The other cup must contain Z/2 (with probably 50%, as you say) or 2Z (also with probability 50%). Right that's not true. If you're presented with 2 cups/envelo

The other cup must contain Z/2 (with probably 50%, as you say) or 2Z (also with probability 50%). Right that's not true. If you're presented with 2 cups/envelopes, whatever, and you're told that one contains double the amount in the other, then you have 2 amounts: Z and Z/2. If you choose a cup, then the other one either contains Z or Z/2. Whether or not you refer to those initial amounts as Z and 2Z, or Z and Z/2 is up to you, but there aren't 3 options, there are only 2. This seems (a) completely different, and (b) unenlightening. It's completely different because I seem to know what all the options were for what you did and the amounts involved. Yep that's my point: in order to attribute an expected value with probabilities for both Z/2 and 2Z assuming the amount you chose was Z, you have to know a bunch of additional information. This explains it much better than I can: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/ http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... But if all you get are two envelopes, and the information that one is double the other, the situation is much simpler: you're choosing between 2 amounts and there is absolutely no difference if you switch.