-13-2006, 04:57 PM
Personally, I think Bayes simply complicates this problem unnescesarily.

There are 3 doors, 2 goats and a car.
You pick a door, Monty gets the other 2 doors.
Would you now agree that you have a 1/3 chance of getting the car and Monty has a 2/3 chance of getting the car?
Now whether Monty shows you anything or not you are given the choice of sticking with your result (the door you picked) or switching to Monty's result (the 2 doors you didn't pick).

Monty doesn't need to show you anything because one of his doors has a goat by definition. He can't not have a goat.

The choice is not between the original door you chose and the door Monty didn't open. It is between success with the door you chose or failure with the door you chose. The odds of the car being behind the door you chose didn't change. By switching you get the odds of you having failed to choose the correct door in the original problem.
One of my equivalent game examples is the following:
1. You choose a door.
2. Monte allows you to switch your one chosen door for both of the other doors, but you owe him the goat that is behind one of them.
Do you switch or keep your chosen door?
The outcomes are the same, since you give Monte the goat that he would have showed you.
This game involves the "collapsing of probabilities to one door" (a big sticking point to our "Multiple Choice Answer E"), but since the probabilities and outcome are the same as the original problem, it better explains why it is better to switch than the keep your original choice.

Gandalf
02-13-2006, 05:21 PM
Has renaissoxx at least agreed that the correct answer to the Monty Hall problem is 2/3 in favor of switching, even if he does not yet agree why it is the correct answer?
I don't think so. He refuses to believe what anyone tells him. He refuses to understand and agree with the five or six explanations on wiki, ones with Venn diagrams and logical path charts. He refuses to play the games on the internet 100 times or so to see empirical evidence that would help him to understand which path is correct. He refuses to recognize equivalent-outcome games that allow the answer to be easier to understand.Then maybe it is time to quit until he tries a simulation, which will establish that 50-50 is wrong even if all of us are clueless about why 50-50 is wrong.

As for the trolldom, I find that impossible cases help my mind create better reasoning and teaching skills. If it were all easy, then we wouldn't learn how to explain things better.The deficiency is not in the explanations. As long as you are explaining with words, he can claim contradictions, other points, confusion, etc.

Let him play the game in a simulated environment. If he claims he won 50% without switching, we can examine the data. (Hence, he should keep a record of what he started with, what door was opened, where the car was. If he wants to play an internet location without recording it, OK if it convinces him, but if he wants to convince us he won 50% he needs to give those specifics. OTOH, there are 99 trials already set for him in this thread. All he's got to do is make and record his initial guesses, and it will be demonstrated that unless his initial guesses were extremely good, he can't win 50% by sticking.)

Alto Reed on a Tenor Sax
02-13-2006, 05:40 PM
I am not trolling on anything. I don't agree with some of you guys' approach to intellectual tasks. Lets me state a few things and see if you guys can see where I am coming from.

1) None of you are responsible for pioneering the vast majority of this knowledge. Just because you are capable of following someone else's path of reasoning and no alarms go off in your mind telling you that there is a problem with it doesn't mean that

A) You did the same thing the person who invented it did. Did you consider all alternatives, did you accept only in absence of any dissenting information, were you open to dissenting information or did you sweep it under the carpet? Did you find that the results stood well in the face of all opposition in the view of others (ie not you)? If not, I don't believe you really understand it or know that

...

2) If you are trying to figure something out with out other people giving you the "answer" it is insanely difficult. Its like trying to get the big picture with tunnel vision, a person can process so many relations (4 at most?) at once. It looks good when you consider these but this doesn't add up, someone who doesn't want to agree could claim this etc. Remembering something once you figure it out is easy. You guys aren't even subjecting the ideas to the level of scrutiny really needed to guarantee its accuracy (Someone told you X and you connected it to some other things you know and said ok without taking the time to look for or consider all dissenting Y's which weren't glaring) and then remembering it and then criticisizing me for not "understanding". Good job! I really value your criticisms alot. (cough cough) I could be wrong but I seriously doubt anyone who reguarly goes through the most rigorous process of determining truth ever does this type of thing, because they know better...

Holy crap. Remind me to hire this guy when it comes time to do our annual statements.

"Sorry boss, I haven't finished yet, because I'm not done completely analyzing from scratch all the spreadsheets involved, as well as the source code for the spreadsheet application that created them, the CPU for the computer I'm using...yeah yeah, I know you want me to just "trust" Excel, but that's not going to help me KNOW that the answer is right...I should be done around 2021, maybe later, since I can only analyze about 4 relations at once..."

Sox, I don't agree with your approach to SCIENCE. Your take is GREAT (and by "great" I mean "pretty lame") to PHILOSOPHY, but in science, you make an observation first, then you form a hypothesis about the data you accumulate.

By the way, how's that simulation coming?

renaissoxx
02-13-2006, 05:49 PM
:roll:
That isn't Bayes Theorem, not even close to it.

Now, lets look at your little example.
What is the pr(A=1 | C = 0) given
pr(C = 0 | B = 1) = 100% and pr(C = 0 | A = 1) = 1/2.

Well, lets look at all the possibilities.
[1 0 0]
[1 0 1]
[0 1 0]
[0 0 1]

each equally likely.

So, pr(A = 1 | C = 0) = pr(A = 1) / pr(C = 0) * pr(C = 0 | A = 1) = (1/2) / (1/2) * 1/2 = 1/2.

Now, compare the scenario above (bolded), to the Monty Hall situation:
[1 0 0]
[0 1 0]
[0 0 1]

Look at the 2 sets. They are fundamentally different. Yet, you're arguing the Bayes Theorem will result in the same set of probabilities. Can you see how that cannot be true?

If I understand you right, you did not understand me. Although, where did you get the trials with more than one outcome?

Second where did you get this? "pr(C = 0 | A = 1) = 1/2"
I never said the probability not outcome 3 given Outcome 1 is 1/2. I said the probability that the information not outcome 3 is given to you, given outcome 1 is 1/2. To be honest I don't understand what you are saying here at all.

If your just saying that you don't have a 1/2 chance to choose the goat after seeing that you didn't choose one, I'll just go through it one more time labeling everything:

Bayes theorem: P(A|B) = P(B|A) * P(A) / P(B), and if B = not outcome 3 with a 2/3 chance and A = Outcome 1 with 1/3 starting chance then this equals:
P(A|B) = 1 * 1/3 / (2/3) = 1/2

Actuary321
02-13-2006, 05:57 PM
Well I meant just 1/3 chance of 3 outcomes, then given one has not occured. (1/3 chance of car, goat 1, goat 2, given not goat 2 for example)

Once you see a goat you have this information. Yeah the solution to the problem hinges on the reaction type thing. Its just I don't agree with any of the explanations of the problem.When he says explanations, he really means definitions.

And when you don't agree with the definition of the problem you will not agree with the solution.

renaissoxx
02-13-2006, 06:09 PM
When he says explanations, he really means definitions.

And when you don't agree with the definition of the problem you will not agree with the solution.

Actually when I say explanation, I mean explanation. As in their way of approaching the problem is utterly wrong and only stumbles upon the solution by sheer coincidence. Its kinda like their gut instinct tells them its true and then they just say whatever sounds technical enough and backs up their claim.

whisper
02-13-2006, 06:43 PM
If I understand you right, you did not understand me. Although, where did you get the trials with more than one outcome?

Second where did you get this? "pr(C = 0 | A = 1) = 1/2"
I never said the probability not outcome 3 given Outcome 1 is 1/2. I said the probability that the information not outcome 3 is given to you, given outcome 1 is 1/2. To be honest I don't understand what you are saying here at all.

If your just saying that you don't have a 1/2 chance to choose the goat after seeing that you didn't choose one, I'll just go through it one more time labeling everything:

Bayes theorem: P(A|B) = P(B|A) * P(A) / P(B), and if B = not outcome 3 with a 2/3 chance and A = Outcome 1 with 1/3 starting chance then this equals:
P(A|B) = 1 * 1/3 / (2/3) = 1/2

:roll:

I see, you just reversed the standard question.

OK. Lets go through derivation

Assume A is selected.
pr(c revealed | A = goat) = 1 - pr(c revealed | A = prize) = 1 - 1/2 = 1/2
pr(c revealed | B = goat) = 1 - pr(c revealed | B = prize) = 1 - 1 = 0
pr(c revealed | C = goat) = 1 - pr(c revealed | C = prize) = 1 - 0 = 1

pr(c revealed) = 1/3 * [1/2 + 0 + 1] = 1/2

pr(A = goat | c revealed) = pr(A goat) / pr(c revealed) * pr(c revealed | A = goat) = (2/3) / (1/2) * (1/2) = 2/3.

Old Actuary
02-13-2006, 06:45 PM
First of all, I don't think Bayes theorem can be used, since the probability that Monty picks a specific door is undefined - unless your version of the problem indicates how Monty picks the door.

Coincidentally, I was cleaning out a closet and happened to find a 1991 magazine (OR/MS Today) which has a discussion of the problem. The analysis was quite different than anyone here has discussed, except perhaps for Soxx.

Some excerpts from the magazine:

However, the answer to the question of whether or not a contestant should switch depends on the assumptions one makes about the motivation of the game show host. . . .

First, many readers are apparently assuming that the game show host is neither benevolent nor malevolent and has simply opened a door at random. If that is the case, there is no benefit to switching. The problem states that "the host, who knows what's behind the doors, opens another door..." This statement [I]doesn't, in fact, preclude the possibility that the door was picked at random. It is not the case that the host has necessarily opened a losing door on purpose; we know only that he knew which was the wining door but he opened a losing door. The host may have known where the car was but still picked a door to open at random, or he may have known where the car was but was somehow instructed by someone backstage to open a door which had been randomly selected.

Second, Ms. vos Savant apparently assuming that the host is benevolent. If we assume that the host is, in fact, trying to be helpful, Ms. vos Savant's logic and conclusion [follow].

Third, though, many people are more cynical. . . .These people feel the game host may be malevolent, trying to trick the contestant. If the host is in fact malevolent, the candidate should [I]not switch. If a malevolent host knows that the contestant has not already picked the winning door, he will not offer a possible switch. If he knows that the contestant has picked the winning door he will open either of the remaining doors and offer a switch. Following Ms. vos Savant's logic, the contestant would accept the switch. Thus, a malevolent host knowing the winning door could ensure that the contestant would never win.

Seems like food for thought.

whisper
02-13-2006, 06:48 PM
Here is another way to look at the problem.

Lets say Monty Hall shows you a door (door C) before you select a door.
The possiblities then are
[1 0 0]
[0 1 0]

In this problem, you have a 50% chance of choosing a goat.

What you're trying to do is argue that the scenario:
[1 0 0]
[0 1 0]
[0 0 1]

has the same probability as the above set.

It should be visually obvious that the two sets are different. So, the results cannot be the same.

whisper
02-13-2006, 06:50 PM
Seems like food for thought.

The game changes depending on the assumptions.

If the game show host has the option of NOT revealing a door, the strategy changes dramatically. Also, the strategy would change if there is at least one good prize, but there could be more. The strategy would change if the host could open the door with the prize as well.

renaissoxx
02-13-2006, 07:02 PM
Has renaissoxx at least agreed that the correct answer to the Monty Hall problem is 2/3 in favor of switching, even if he does not yet agree why it is the correct answer?

When I originally saw the problem saying

"There are three possible scenarios, each with equal probability (1/3):

The player picks goat number 1. The game host picks the other goat. Switching will win the car.
The player picks goat number 2. The game host picks the other goat. Switching will win the car.
The player picks the car. The game host picks either of the two goats. Switching will lose"

I said thats wrong, you have a 1/2 chance to get each remaining outcome once you see that one of them is impossible having seen a goat, and it is at this time you must decide. This is weather you use Bayes' theorem which as far as I had used it (but not any more) did not consider how you got information, or just the common sense that you cannot have chosen goat 1 if Monty does so you should cross that possibility off. Any argument they make against this would be circular as all the explanations depend on saying that you had a 2/3 chance to choose a goat even though at the time a goat is revealed you know that is was etc not the case. Just like if you flip a coin and get heads you don't say the probability I got tails was .5, you say it is was 0.

The only argument which could be succesful in explaining this problem is one which takes into account the information that is given when you see that you did not choose the goat, and I created such an explanation for myself. I was never commiting to the claim that the answer was 1/2. I was saying that the explanations were wrong, and you had a 1/2 chance to have chosen a goat once you reveal one.

The only explanation for this problem that I think is, if not correct, is a good explanation is this one:

(There is a 50% chance of having chosen the car normally when given you didn't choose goat 2), but since you only know you chose goat 2 50% of the time you chose the car, but you know that 100% of the time you choose the other goat, the chance is one third.

Another words:

P(Car|Not Goat 2) = (.5 * .5) / (.5 * .5 + .5 * 1) = .333333

:roll:

I see, you just reversed the standard question.

OK. Lets go through derivation

Assume A is selected.
pr(c revealed | A = goat) = 1 - pr(c revealed | A = prize) = 1 - 1/2 = 1/2
pr(c revealed | B = goat) = 1 - pr(c revealed | B = prize) = 1 - 1 = 0
pr(c revealed | C = goat) = 1 - pr(c revealed | C = prize) = 1 - 0 = 1

pr(c revealed) = 1/3 * [1/2 + 0 + 1] = 1/2

pr(A = goat | c revealed) = pr(A goat) / pr(c revealed) * pr(c revealed | A = goat) = (2/3) / (1/2) * (1/2) = 2/3.

None of these have anything to do with my posts. Read them or don't read them, this straw man stuff is getting old. You are still claiming a 1/3 probability for each outcome at a point where you can see that it is not a 1/3 probability.

whisper
02-13-2006, 07:18 PM
When I originally saw the problem saying

"There are three possible scenarios, each with equal probability (1/3):

The player picks goat number 1. The game host picks the other goat. Switching will win the car.
The player picks goat number 2. The game host picks the other goat. Switching will win the car.
The player picks the car. The game host picks either of the two goats. Switching will lose"

:roll:
Do you think about what you write?

Basic statistics:

If there are 3 scenarios, each equally likely, and you win 2 out of the 3 times switching - what is the probability of winning if you switch?

renaissoxx
02-13-2006, 07:31 PM
:roll:
Do you think about what you write?

Basic statistics:

If there are 3 scenarios, each equally likely, and you win 2 out of the 3 times switching - what is the probability of winning if you switch?

I don't understand you. What is the point of your posting here? You never read a word I say, then post straw man after straw man as if you understand what I wrote and think it is silly. If you don't want to read my posts, who is making you?

When you get to the point in any of those 3 setences, where it says either "The Game host picks the other goat" or "The game host picks either of the two goats" you stop and cross off the setence starting with "The player picks" the goat that the Host has shown. This is what I am claiming. Ok?

Gandalf
02-13-2006, 07:41 PM
Everyone has the outcome that matches experimental results. Everyone has an approach that would match experimental results if you change it to one car and n goats. But no one is right.

Right. :roll:

Within the constraints of the original problem (that Monte will always show a goat, but Monte may have a strategy that you don't know for choosing which goat if you have chosen the car), it is possible that the odds are only 50-50 to switch when if you see Goat 1. However, if the odds on switching when you see Goat 1 are only 50-50, the odds when you see Goat 2 become 100%, and it must remain the case that the odds on switching when you see goat are 2/3.

renaissoxx
02-13-2006, 08:16 PM
I spent some time earlier coming up with an example of a situation where the types of explanations used on the wiki wouldn't work but my understanding of the situation would. But then I realized you would just say it wasn't the same problem, and it would probably take you a long time to solve the problem without any clear alogrithm for doing them.

What If 1 door had a goat, 1 door had 500 dollars, and one door had 1000 dollars, and when you chose the 1000 Monty rolled a 3 sided die and picked the goat if 2 or 3 rolled, 500 if 1 rolled, and if you picked one of these two he just picked the other. Whats expected value of switching if Monty reveals the goat?

Using the information you have at this point in the game and nothing more:

You see the goat.
Prob(1000|goat) = 1/2 * 2/3 / (1/2 * 2/3 + 1/2 * 1) = 2/6 / (2/6 + 1/2) = 2/5
Prob(500|goat) = 1/2 / (2/6 + 1/2) = 3/5

500 * 2/5 + 1000 * 3/5 = 800 Switching
500 * 3/5 + 1000 * 2/5 = 700 Staying

I don't see how you are going to answer any question where it doesn't happen to occur that the relative frequency with which he shows you a particular goat doesn't by sheer luck match the starting probability of choosing any goats.

Ailing Factuary
02-13-2006, 08:30 PM
Mods, can you ban this guy?

Westley
02-13-2006, 08:38 PM
Mods, can you ban this guy?You could always try ignoring him. :lol:

Seriously, there's a thread in political about Riscalculator, and Titania mentions he's a good troll, and that thread is a page and a half. This guy has gone eight full pages just saying "But what if two plus two equals five - did you consider that? How can you be sure? If I take two cups, and add two more cups, and there's less than five, how do I know that my 'cup model' isn't flawed and the real answer is five?"

You guys are smarter than this.

Westley
02-13-2006, 08:49 PM
And besides, when you take two cups and add two more cups, maybe it gives you four cups this time, but how would you prove that it would give you four cups every time? Shouldn't we have some kind of probabilistic model for how many cups there will be?

Gandalf
02-13-2006, 08:50 PM
I don't see how you are going to answer any question where it doesn't happen to occur that the relative frequency with which he shows you a particular goat doesn't by sheer luck match the starting probability of choosing any goats.
Right. You don't see. That is clear.

As I said just 5 posts earlier, if you were given more information beyond "Monte will show you a goat", then seeing Goat 1 could change the odds, in the extreme driving them to 50-50. We could solve that problem, but that isn't the problem.

Likewise if Monte were to also open the door you had chosen as well as show you a goat elsewhere, the odds in favor of switching would change dramatically, depending on what you saw. Does that mean there is no answer to the case where Monte doesn't open the door you chose? We could solve that one; I don't know if you could figure out the algorithm.

:roll:

whisper
02-13-2006, 11:12 PM
OK, last time I'm going to try to explain this. I'm going to set up two problems. Here is the 1st.

There are three doors A B C.
Behind each door, there is an object 0 1 2.

Example 1:
In this example the rules are as follows:
The Host can never reveal 2.
If the Host can reveal 0, he must reveal 0. Otherwise he will reveal 1.

So, the universe of possible outcomes is this:
[2 1 0] U
[2 0 1] V
[1 2 0] W
[1 0 2] X
[0 2 1] Y
[0 1 2] Z

pr(u) = pr(v) = pr(w) = pr(x) = pr(y) = pr(z) = 1/6

What is pr(A=2 | C =0) using Baysian analysis.

pr(C = 0 | A = 2) = 1/2 * (1 + 0) = 1/2 (as you can see by looking at U&V)
pr(C = 0 | B = 2) = 1/2 * (1 + 0) = 1/2 (as you can see by looking at W&X)
pr(C = 0 | C = 2) = 1/2 * (0 + 0) = 0 (as you can see by looking at Y&Z)

pr(C = 0) = 1/3 * (1/2 +1/2 + 0) = 1/3

So, using Bayes Theorem, we have
pr(A=2 | C = 0) = pr(A=2)/pr(C=0)*pr(C=0|A=2) = (1/3)/(1/3)*(1/2) = 1/2

Example 2:
In this example the rules are as follows:
The Host can never reveal 2.
If the Host can reveal 0 or 1, with equal probability.

So, the universe of possible outcomes is this:
[2 1 0r] S
[2 1r 0] T
[2 0r 1] U
[2 0 1r] V
[1 2 0r] W
[1 0r 2] X
[0 2 1r] Y
[0 1r 2] Z

pr(s) = pr(t) = pr(u) = pr(v) = 1/12
pr(w) = pr(x) = pr(y) = pr(z) = 1/6

What is pr(A=2 | C = 0r) using Baysian analysis.

pr(C = 0r | A = 2) = 1/4 * (1 + 0 + 0 + 0) = 1/4 (as you can see by looking at S&T&U&V)
pr(C = 0r | B = 2) = 1/2 * (1 + 0) = 1/2 (as you can see by looking at W&X)
pr(C = 0r | C = 2) = 1/2 * (0 + 0) = 0 (as you can see by looking at Y&Z)

pr(C = 0r) = 1/3 * (1/4 +1/2 + 0) = 1/4

So, using Bayes Theorem, we have
pr(A=2 | C = 0r) = pr(A=2)/pr(C=0r)*pr(C=0r|A=2) = (1/3)/(1/4)*(1/4) = 1/3

_________________________________________

Notice, I made no restrictions on what 2, 1 and 0 are. All I specified is what the host can reveal.

Now, Example 1 is your standard conditional probability. Example 2 is Monty Hall problem.

You're analysis is flawed, because you're discounting the fact when A=2, the host can reveal C or B - he's not forced to only reveal C.

Cloister
02-13-2006, 11:44 PM
You're analysis is flawed, because you're discounting the fact when A=2, the host can reveal C or B - he's not forced to only reveal C.

Exactly. Why no one has pointed this out so far (that I've noticed) is surprising.

His belief (assuming not a troll) is that: (There is a goat behind door 2) is equivalent to (Monte shows you a goat behind door 2). They aren't the same at all - Monte showing you a goat imparts more information than simply that there's a goat, because his options are restricted.

Old Actuary
02-14-2006, 12:19 AM
pr(u) = pr(v) = pr(w) = pr(x) = pr(y) = pr(z) = 1/6

You state that equation as a fact, but it's only an assumption. You are implicitly assuming that the objects are distributed randomly.

My question is: does the answer depend on the strategy used by the host? I believe all the Bayesian formulas assume that the strategy is known and that it is random. That may not be true.

JMO
02-14-2006, 07:13 AM
Mods, can you ban this guy?
Not necessary. DNFTT.

Alto Reed on a Tenor Sax
02-14-2006, 08:43 AM
You could always try ignoring him. :lol:

Naaah. I think most of us know somebody like this in real life, who is NOT a troll, just an idiot. Maybe we are getting some kind of satisfaction whaling onSox as the avatar of every single idiot who has ever argued with us, getting angrier and angrier despite being dead, dead wrong.

whisper
02-14-2006, 09:39 AM
You state that equation as a fact, but it's only an assumption. You are implicitly assuming that the objects are distributed randomly.

I'm explicitly assuming that the objects are distributed randomly. That's why I had that formula in my analysis. :)

My question is: does the answer depend on the strategy used by the host? I believe all the Bayesian formulas assume that the strategy is known and that it is random. That may not be true.

Yes, the results depend on the strategy of the host and the game.
The approach will show you which is the best strategy given whatever assumptions you wish to add.

MathGeek92
02-14-2006, 09:59 AM
After changing my pants due to p*ssing from laughter ....

One thing is perfectly clear. Sox does not believe items that have been proven as fact. How much proof does Sox need to prove facts?

The only sentence I agree with is we did not "invent" this math stuff, but it's clear we understand it. This whole string is like arguing with my wife that you don't have a 50-50 chance of winning the lottery (she states either you win or you don't).

Oops need to change my pants again.

renaissoxx
02-14-2006, 11:17 AM
OK, last time I'm going to try to explain this. I'm going to set up two problems. Here is the 1st.

There are three doors A B C.
Behind each door, there is an object 0 1 2.

Example 1:
In this example the rules are as follows:
The Host can never reveal 2.
If the Host can reveal 0, he must reveal 0. Otherwise he will reveal 1.

So, the universe of possible outcomes is this:
[2 1 0] U
[2 0 1] V
[1 2 0] W
[1 0 2] X
[0 2 1] Y
[0 1 2] Z

pr(u) = pr(v) = pr(w) = pr(x) = pr(y) = pr(z) = 1/6

What is pr(A=2 | C =0) using Baysian analysis.

pr(C = 0 | A = 2) = 1/2 * (1 + 0) = 1/2 (as you can see by looking at U&V)
pr(C = 0 | B = 2) = 1/2 * (1 + 0) = 1/2 (as you can see by looking at W&X)
pr(C = 0 | C = 2) = 1/2 * (0 + 0) = 0 (as you can see by looking at Y&Z)

pr(C = 0) = 1/3 * (1/2 +1/2 + 0) = 1/3

So, using Bayes Theorem, we have
pr(A=2 | C = 0) = pr(A=2)/pr(C=0)*pr(C=0|A=2) = (1/3)/(1/3)*(1/2) = 1/2

Example 2:
In this example the rules are as follows:
The Host can never reveal 2.
If the Host can reveal 0 or 1, with equal probability.

So, the universe of possible outcomes is this:
[2 1 0r] S
[2 1r 0] T
[2 0r 1] U
[2 0 1r] V
[1 2 0r] W
[1 0r 2] X
[0 2 1r] Y
[0 1r 2] Z

pr(s) = pr(t) = pr(u) = pr(v) = 1/12
pr(w) = pr(x) = pr(y) = pr(z) = 1/6

What is pr(A=2 | C = 0r) using Baysian analysis.

pr(C = 0r | A = 2) = 1/4 * (1 + 0 + 0 + 0) = 1/4 (as you can see by looking at S&T&U&V)
pr(C = 0r | B = 2) = 1/2 * (1 + 0) = 1/2 (as you can see by looking at W&X)
pr(C = 0r | C = 2) = 1/2 * (0 + 0) = 0 (as you can see by looking at Y&Z)

pr(C = 0r) = 1/3 * (1/4 +1/2 + 0) = 1/4

So, using Bayes Theorem, we have
pr(A=2 | C = 0) = pr(A=2)/pr(C=0r)*pr(C=0r|A=2) = (1/3)/(1/4)*(1/4) = 1/3

_________________________________________

Notice, I made no restrictions on what 2, 1 and 0 are. All I specified is what the host can reveal.

Now, Example 1 is your standard conditional probability. Example 2 is Monty Hall problem.

You're analysis is flawed, because you're discounting the fact when A=2, the host can reveal C or B - he's not forced to only reveal C.

0r is the event that C is 0, >>AND<< it is revealed by the host. On the last line of your post, where you write:

"pr(A=2 | C = 0) = pr(A=2)/pr(C=0r)*pr(C=0r|A=2) = (1/3)/(1/4)*(1/4) = 1/3"

That is totally wrong using your definition of bayes' formula because the events C=0, and C=0 and it is revealed (0r) are NOT the same. If you calculate the actual probability A=2 | C = 0, you get 1/2. If you look at your tables, the top 2 1/12 probability items are both C = 0 as well as one of the 1/6 probability items which puts them on equal ground.

I am not sure what direction you are trying to go with your straw men arguments now. It looks like you are trying to recreate my argument worded a different way and then take credit for it. People like you who try and decieve readers to try and make them falsely believe you are correct about something and someone else is wrong when you don't know or care if its the case make me sick. Perhaps this is what Dr.T meant about "good troll". They come and figure something out, and then you cover it up by constantly insulting them and trying to drown out their posts and then reword their argument and claim you came up with it.

But in this case youve only managed to make my point even clearer. As prob(A=2|C=0) = 1/2, and prob(A=2|C=0 Intersection C is revealed) = 1/3, the problem with bayes' theorem should become clear. Simply being given a piece of information is not enough. You have to know if there was a correlation between the information being given to you and the outcome.

whisper
02-14-2006, 11:32 AM
0r is the event that C is 0, >>AND<< it is revealed by the host. On the last line of your post, where you write:

"pr(A=2 | C = 0) = pr(A=2)/pr(C=0r)*pr(C=0r|A=2) = (1/3)/(1/4)*(1/4) = 1/3"

That is totally wrong using your definition of bayes' formula because the events C=0, and C=0 and it is revealed (0r) are NOT the same.

You're right, there is a typo in my explanation, which I've fixed.
It should have read: pr(A=2 | C = 0r) = pr(A=2)/pr(C=0r)*pr(C=0r|A=2) = (1/3)/(1/4)*(1/4) = 1/3

I am not sure what direction you are trying to go with your straw men arguments now. It looks like you are trying to recreate my argument worded a different way and then take credit for it.

:roll:
These aren't strawman arguments.

People like you who try and decieve readers to try and make them falsely believe you are correct about something and someone else is wrong when you don't know or care if its the case make me sick.

I've adequately demonstrated the math, the scenarios, and the reasoning behind it. If you want to look at this, say "You're still wrong!" than that is your perogative. But, you've not demonstrated that I'm wrong.

You're basic belief is wrong.

Look at this way. There are 2 contestants, they are unaware of each other.

The 1st contestant comes in and is faced with the following set of possibilities and he has to guess where the one is.
[1 0 0]
[0 1 0]
[0 0 1]

After he selects, a losing possibliity is revealed to him.

You're asserting he has the exact same probability of winning as the 2nd contestant who is unaware of the 1st selection, but is aware of the revealed answer and is faced with the set:
[1 0 0]
[0 1 0]

If you can't understand why the probability cannot be the same, than it is your lack of comprehension of math and statistics.

renaissoxx
02-14-2006, 11:49 AM
You're right, there is a typo in my explanation, which I've fixed.
It should have read: pr(A=2 | C = 0r) = pr(A=2)/pr(C=0r)*pr(C=0r|A=2) = (1/3)/(1/4)*(1/4) = 1/3

:roll:
These aren't strawman arguments.

I've adequately demonstrated the math, the scenarios, and the reasoning behind it. If you want to look at this, say "You're still wrong!" than that is your perogative. But, you've not demonstrated that I'm wrong.

You're basic belief is wrong.

Look at this way. There are 2 contestants, they are unaware of each other.

The 1st contestant comes in and is faced with the following set of possibilities and he has to guess where the one is.
[1 0 0]
[0 1 0]
[0 0 1]

After he selects, a losing possibliity is revealed to him.

You're asserting he has the exact same probability of winning as the 2nd contestant who is unaware of the 1st selection, but is aware of the revealed answer and is faced with the set:
[1 0 0]
[0 1 0]

If you can't understand why the probability cannot be the same, than it is your lack of comprehension of math and statistics.

I've already explained why the probabilities are not the same, as well as why arguments to that effect which depend on a 2/3 probability to have chosen a goat after a goat has been revealed do not make sense. >>I<< gave arguments which do not depend on this type of reasoning, as well as the reasoning behind them. And now you simply recreate my argument worded slightly differently and claim that I was claiming something else. (Although less clearly imo, as you didn't even specify which door was chosen) Yes that involves a strawman argument.

Furthermore I believe you know that, and are just trying one underhanded tactic after another to hide this fact. You wouldn't have went down this path if you didn't partially understand that this is what I was doing. You are completely amoral.

whisper
02-14-2006, 12:05 PM
I've already explained why the probabilities are not the same, as well as why arguments to that effect which depend on a 2/3 probability to have chosen a goat after a goat has been revealed do not make sense. >>I<< gave arguments which do not depend on this type of reasoning, as well as the reasoning behind them. And now you simply recreate my argument worded slightly differently and claim that I was claiming something else. (Although less clearly imo, as you didn't even specify which door was chosen) Yes that involves a strawman argument.

:yawn:
I was going back to 1st principles, and show the conditions that would have to hold true for your assertion (that the probability is 1/2) would have to be for that to be true. I showed the conditions that are in the problem and why they result in a different answer.

So, no I wasn't recreating your arguments worded slightly different. I was showing you the assumptions necessary to create each event.

Furthermore I believe you know that, and are just trying one underhanded tactic after another to hide this fact. You wouldn't have went down this path if you didn't partially understand that this is what I was doing. You are completely amoral.

:lol:
Yeah, me and my nefarious use of math and statistics!
Only the ammoral use mathematics!

SharksFan08
02-14-2006, 12:22 PM
I was never commiting to the claim that the answer was 1/2. I was saying that the explanations were wrong, and you had a 1/2 chance to have chosen a goat once you reveal one.

Can you elaborate on the first part of this? Are you saying that you know 1/2 is not the right answer, or that you just aren't saying definitively that it is the right answer?

For your second sentence, you'd be right if you said you have a 1/2 chance to choose a goat once you reveal one, but not that you have a 1/2 chance to have chosen a goat once you reveal one.

Chubbs
02-14-2006, 12:52 PM
Ok. Here is my question.

You have 3 possible outcomes equally likely.

What is the probability of the first two outcomes given the third outcome has not occured?

1/2 each Correct?

A: Outcome 1 occurs
B: Outcome 3 does not occur

(Prob(B|A)*Prob(A))/Prob(B) = (1*1/3)/(2/3) = 1/2

Also if you did an experiment with 100 trials, you should have close to 33 of each outcome right? What happens when you find out you didn't choose outcome 3? You cross all the outcome 3 results off your list and are left with about 33 1, 33 2.

So the next question is, is that dependent on how you got the information? If it is, I was never taught so in school or determined it was.

It is most certainly depenpent on how you got your information!!
There are three doors. One has a car two have goats. Pick your door, no Monty Hall game, just plain up pick your door once and for all.

Now the game is over, whats the prob you got the car. I think we all agree its 1/3.

Well now let me show you a goat behind one of the doors you didn't pick. Was your prob retroactively changed to 1/2. But if I showed you a goat before the game, your prob would have been 1/2. So clearly how you got the info is very important.

Gandalf
02-14-2006, 12:52 PM
I was never commiting to the claim that the answer was 1/2. I was saying that the explanations were wrong, and you had a 1/2 chance to have chosen a goat once you reveal one.
Can you elaborate on the first part of this? Are you saying that you know 1/2 is not the right answer, or that you just aren't saying definitively that it is the right answer?

Westley explained the first part in post 201 above. Renaisoxx is saying"But what if two plus two equals five - did you consider that? How can you be sure? If I take two cups, and add two more cups, and there's less than five, how do I know that my 'cup model' isn't flawed and the real answer is five?"
If you need further elaboration, see post 202.

Chubbs
02-14-2006, 12:57 PM
This guy is either a complete dolt, or an evil genius who is capable of psuedo proving and semi logical arguements to show just about anything.

Westley
02-14-2006, 12:58 PM
This guy is either a complete dolt, or an evil genius who is capable of psuedo proving and semi logical arguements to show just about anything.But, here's the thing: you are taking two cup and adding the other two cups and you say it's four. But, what if you take the other two cups and then add the first two cups? How can that be four as well - you got four by doing it the first way, and now you did something different, so how can the result possibly be the same?

renaissoxx
02-14-2006, 12:59 PM
:yawn:
I was going back to 1st principles, and show the conditions that would have to hold true for your assertion (that the probability is 1/2) would have to be for that to be true. I showed the conditions that are in the problem and why they result in a different answer.

So, no I wasn't recreating your arguments worded slightly different. I was showing you the assumptions necessary to create each event.

:lol:
Yeah, me and my nefarious use of math and statistics!
Only the ammoral use mathematics!

I already laid out the problem the correct way using math. You simply reworded my explanation. You are a liar. I never asserted that the probability is 1/2. I said that P(A=2|C=0) is 1/2. That is true.

It is most certainly depenpent on how you got your information!!
There are three doors. One has a car two have goats. Pick your door, no Monty Hall game, just plain up pick your door once and for all.

Now the game is over, whats the prob you got the car. I think we all agree its 1/3.

Well now let me show you a goat behind one of the doors you didn't pick. Was your prob retroactively changed to 1/2. But if I showed you a goat before the game, your prob would have been 1/2. So clearly how you got the info is very important.

This is exactly what I have clearly shown mathematically in my posts. The point of the line that you oh so deviously took completely out of context was to show that >>>bayes' theorem<<< does not take into account how you got information. I was not taught that you can only use bayes' theorem if there is no correlation between when the given information is given and the outcome. And yet that is the case here.

Anotherwords, what if you did not know how Monty was choosing to open which door? All you would know is that
the door he shows you didn't have a goat, and using bayes formula you would be arrive at a 1/2 probability that switching would get you the car. But the results would show this was wrong.

What went wrong in this case? You didn't know or account for that there was a correlation between how often the given information was given and what was behind your door. That is, bayes' formula is not complete or needs to only be used with the assumption that there is no correlation between frequency that given information is given and the outcome.

JMO
02-14-2006, 01:02 PM
This guy is either a complete dolt, or an evil genius who is capable of psuedo proving and semi logical arguements to show just about anything.

renaissoxx = griffin

or

renaissoxx = a troll

Take your pick. I like this thread, though, because of all the different ways of looking at the problem get the same answer for everyone else. :yawn:

Westley
02-14-2006, 01:12 PM
renaissoxx = griffin

or

renaissoxx = a troll
Why would you use "or" for that statement?

whisper
02-14-2006, 01:13 PM
I already laid out the problem the correct way using math. You simply reworded my explanation. You are a liar. I never asserted that the probability is 1/2. I said that P(A=2|C=0) is 1/2. That is true.

That is true, if and only if, you know for a fact C = 0. I showed that has to be the correct answer if the host is obligated to reveal C if C=0.

But, in the game, the contestant doesn't know that C = 0 for a fact. He only knows that the host revealed C and C = 0.

The two are two different assertions.

Cloister
02-14-2006, 01:16 PM
I already laid out the problem the correct way using math. You simply reworded my explanation. You are a liar. I never asserted that the probability is 1/2. I said that P(A=2|C=0) is 1/2. That is true.

That is true. What isn't true is:

P(A=2|(Contestent Chose A) and (C Revealed after Contestant Chose A) and (C=0)) = 1/2

What the solution to the Monte Hall problem says is:

P(A=2|(Contestent Chose A) and (C Revealed after Contestant Chose A) and (C=0)) = 1/3

and

P(B=2|(Contestent Chose A) and (C Revealed after Contestant Chose A) and (C=0)) = 2/3

Based on all the discussion, you don't seem to be able to accept the fact (and it is a fact, and has been demonstrated mathematically be several people) that, under the assumptions regarding Monte's behaviour:

P(A=2|C=0) is not equal to P(A=2|(Contestent Chose A) and (C Revealed after Contestant Chose A) and (C=0))

JMO
02-14-2006, 01:18 PM
Why would you use "or" for that statement?
Sorry. I left out the goats.

whisper
02-14-2006, 01:27 PM
Here is another form of the question:

1% of woman over age 40 have breast cancer.
80% of the woman with breast cancer will test positive for breast cancer.
9.6% of woman without breast cancer will test positive for breast cancer.

A woman over the age 40 tests positive for breast cancer.

What is the probability the woman has breast cancer?

http://yudkowsky.net/bayes/bayes.html

Chubbs
02-14-2006, 01:38 PM
I already laid out the problem the correct way using math. You simply reworded my explanation. You are a liar. I never asserted that the probability is 1/2. I said that P(A=2|C=0) is 1/2. That is true.

This is exactly what I have clearly shown mathematically in my posts. The point of the line that you oh so deviously took completely out of context was to show that >>>bayes' theorem<<< does not take into account how you got information. I was not taught that you can only use bayes' theorem if there is no correlation between when the given information is given and the outcome. And yet that is the case here.

Anotherwords, what if you did not know how Monty was choosing to open which door? All you would know is that
the door he shows you didn't have a goat, and using bayes formula you would be arrive at a 1/2 probability that switching would get you the car. But the results would show this was wrong.

What went wrong in this case? You didn't know or account for that there was a correlation between how often the given information was given and what was behind your door. That is, bayes' formula is not complete or needs to only be used with the assumption that there is no correlation between frequency that given information is given and the outcome.

I am certain that the assumptions for the monty hall problem are that he will always show you a goat. If that is true, then most of what you have said is irrelevant.

New question for you, to see if you really are a dolt, or just an evil genius.

You are on a very precise scale and your weight reads "X". You let a fart. After the propulsive forces have been expended and the scale returns to an equilibrium point, your weight reads "Y". Which of the following is true?

X=Y
X>Y
X<Y

Actuary321
02-14-2006, 03:44 PM
0r is the event that C is 0, >>AND<< it is revealed by the host. On the last line of your post, where you write:

"pr(A=2 | C = 0) = pr(A=2)/pr(C=0r)*pr(C=0r|A=2) = (1/3)/(1/4)*(1/4) = 1/3"

That is totally wrong using your definition of bayes' formula because the events C=0, and C=0 and it is revealed (0r) are NOT the same. If you calculate the actual probability A=2 | C = 0, you get 1/2. If you look at your tables, the top 2 1/12 probability items are both C = 0 as well as one of the 1/6 probability items which puts them on equal ground.

I am not sure what direction you are trying to go with your straw men arguments now. It looks like you are trying to recreate my argument worded a different way and then take credit for it. People like you who try and decieve readers to try and make them falsely believe you are correct about something and someone else is wrong when you don't know or care if its the case make me sick. Perhaps this is what Dr.T meant about "good troll". They come and figure something out, and then you cover it up by constantly insulting them and trying to drown out their posts and then reword their argument and claim you came up with it.

But in this case youve only managed to make my point even clearer. As prob(A=2|C=0) = 1/2, and prob(A=2|C=0 Intersection C is revealed) = 1/3, the problem with bayes' theorem should become clear. Simply being given a piece of information is not enough. You have to know if there was a correlation between the information being given to you and the outcome.
You keep using that word? I do not think it means what you think it means.

I think we have finally discovered who renaissoxx really is.

Man in black: [turning his back, and adding the poison to one of the goblets] Alright, where is the poison? The battle of wits has begun. It ends when you decide and we both drink - and find out who is right, and who is dead.
Vizzini: But it's so simple. All I have to do is divine it from what I know of you. Are you the sort of man who would put the poison into his own goblet or his enemy's? Now, a clever man would put the poison into his own goblet because he would know that only a great fool would reach for what he was given. I am not a great fool so I can clearly not choose the wine in front of you...But you must have known I was not a great fool; you would have counted on it, so I can clearly not choose the wine in front of me.
Vizzini: [happily] Not remotely! Because Iocaine comes from Australia. As everyone knows, Australia is entirely peopled with criminals. And criminals are used to having people not trust them, as you are not trusted by me. So, I can clearly not choose the wine in front of you.
Man in black: Truly, you have a dizzying intellect.
Vizzini: Wait 'till I get going!! ...where was I?
Man in black: Australia.
Vizzini: Yes! Australia! And you must have suspected I would have known the powder's origin, so I can clearly not choose the wine in front of me.
Man in black: You're just stalling now.
Vizzini: You'd like to think that, wouldn't you! You've beaten my giant, which means you're exceptionally strong...so you could have put the poison in your own goblet trusting on your strength to save you, so I can clearly not choose the wine in front of you. But, you've also bested my Spaniard, which means you must have studied...and in studying you must have learned that man is mortal so you would have put the poison as far from yourself as possible, so I can clearly not choose the wine in front of me!
Man in black: You're trying to trick me into giving away something. It won't work.
Vizzini: It has worked! You've given everything away! I know where the poison is!
Man in black: Then make your choice.
Vizzini: I will, and I choose...[pointing behind the man in black] What in the world can that be?
Man in black: [turning around, while Vizzini switches goblets] What?! Where?! I don't see anything.
Vizzini: Oh, well, I...I could have sworn I saw something. No matter. [Vizzini laughs]
Man in black: What's so funny?
Vizzini: I...I'll tell you in a minute. First, lets drink, me from my glass and you from yours.
[They both drink]
Man in black: You guessed wrong.
Vizzini: You only think I guessed wrong! That's what's so funny! I switched glasses when your back was turned! Ha ha, you fool!! You fell victim to one of the classic blunders. The most famous is never get involved in a land war in Asia; and only slightly less well known is this: Never go in against a Sicilian, when death is on the line!
[Vizzini continues to laugh hysterically. Suddenly, he stops and falls right over. The Man in black removes the blindfold from the princess.]

tymesup
02-14-2006, 03:58 PM
Suppose Monty has discretion whether to offer the contestant a switch. Further suppose Monty does not want to give away the car.

How often should he offer a switch when the contestant picked the car? How often when the contestant picked a goat?

Gandalf
02-14-2006, 05:03 PM
Suppose Monty has discretion whether to offer the contestant a switch. Further suppose Monty does not want to give away the car.

How often should he offer a switch when the contestant picked the car? How often when the contestant picked a goat?
The problem is not completely defined:
1. Does the contestant know that Monty has the option?
2. Does the contestant know Monty's strategy, or even that Monty's strategy is to avoid giving away the car?
3. If Monty gives a choice, will he still show a goat?

This solution may not be unique, but assuming the contestant knows Monty has discretion and will try to minimize his chances of giving away the car, then a best strategy for Monty is to never give the option to switch if a goat has been chosen. Further, assuming he will never give the option when a goat is chosen, he should always give the option if the car is chosen, but this should not matter.

Perhaps other strategies may be equally good for Monty, but none are better against a contestant who plays to maximize his chances of winning the car.

If Monty will not show a goat when he offers a choice, Monty's definitely has other strategies that are as good (but not better) than never letting a contestant who has chosen a goat switch, against a strong contestant.

Dr T Non-Fan
02-14-2006, 06:24 PM
Well, that problem is more about game theory.
If Monte is trying to minimize his risk of ruin, then he always offers the switch when the contestant picks the car, and he never offers the switch when the contestant picks the goat.
If contestants pick up on this, then no one ever switches when given the choice to switch, and they go away empty-handed when not given the choice to switch.
Monte wins 2/3 of the time, contestant wins 1/3 of the time. It's as if Monte never really had the option of giving a choice to switch.

Griffin
02-15-2006, 12:50 PM
Why would you use "or" for that statement?Because JMO knows how easily you are intimidated.

renaissoxx
02-16-2006, 10:22 AM
Time for me to sum up...

My two main points in this thread are as follows:

The explanations of the problem on Wiki are screwed up. Here I will explain why I think this is the case, as well as a disagreement with those of you saying things like "You knew monty was going to show you a goat so it doesn't give you any new information when he does".

Here are 2 different ways of looking at the problem, the second is the way I look at the problem. The first is the way the wiki and most of you look at the problem.

The probability that Monty shows you (Goat 1 OR Goat 2) is 1.
A: Monty reveals goat 1 or goat 2
Prob(A) = 1
Prob(B) = 1/3, Prob(B|A) = (1*1/3) / (1) = 1/3
Prob(C) = 2/3, Prob(C|A) = (1*2/3) / (1) = 2/3
Event A is logically independent of event B and C.

This is the way I do the problem:
A: Monty reveals goat 1
B: Monty reveals goat 2
D: You chose goat 1
E: You chose goat 2
Prob(A) = Prob(B) = 1/2
Prob(C) = Prob(D) = Prob(E) = 1/3
Prob(C|A) = Prob(C|B) = (1/2 * 1/3) / (1/2) = 1/3

So, in the strictly mathematical sense, the chance of having chosen the car is independent of the information given when he reveals you didn't choose that goat. But logically it obviously is not independent: The chance that you chose the car is raised by knowing that you did not choose the shown goat, but just happens to be offset by the lowered chance of having chosen the car given that its less likely he would have shown that goat if you chose the car than if you had chosen a goat.

And to show this we only need to change the chance monty chooses goat 1 to 2/3 and goat 2 to 1/3 when you chose the car. The first thing to recognize in this case, is that the first way of looking at the problem gives you the exact same answer 1/3... which is wrong in this case. The second way gives you:

A: Monty reveals goat 1
B: Monty reveals goat 2
D: You chose goat 1
E: You chose goat 2
Prob(A) = 1/3 + (1/3)*(2/3) = 5/9
Prob(B) = 1/3 + (1/3)*(1/3) = 4/9
Prob(C) = Prob(D) = Prob(E) = 1/3
Prob(C|A) = ((2/3) * (1/3)) / (5/9) = 2/5
Prob(C|B) = ((1/3) * (1/3)) / (4/9) = 1/4

So, that the first "method" works for solving the problem in the original case is dependent not only on the fact that all you care about is weather or not you won the car... not only on the fact that the two goats are identical, but also on the fact that the chance he reveals each one is identical as well.

To sum up my problem with the wiki explanations, they simply do not use all the information. When Monty reveals a goat hes revealing one goat or the other which is more information than the fact that he reveals one of the two goats which you knew to begin with. If the wiki entry had cited dependency on all the things I mentioned above for their explanations to be a valid way of approaching the problem, ok but its still silly. Why use a method that only works in one case that you have to make sure works using the general method first when you can just use the general method to begin with?

But without citing these dependencies its like saying "Oh multiplication is just adding the two numbers see 2*2=4"....

The second main point I was making in this thread is regarding the Bayes' theorem. I can sum this point up fairly easily just by looking at this scenario:

What if the game was the same but there was no Monty and you were just revealed a goat after making your choice. The goat was revealed according the previous rules, BUT you didn't know that. Without any knowledge of how the goat was revealed you would only have the information that you did not choose whichever goat you were shown, which would lead you to believe that you had a 1/2 chance of winning by switching doors. But that would shown to be wrong by experimentation.

Bayes' formula would lead you to the wrong answer in this case. The result you would obtain by bayes' formula would be shifted by the factor (Prob(B was given|A)) / (Prob(B was given|A) + Prob(B was given|A compliment)) which multiplied by Bayes' Theorem gives you the formula I put up before.

So of course the next question is, any time you use bayes' formula how can you know there isn't a correlation between the likelihood you were given the information you were given and the outcome? You can't really... So Bayes' theorem is dependent on this assumption which is not always true.

Thats all for me folks, Have a nice day.

Westley
02-16-2006, 10:42 AM
Time for me to sum up...

If you take two yellow cups, and add two more yellow cups and get four yellow cups, you can't possibly be crazy enough to think that you can just make this wild assumption that two blue cups plus two more blue cups will give you four blue cups, because when you add two plus two, you get four yellow cups, I already showed you that.

Go ahead and try to explain it to him guys, I'm sure that this time he will actually read what you write.

JMO
02-16-2006, 10:47 AM
Go ahead and try to explain it to him guys, I'm sure that this time he will actually read what you write.
Lewis Carroll's little gem, Achilles and the Hare.

Westley
02-16-2006, 10:50 AM
Lewis Carroll's little gem, Achilles and the Hare.

?

Not familiar with the reference.

campbell
02-16-2006, 10:59 AM
?

Not familiar with the reference.

It's an amusingly infuriating dialogue between Achilles and the Tortoise, and it shows up in the book Godel, Escher, Bach by Douglas Hofstadter.

It goes something like this.

Tortoise: If I know the statements "If A, then B" and "A" to be true, then "B"?

Achilles: Yes.

Tortoise: Explain that to me.

Achilles: Isn't it obvious? If you've got "If A, then B" and "A", then B must be true!

Tortoise: Ah, so you mean 'If ("If A, then B" and "A") then "B"', right?

Achilles: Yes.

Tortoise: So why is B true? We've got three things now: "If A, then B", "A", and "If ('if A, then B' and 'A') then B"... I don't see that B necessarily follows...

and the Tortoise forces Achilles to ever longer and longer nested logical statements, but never allows Achilles to reach B.

It's a logical deduction rendition of Zeno's paradox, where Achilles gives the Tortoise 10 rods (or whatever) head start, and Zeno argues that Achilles will never be able to catch up to the Tortoise.

campbell
02-16-2006, 11:00 AM

Chubbs
02-16-2006, 11:00 AM
He never ansers my question. He won't gamble with me and He won't answer the fart riddle.

My vote is changing from evil genious to dolt.

JMO
02-16-2006, 11:10 AM
Thanks!

whisper
02-16-2006, 11:37 AM
What if the game was the same but there was no Monty and you were just revealed a goat after making your choice. The goat was revealed according the previous rules, BUT you didn't know that.

Without any knowledge of how the goat was revealed you would only have the information that you did not choose whichever goat you were shown, which would lead you to believe that you had a 1/2 chance of winning by switching doors. But that would shown to be wrong by experimentation.

:roll:
You don't understand the fundamental aspect of Bayes theorem. You're assertion right here shows that you've no clue.

Bayes theorem allows you to systematically include new information into your probabilities. If you didn't know what the reveal strategy is, using a Baysian approach, you wouldn't assume what the reveal strategy was and use that. In a Baysian approach, you would think of all the reveal strategies were, assign a probability to each, and figure out what the best solution was. Once you played the 1st game, you would update the probabilities of the outcome, and repeat. As the attempts increase, you'd find the optimal solution.

whisper
02-16-2006, 12:23 PM
Here are 2 different ways of looking at the problem, the second is the way I look at the problem. The first is the way the wiki and most of you look at the problem.

And to show this we only need to change the chance monty chooses goat 1 to 2/3 and goat 2 to 1/3 when you chose the car. The first thing to recognize in this case, is that the first way of looking at the problem gives you the exact same answer 1/3... which is wrong in this case. The second way gives you:

So, that the first "method" works for solving the problem in the original case is dependent not only on the fact that all you care about is weather or not you won the car... not only on the fact that the two goats are identical, but also on the fact that the chance he reveals each one is identical as well.

:roll:
The way that we do the problem wouldn't reveal the answer of 1/3.
Here is what Bayes Theorem would produce:

Here is the set of all cases:
[2 1 0r]: S = 1/9
[2 1r 0]: T = 1/18
[2 0r 1]: U = 1/9
[2 0 1r]: V = 1/18
[1 2 0r]: W = 1/6
[1 0r 2]: X = 1/6
[0 2 1r]: Y = 1/6
[0 1r 2]: Z = 1/6

pr(C = 0r | A = 2) = 1/9 / (1/9+1/18+1/9+1/18) = 1/3
pr(C = 0r | B = 2) = 1/6 / (1/6 + 1/6) = 1/2
pr(C = 0r | C = 0) = 0 / (1/6 + 1/6) = 0

pr(C = 0r) = 1/3*(1/3 + 1/2) = 5/18

Pr(A=2 | C=0r) = pr(A=2) / pr(C=0r) * pr(C=0r | A=2) = 1/3 / (5/18) * (1/3) = 2/5

Which, unsurprisingly, means the contestant should change his selection when the door is revealed - because he has a 3/5 chance of winning.

-------------------

Lets take it one step further. Lets say, we don't know if pr(S) = 1/9 or pr(S) = 1/12 or pr(S) = 1/6. So we'll assume that they are equally likely to be true.

What strategy should the player employ?

If pr(s) = 1/6, the pr(A=2 | C=0r) = 1/2
If pr(s) = 1/12, the pr(A=2 | C=0r) = 1/3
If pr(s) = 1/9, the pr(A=2 | C=0r) = 2/5

then pr(A=2 | C=0r) = 1/3 * (1/2 + 1/3 + 2/5) = 37/90.

So, the player should still switch.

Now, here is the big finale - as the game continues, we can use the results of the game to update the bolded assumption using - Bayes Theorem!

I'll let you work on that problem.

Old Actuary
02-16-2006, 12:32 PM
This is the way I do the problem:
A: Monty reveals goat 1
B: Monty reveals goat 2
D: You chose goat 1
E: You chose goat 2
Prob(A) = Prob(B) = 1/2
Prob(C) = Prob(D) = Prob(E) = 1/3
Prob(C|A) = Prob(C|B) = (1/2 * 1/3) / (1/2) = 1/3

Actually, there is nothing in the problem that says Prob(A) = Prob(B) = 1/2. The most that can be said is Prob(A)+Prob(B)=1

Then the solution is:

Prob(C| A or B)

Since Prob(A or B) = 1, then Prob (C| A or B) = Prob(C) = 1/3

But logically it obviously is not independent: The chance that you chose the car is raised by knowing that you did not choose the car, but just happens to be offset by the lowered chance of having chosen the car given that its less likely he would have shown that goat if you chose the car than if you had chosen a goat.

I wish I knew what that meant. It sounds very deep.

Gandalf
02-16-2006, 12:34 PM
I wish I knew what that meant. It sounds very deep.
As in, "PhD = Piled high and deeper"?

VincentRupp
02-16-2006, 02:48 PM
The Monty Hall Problem demonstrates that probability is often a matter of where you're standing. From the contestant's (your) perspective, the probability of originally selecting the car is 1/3 because you have no information. Monty Hall, on the other hand, knows your probability of selecting the car, after you've chosen a door, is either 0 or 1. He then has two doors from which to choose and makes a decision given his perfect information. Given no information about your chosen door, the probability it hides the car is still 1/3. However, because his decision is not random (and you know it's not random), you have more information about the remaining door, which does affect its probability of hiding the car. If Monty opened a door at random, then your odds of winning the car, conditional on his not revealing the car, are 50-50.

Chubbs
02-16-2006, 03:11 PM
The Monty Hall Problem demonstrates that probability is often a matter of where you're standing. From the contestant's (your) perspective, the probability of originally selecting the car is 1/3 because you have no information. Monty Hall, on the other hand, knows your probability of selecting the car, after you've chosen a door, is either 0 or 1. He then has two doors from which to choose and makes a decision given his perfect information. Given no information about your chosen door, the probability it hides the car is still 1/3. However, because his decision is not random (and you know it's not random), you have more information about the remaining door, which does affect its probability of hiding the car. If Monty opened a door at random, then your odds of winning the car, conditional on his not revealing the car, are 50-50.

Uh no. If monty shows a goat at random, you prop is still 2/3 to switch because you are trading 1 door for 2. The only thing that changes about Monty showing a random door is that if he shows the car, you prob of winning car just went to zero.

Alto Reed on a Tenor Sax
02-16-2006, 03:41 PM
Uh no. If monty shows a goat at random, you prop is still 2/3 to switch because you are trading 1 door for 2. The only thing that changes about Monty showing a random door is that if he shows the car, you prob of winning car just went to zero.

No I think he was saying that Monty opens a DOOR at random, meaning sometimes he opens a door with a goat, and sometimes he opens a door with a car. In that case, I think your odds of switching are indeed 50/50, right?

renaissoxx
02-16-2006, 03:51 PM
:roll:
You don't understand the fundamental aspect of Bayes theorem. You're assertion right here shows that you've no clue.

Bayes theorem allows you to systematically include new information into your probabilities. If you didn't know what the reveal strategy is, using a Baysian approach, you wouldn't assume what the reveal strategy was and use that. In a Baysian approach, you would think of all the reveal strategies were, assign a probability to each, and figure out what the best solution was. Once you played the 1st game, you would update the probabilities of the outcome, and repeat. As the attempts increase, you'd find the optimal solution.

No matter what I say you still choose to make up some superficial counter argument. On one hand, your trying to reword my argument again and claim I'm saying something else, on the other your missing the value of my specific way of stating the issue with Bayes' theorem.

Suppose you DID know how Monty was revealing the doors. Or at least you thought you knew. And you used Bayes in conjunction with the event that he revealed the door as opposed to just the event that you did not choose the goat he shows. Guess what you still don't know? Whether or not there is a correlation between how you got THAT information and the outcome. Meaning maybe Monty tells you and chooses the car one way if you choose a goat and a different way if you choose a car. Maybe he just lies about his strategy. Maybe if you win the car he goes back in time and alters his strategy... etc

There are infinite possible causes of correlations between the frequency with which information is given and the outcome, and only one way to make Bayes' theorem useful in the face of them. Recognizing that it is useless without the assumption that none of these correlations exist.

Furthermore, you seem to be reverting back to the Monty example when the whole point of the example you quoted was that it was not the same situation... IE think of any situation instead where you are given information and have absolutely no way of knowing if there is a correlation between how frequently the given information was given and the outcome. The extent of your reasoning on this matter seems to be that if theres a person there they might be reacting to what you are doing rather than just revealing a door randomly. Knowing that this was probably the case is exactly why I said to imagine there was no host in addition to just not knowing what he was doing.
How hard would it be to come up with a situation that didn't involve a gameshow host where this assumption would fail?

The whole point of the formula I used, and the factor by which bayes formula is multiplied in case of a correlation between the frequency with which information is given and the outcome is so if you get the wrong answer with a conditional probability you can look at it to try and figure out what the correlation is.

The ability to use the traditional Bayes' theorem once you already know what the correlation is by simply treating the fact that information is revealed to you as an event instead of just the information itself is not something I ever disputed. The issue is that Bayes' formula doesn't come with a warning that says: Does not work if there is a correlation between the frequency with which information is given and the outcome. Just because in this one case you can see that there probably is one and perhaps figure it out doesn't mean that you shouldn't worry about it other than here.

:roll:
The way that we do the problem wouldn't reveal the answer of 1/3.
Here is what Bayes Theorem would produce:

Here is the set of all cases:
[2 1 0r]: S = 1/9
[2 1r 0]: T = 1/18
[2 0r 1]: U = 1/9
[2 0 1r]: V = 1/18
[1 2 0r]: W = 1/6
[1 0r 2]: X = 1/6
[0 2 1r]: Y = 1/6
[0 1r 2]: Z = 1/6

pr(C = 0r | A = 2) = 1/9 / (1/9+1/18+1/9+1/18) = 1/3
pr(C = 0r | B = 2) = 1/6 / (1/6 + 1/6) = 1/2
pr(C = 0r | C = 0) = 0 / (1/6 + 1/6) = 0

pr(C = 0r) = 1/3*(1/3 + 1/2) = 5/18

Pr(A=2 | C=0r) = pr(A=2) / pr(C=0r) * pr(C=0r | A=2) = 1/3 / (5/18) * (1/3) = 2/5

Which, unsurprisingly, means the contestant should change his selection when the door is revealed - because he has a 3/5 chance of winning.

-------------------

Lets take it one step further. Lets say, we don't know if pr(S) = 1/9 or pr(S) = 1/12 or pr(S) = 1/6. So we'll assume that they are equally likely to be true.

What strategy should the player employ?

If pr(s) = 1/6, the pr(A=2 | C=0r) = 1/2
If pr(s) = 1/12, the pr(A=2 | C=0r) = 1/3
If pr(s) = 1/9, the pr(A=2 | C=0r) = 2/5

then pr(A=2 | C=0r) = 1/3 * (1/2 + 1/3 + 2/5) = 37/90.

So, the player should still switch.

Now, here is the big finale - as the game continues, we can use the results of the game to update the bolded assumption using - Bayes Theorem!

I'll let you work on that problem.

I wasn't talking about you whisper, I was talking about the first strategy I outlined, which happens to be what both the wiki explanation of the problem and many forum posters here were alluding to when they say things like "There is a 2/3 probability the 2 doors you didn't choose have the car and it's carried by the remaining door when one is opened" and "The information that is given to you when the door is opened is irrelevant... you already knew he was going to show you a goat" In fact didn't you say some of these things before? Both of these only coincide with a strategy that WOULD lead you to the wrong answer in the scenario I outlined. And there is no reason to believe you would have stopped using it in the new scenario, because noone cited dependency on the goats being exactly the same and him having the exact same probability of revealing each.

But hey thanks for writing out my working of the problem all over again. Never admit your wrong just drown out the other person with post spam and copy their arguments and working out of problems as if they were yours right? Why would I need to work on a problem that I just worked out 2 posts before you did?

And once again, if you can ever get the wrong answer with Bayes' Theorem then you don't have a single thing to say about my very simple claim that Bayes' theorem is dependent on there not being a correlation between the frequency with which information is given and the outcome, do you?

Actually, there is nothing in the problem that says Prob(A) = Prob(B) = 1/2. The most that can be said is Prob(A)+Prob(B)=1

Then the solution is:

Prob(C| A or B)

Since Prob(A or B) = 1, then Prob (C| A or B) = Prob(C) = 1/3

I wish I knew what that meant. It sounds very deep.

"An unambiguous statement of the problem, with explicit constraints on the host as described by Mueser and Granberg:

Behind each of three doors is either a goat or a car (two goats, one car), with the car behind each door with equal probability.
The player picks one of the three doors. The contents are not revealed.
The game host knows what is behind each door.
The game host must open one of the remaining doors and must make the offer to switch.
The game host will always open a door with a goat.
That is, if the player picks a door with a goat, the game host picks the other door with a goat.
And if the player picks the door with a car, the game host randomly picks either of the two doors with a goat.
The host offers the player the chance to either claim what is behind the originally-chosen door, or to switch, claiming what is behind the one remaining door. "

http://en.wikipedia.org/wiki/Monty_Hall_problem

Westley
02-16-2006, 03:54 PM
Maybe if you win the car he goes back in time and alters his strategy... etc This is pure brilliance. Try arguing against this one using Bayes Theorem!

Gandalf
02-16-2006, 03:57 PM
In this specific case, there seems to be a high correlation between the troll's various posts. Empirically, the result seems to be: if you respond, he will post something else silly; perhaps quoting your post to suggest he read it but giving no indication he tried to understand it.