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Probability Old Exam P Forum

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  #1  
Old 05-08-2019, 09:27 PM
actuaryxjh actuaryxjh is offline
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Default meaning of expected value

Hi, let's consider this case:

A small quiz has three multiple choices with each question three choices.

so, using np formula, we get the expected value be 3*(1/3)=1

What does this "1" mean? how does it represent the "average", "expect" in this case?

What I am asking is, what does this "one" mean back into the situation, not in theoretical sense.
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  #2  
Old 05-08-2019, 09:52 PM
Academic Actuary Academic Actuary is offline
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Repeat the test n times selecting answers at random. Sum the scores and divide by n, That is your sample average. Specify an epsilon and a probability. You can determine an n such that the average will be within epsilon of 1, with the specified probability.
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  #3  
Old 05-12-2019, 01:08 AM
actuaryxjh actuaryxjh is offline
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Originally Posted by Academic Actuary View Post
Repeat the test n times selecting answers at random. Sum the scores and divide by n, That is your sample average. Specify an epsilon and a probability. You can determine an n such that the average will be within epsilon of 1, with the specified probability.
.
Oh, don't quite get.
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  #4  
Old 05-12-2019, 02:12 AM
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hostess hostess is online now
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Quote:
Originally Posted by actuaryxjh View Post
Hi, let's consider this case:

A small quiz has three multiple choices with each question three choices.

so, using np formula, we get the expected value be 3*(1/3)=1

What does this "1" mean? how does it represent the "average", "expect" in this case?

What I am asking is, what does this "one" mean back into the situation, not in theoretical sense.
honestly im not sure if there is a practical explanation that holds up to all cases. i think its just a weighted sum.

think about the following case:

There is a multiple choice test with 3 questions, each with 4 answer choices and you randomly guess each question. The expected value is 3/4.

Is there really a practical explanation of wht 3/4 is? Surely you can't get 3/4 of a question right.

imo in the continous case you can make an argument to say that a practical explanation does exist. but when you discretise events, your expected value might not even be in your support so it's harder to come up with a practial explanation.
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  #5  
Old 05-12-2019, 10:39 AM
Chuck Chuck is offline
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Doesn't it just mean if you take the exam a gazillion times, an average you will get one right per time?
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  #6  
Old 05-12-2019, 11:02 AM
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Vorian Atreides Vorian Atreides is offline
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Quote:
Originally Posted by actuaryxjh View Post
Hi, let's consider this case:

A small quiz has three multiple choices with each question three choices.

so, using np formula, we get the expected value be 3*(1/3)=1

What does this "1" mean? how does it represent the "average", "expect" in this case?

What I am asking is, what does this "one" mean back into the situation, not in theoretical sense.
What's the minimum number of items you could get correct?

What's the maximum number of items you could get correct?

What's the answer if you have 5 items?

Work through the answers of the first two questions with 5 items and compare them to your answer to the previous question.


Rework the min, max, and expected value for differing values of total number of items available; then look at connections to the data you've just collected.
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  #7  
Old 05-12-2019, 01:43 PM
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Originally Posted by Chuck View Post
Doesn't it just mean if you take the exam a gazillion times, an average you will get one right per time?
So in my example youíll get 3/4 of a question rights on average?
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Old 05-12-2019, 01:49 PM
Academic Actuary Academic Actuary is offline
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So in my example you’ll get 3/4 of a question rights on average?
Except saying on average has no precise meaning because your realized average will very rarely equal to true expected value. In fact as you increase n the probability that the average exactly equals the true expected value goes to zero (for finite n).

Last edited by Academic Actuary; 05-12-2019 at 02:00 PM..
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  #9  
Old 05-12-2019, 03:32 PM
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Originally Posted by Academic Actuary View Post
Except saying on average has no precise meaning because your realized average will very rarely equal to true expected value. In fact as you increase n the probability that the average exactly equals the true expected value goes to zero (for finite n).
this is only for the discrete case correct?
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Old 05-12-2019, 03:43 PM
Academic Actuary Academic Actuary is offline
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this is only for the discrete case correct?
In the continuous case the average will never equal the expected value. In the discrete case it can but the probability of equality decreases. Consider flipping a fair coin. With 2 flips there is a 50% chance the average number of heads equals .5. With 100,000,000,000 flips the probability of 50,000,000,000 heads is very small, but the probability distribution for the average is collapsing around .5.

What you could say is the average converges stochastically to the expected value.
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