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Nowhere in the original problem is it stated that the value you're shown influences the probability that the other value will be higher or lower. It simply states that you had 50% chance to pick either N or 2N, and that now that you know one value, you have a 50% chance that you hold either N or 2N. Reasoning on from there they then state that your EV is 50% * value/2 + 50% * value*2.


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Hi Naobisdad,

All we are told in the original problem is that one of the envelopes contains twice the other. This tells us that once we open one envelope, that the other contains either half as much or twice as much. It tells us nothing about the chances that the other envelope contains twice as much, and to simply assume that these chances are 50% is not correct.

As an example, suppose that the designer of the game opted to put 50 and 100 in the envelopes (but that we don't know this). Clearly, it is -EV to switch in this case if we open the envelop with 100, as there would be a zero percent chance that it would be 200, and a 100 percent chance that it would be 50. Basically, the two envelopes contain X and 2X. The EV of switching no matter what will be -X if we chose the 2X envelope, and X if we chose the 2X envelope. Since there's a 50% chance of chossing either envelope, the net EV of switching no matter what is zero.

As a more involved example, suppose that we're told that the numbers were chosen by randomly selecting some integer k between 0 and K (where K is some positive integer that we are not told), and then placing $2^k in one envelop and $2^(k+1) in the other. Then, clearly, we should swtich if our envelope contains $1. But, what about for all other values? Suppose our envelope contains $2 or more, and we've decided to switch no matter what. Then, if our envelope contained $2^K, we would automatically lose $2^(K-1) by switching. If our envelope contained less than $2^K but more than $1, chances would be 50-50 that the other envelope contained more, so we would expect to gain on average 1/2 the amount in our envelope by switching (i.e., argument #1 from the original post applies for those cases). So, the total EV of switching no matter what would be: Sum(i=1..K-1){1/2*2^i} - 2^(K-1) + 1 (the +1 at the end is because we always gain $1 when we see $1 in the envelope and switch). The first term is a geometric series (minus the i=0 term); the summation is: 1/2*(1-2^K)/(-1)-1/2, or 2^(K-1)-1. So the total EV of a switch = 2^(K-1)-1 + 2^(K-1) - 1, with equals zero. So, actually, in this setup, we should switch if the envelope we open contains $1 (since the EV for switching in this case is +$1), and we should keep our envelope otherwise (since the net EV for all other cases is -$1). Note that in this example, argument #1 (50% gain by switching) holds for most of the cases, but this is offset by the 50% loss which happens when switch the largest possible amount, which offsets the 50% gains in the cases of the smaller amounts.

To prove that, without inferring any other information about the problem, that switching is EV neutral, consider this: Suppose the deal is that we're splitting the money with another player, with whom we were not allowed to discuss a strategy beforehand. After we choose, the envelopes will be resealed and presented to the other player. He will be handed the envelope that we chose, will open it, and will decide which envelope to keep. If it were +EV to always switch, then we would of course have switched. But from his perspective, it would also be +EV to switch. Except that clearly it cannot be +EV to switch twice, since you end up back at the same envelope. So, absent any other information, switching must be EV neutral.