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The Two Envelope Problem Sept. 16, 2013, 6:27 p.m.
Last Sunday, Szym and I spent nearly the entirely afternoon arguing about the Two Envelope Paradox. Mostly, I thought it might be a good to do a blog post on it because we spent so long on it. Also, I think there might be a message here that just shows how unorganized data on the Internet is.
We came across the two envelope paradox on Wikipedia. Go take a look.
You have two indistinguishable envelopes that each contain money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. You pick at random, but before you open the envelope, you are offered the chance to take the other envelope instead.
"It can be argued that it is to your advantage to swap envelopes by showing that your expected return on swapping exceeds the sum in your envelope. This leads to the absurdity that it is beneficial to continue to swap envelopes indefinitely."
The absurd argument can be illustrated through an example.
"Assume the amount in my selected envelope is $20. If I happened to have selected the larger of the two envelopes, that would mean that the amount in my envelope is twice the amount in the other envelope. So in this case the amount in the other envelope would be $10. However if I happened to have selected the smaller of the two envelopes, that would mean that the amount in the other envelope is twice the amount in my envelope. So in this second scenario the amount in the other envelope would be $40. The probability of either of these scenarios is one half, since there is a 50% chance that I initially happened to select the larger envelope and a 50% chance that I initially happened to select the smaller envelope. The expected value calculation for how much money is in the other envelope would be the amount in the first scenario times the probability of the first scenario plus the amount in the second scenario times the probability of the second scenario, which is $10 * 1/2 + $40 * 1/2. The result of this calculation is that the expected value of money in the other envelope is $25. Since this is greater than my selected envelope, it would appear to my advantage to always switch envelopes."
A classic case of bad modeling!
Math is a language. It's supposed to be used to describe natural phenomenon.
The biggest flaw in the argument which results in the false conclusion that it *always* makes sense to switch envelopes (btw, this problem is not to be confused with the Monty Hall problem) is the poor modeling of the scenario. It's important not to divorce the mathematical description from what's naturally happening. "Wait a minute! Once I've switched envelopes once, I know what's in the other envelope and there is no need to keep switching!"
Does the mathematical description reflect that phenomenon? The paradox only seems to occur because the math does not accurately describe the situation. To illustrate that point, consider a completely different game:
You're happy with an envelope of money on your desk. Everyday, a man shows up at your door with two indistinguishable envelopes that each contain money. One contains twice as much as what you already have. The other contains half as much as what you already have. You can sit on the money you have. Or, you can switch your envelope with the man.
It would appear that I have either the option of sticking with my $20. Or, I have a chance of getting $10 * 1/2 + $40 * 1/2 = $25. The mathematical conclusion is that I should always choose from the man at my door. There doesn't seem to be any paradox.
So then it seems that the same math is describing two completely different scenarios. That would be bad because math is supposed to be a great language.
As an exercise, the first game should be modeled exactly according to the decision being made. There are two envelopes containing $a$ and $2a$ dollars. After taking an envelope, I either have $a$ or $2a$ in my hand. Switching envelopes can be modeled as what one can gain. Half the time, I'll have a chance of gaining ($2a$-$a$). Half the time, I'll have a chance of losing ($a$-$2a$). Mathematically, switching can be modeled as: 0.5*$a$ + 0.5*(-$a$) = 0. It doesn't matter if I switch or not.
Now go back to that Wikipedia article and learn some Bayesian probability!
Edit: Apparently, the viewpoint of gaining vs. not gaining is the non-probabilistic approach that Raymond Smullyan took. Even neater.