conditional expectation question

[pjb028]

I ran into this in a practice exam and completely forgot how to do it.

Let X be the number of rolls of a fair die before getting a 6, and let Y be the number of rolls before getting the first even number. Find E[X l Y=5].

I have the solution, but I cannot follow the work that is presented with it. Can someone please help me?

[joni308]

I ran into this in a practice exam and completely forgot how to do it.

Let X be the number of rolls of a fair die before getting a 6, and let Y be the number of rolls before getting the first even number. Find E[X l Y=5].

I have the solution, but I cannot follow the work that is presented with it. Can someone please help me?

I have zero confidence that any manual you are holding or solution can give you the sleek solution you deserve. Probability is art. Treat it like art. When you see a new problem what's the first thought that comes to your mind? That you want to make love to it? Then do it. When you make love do you pick up a manual?

Anyhow, I'll give you Joni's solution that noone else will give you.

Solution

You are conditioning on the first time you see an even being 5. That means there is no way you can see a 6 before time 5. So what does this mean? The first 4 tosses are wasted. So let's fast forward to time 4. Looking ahead, if I asked you what the average time to get a 6 is, without the condition, it would be 6. This is a geometric distribution. But darn it. You have this condition. So what? You are going to quit in the middle of making love and leave your partner unsatisfied? Relax and think! You are conditioning on the next roll being an even. It would be nice it it was conditioned on being an odd. Why? Because then the expected number of rolls to 6 would be 7 since he next one cannot be a 6. OK, so you know the UNconditional expectation and the conditional on half the space; namely next one being odds.

So the problem is solved in half a line

6=7/2 +X/2 so X=5. Add to it the 4 first stems and get 9.

You did not provide the solution. Is this the answer? I certainly hope so otherwise I have to assume those were fake orga*ms.

[Kaner3339]

atta boy Joni.

[alexcrewe]

I ran into this in a practice exam and completely forgot how to do it.

Let X be the number of rolls of a fair die before getting a 6, and let Y be the number of rolls before getting the first even number. Find E[X l Y=5].

I have the solution, but I cannot follow the work that is presented with it. Can someone please help me?

Thought #1: First even number is 5 rolls in. There is a 1/3 chance it is a 6, 2/3 chance it is not. That gives us the following probabilities:

First 6 (Probability)
1 (0)
2 (0)
3 (0)
4 (0)
5 (1/3)
6+ (2/3)

We just have to figure out to do with the 6+ category. 2/3 of the time, it will take more than 5 rolls. How many? As Joni said, we are working with a geometric distribution here, so we'd expect to take 6 more rolls. That means we are looking at an expected value of 11 rolls total during that 2/3 of the time that we didn't roll a 6 on roll #5. Put it all together, and what do you find?
P(X|Y=5) = 1/3 (5) + 2/3 (11) = 9 rolls

[joni308]

You can also solve it like the masses would. All you need to know in order to calculate an expectation is the support and their respective probabilities. What's the support? If Y=5 then X can only be 5,6,7,8,....

What are the probabilities? Pretty straight forward to calculate

P(x=5|Y=5)=P(X=5 AND Y=5)/P(Y=5)
P(x=6|Y=5)=P(X=6 AND Y=5)/P(Y=5)
P(x=7|Y=5)=P(X=7 AND Y=5)/P(Y=5)
.....

Find all these probabilities and then take the sum

5.P(x=5|Y=5)=P(X=5 AND Y=5)/P(Y=5) +
6.P(x=6|Y=5)=P(X=6 AND Y=5)/P(Y=5) +
7.P(x=7|Y=5)=P(X=7 AND Y=5)/P(Y=5) +
....

And then you do get the answer.

This is the safe way to solve the problem but it's clearly for LOSERS.

[zzzsilver]

The statement: "Let Y be the number of rolls before getting an even number" and then Y being equal to 5 -
to me means that we can not get an even number until roll 6. Some of the explanations above seem to imply that you get your first even number on roll 5.

Can someone verify for me what the condition that Y=5 really means. Is it that you get your first even number on roll 5 or roll 6 ????Thanks.

[zzzsilver]

bump

[Gandalf]

If Y=5, then the first even number actually occurs on the 6th roll. So that's the first possible 6. On average then, you'd expect the first 6 to occur in this situation on the 11th roll, making the answer E[X|Y=5] = 10. However, 10 is not one of the answer choices.

Your general reasoning seems to be "If Y = 5, then all we know is that there is no 6 before the 6th roll". If fact, Y=5 tells you two things:
1. There is no 6 before the 6th roll.
2. The 6th roll is even (thus the probability that the 6th roll is a 6 is 1/3, not 1/6).

Minor digression: I am not positive you are using "before" the same way they intend. I used it the way you did. Their intent might be that Y = 5 means the first even number occurs on the 5th roll (i.e., is it possible to get an even number before you roll the dice once? Many would say "of course not", in which case Y=1 if the first roll is even.) It doesn't matter, in this problem, which definition is used, as long as you use the same definition for "before the first even number" and "before the first 6".
Hopefully (even though you get the same answer under either interpretation here) a real exam question would be clearer. For example, SOA sample question 136 is somewhat similar, and avoids "before".

[zzzsilver]

ok thanks!