The Setup:
You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other (i.e., ten dollars and twenty dollars). The player may select one envelope and keep whatever amount it contains, but upon selection, is offered the possibility to take the other envelope instead.
The Problem:
I open up enevelope #1 and it contains ten dollars. There is a 50% chance the other envelope contains twenty dollars, 50% it contains five. The average value of the other envelope is twelve and a half dollars ((20 + 5) / 2) so it is always worth it to swap--I don't even have to open my envelope to know this. In fact, if I don't open my envelopes its always +ev (expected value) to keep swapping envelopes back and forth forever.
The logician Raymond Smullyan expressed the problem in a way which doesn't involve probabilities. The following plainly logical arguments lead to conflicting conclusions:
1. Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss.
2. Let the amounts in the envelopes be Y and 2Y. Now by swapping, the player may gain Y or lose Y. So the potential gain is equal to the potential loss.
Status: unsolved.
A convoluted explanation to a simple answer:
ReplyDeletePerhaps there it is fallacious for us to think that switching envelopes has equal probability of one of two outcomes. The idea of switching the envelope being a variable is the assumption on which the net +ev calculation is grounded. But, I think it might be helpful to tease out the different possible frameworks in which this case can be examined so that we might discover the proper (and improper) ways of calculation.
The second envelope as a variable: This one will modify the original setup such that it will comply to the ev calculation. Imagine you are at a carnival and there is a booth in which you buy a tin can for $10. The tin can contains one of two amounts, each with equal probability: $20 or $5. If, here, we are to assume that the amounts in the tin can are distributed at true randomness, we can calculate ((20+5)/2) - 10 = 2.5. Here, it will ALWAYS be profitable to buy the tin can (the equivalent of switching the envelope).
If you compare that carnival game to the two envelope situation, you are bound to find the subtle, but fundamental, difference in their setup. In the carnival game, the amount in the tin can is a true variable. However, in the two envelope case, it's not necessarily so (perhaps, it's necessarily not so). Consider the envelope case under this framework: you are either given the lower or the higher of the two amounts possible. Depending on which you are given, the second envelope's amount is determined: it is no longer a variable.
The probability, then, pertains not to the amount in the envelope but which fixed set we are in. Allow me to clarify: there are two possible setup scenarios. You have $10 and the other envelope has $20, and, you have $10 and the other envelope has $5. I will hereforth refer to the former setup as Set L the latter as Set H, the initial envelope as EI, and the other envelope as ES. The probability of you being in Set L is 50% and the probability of you being in Set H is 50%. Consider the implications of this case in the framework of fixed sets: once you are given one envelope, the fixed set in which you are in is determined. Thus, the amount in second envelope is not a variable--there is no probability of getting $20 or $5 as both amounts in the envelopes are determined--so, the paradox of switching envelopes infinitely many times for infinite +ev is averted.
The probability is not eliminated but simply shifted to another aspect of the case. Probability no longer affects the amount in the envelope, but now affects which fixed set we are in. The dilemma, then, is no longer "will the other envelope have $20 or $5", but rather "am I in Set L or Set H?." But does this rephrasing eliminate the fact that switching the envelope ONCE is always +ev? Nope. That switch, however, isn't +ev because of it's inherently profitable to switch but rather the result of the incongruity of amounts between Set L and Set H. The amounts in Set L and Set H are, in essence, different sets of amounts. Imagine this: if Set L EI = Set H ES and Set L ES = Set H EI, then we can reduce the scenarios down into a single set (since Set L is now the inverse of Set H) and ask the question "did I get the higher amount or the lower amount?" Here, of course, the ev of the switch is 0.
So the original envelope paradox has been completely averted by presenting it in a framework such that the probability is set in terms of "world of possibilities" instead of "probability of outcomes." We should be asking "what world am I in" instead of "what are the chances of that envelope being X." The answer is so simple. =)