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Old 06-28-2006, 05:34 AM
jason1990 jason1990 is offline
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Join Date: Sep 2004
Posts: 932
Default Re: The envelope problem, and a possible solution

It occurred to me that there is an interesting connection between this Paradox and confidence intervals.

Suppose I am about to play 10,000 hands of poker and that I know my standard deviation is 15 BB/100, so that my standard error over those 10,000 hands will be 1.5. Let m be my true winrate and M the winrate I will observe over those 10,000 hands. The number m is a fixed, but unknown quantity. The number M is a random variable.

Before I play, I can say that

P(M - 3 < m < M + 3) = 0.95.

This is a meaningful statement and it is true. Now, I go play and I observe that M=4. It is tempting to plug this into the above statement and conclude that

P(1 < m < 7) = 0.95.

But I can't do that. That's meaningless. The number m is fixed. Either 1<m<7 is true or it's not. There's no probability involved. This is a well known problem and it is why [1,7] is called a "confidence interval". We don't say that m is in [1,7] with 95% probability. We say that we are 95% confident that m is in [1,7]. This 95% is not a probability and we cannot work with it as though it were a probability. For example, it makes no sense to ask about the expected value of m.

On the other hand, if I were a Bayesian, then I could regard m as a random variable, assign it some probability distribution, and talk about actual probabilities involving m. I could talk about m's expected value, and about its conditional expectation given M.

Okay, now back to the Paradox. Let T be the total amount in the envelopes and X the amount in the one I choose. Suppose I am a frequentist, so that T is a fixed, but unknown quantity, and X is a random variable. Before I open that envelope, I can say that P(X=T/3)=0.5. This is a meaningful and true statement. Now I open the envelope and find that X=100. It is tempting to plug this in and conclude that P(T=300)=0.5. But I can't do that. That's meaningless. The number T is fixed. Either T=300 or it does not. There's no probability involved. I might say that I am 50% "confident" that T=300. Hence, I am 50% "confident" that the other envelope contains 200. Similarly, I am 50% "confident" the other envelope contains 50. But I can't work with "confidences" the way I can with probabilities. That's why we gave them a different name, so we wouldn't get confused and accidentally do things like compute expectation.

But if I were a Bayesian, then I could regard T as a random variable, assign it some probability distribution, and talk about actual probabilities involving T, and hence, involving the other envelope. Then, as a Bayesian, I could actually compute the conditional expectation of that other envelope, given X=100.
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