The Envelope Problem:
Omega (who never lies) walks up to you and states "these envelopes contain the values X and 2X" He then SHUFFLES them up [this informs us our subsequent switching strategy should be irrelevant], slides one over to you. He states "I have given you an envelope containing either X or 2X. You can open it, and then decide whether you'd like to switch for the other envelope." You open it up, see a $100 bill. Do you switch? Some decision theory has to take you from probabilistic knowledge about the world to an action you should take, if you care about acting (VNM) rationally.
Premises and Conclusion of EV Calculus Using the Same EV Calculus Rules We Normally Use:
1) Your envelope contains $100.
2) the other envelope contains $50 with .5 probability, and $200 with .5 probability
3) Half the time, switching costs me $50. Half the time, switching gains me $100
4) therefore, I stand to gain an average of $25 by switching.
Which numbered part do you think is false?
If you think it's 2) because the state of the world is already fixed by Omega, rather than variable, then I offer this: when something has a fixed state, but your knowledge of it is probabilistic, ev calculus treats it as if it is not determined but will be at the stated probabilities. If you disagree, then your next task is to explain how we calculate the odds of being dealt an ace from a shuffled but now fixed-state deck of playing cards.
We can solve this using logic other than EV calculus, namely an observation that since the envelopes were shuffled and fixed, switching can't change our actual EV unless we know something new about the other envelope, and we do not. We know as much as we did before our envelope was opened. But why are steps 1-4 above so persuasive under the same logic we correctly apply to cards being dealt from a shuffled deck, despite leading to the wrong conclusion (that switching is +ev)?
"When we switch we stand to finish with 1.5x, if we don't switch, we stand to finish with 1.5x" is incomplete work. You observe $100 in your envelope. So at the very least you should be able to redo your math and get us an actual dollar amount EV. It has to be $100 since your premise is switching doesn't matter. That means X = $66.66 in the earlier formulation of "we are guaranteed 1.5x." Well, if X = $66.66, your envelope must contain either $66.66 or $123.33 by reference to the rules of the game. So your solution is not well formed, it is not coherent.
[EDIT 1 Below]
Stolzman is arguing that EV calculus doesn't apply here, but that's not an issue. Well, it's a major issue if EV calculus gives you neither the correct answer nor an indication as you work through steps 1-4 that you shouldn't be using this method. In other words, it doesn't just fail to return an answer, it returns the wrong answer! Only because we have access to other logic do we even realize it has failed. This is a big problem.
[EDIT 2 Below]
Well, I correctly predicted Step 2 is where the action is, but people don't seem to agree with my note on it above. Yes, the state of the universe is fixed. No, your knowledge about the state of the universe does not therefore use only 1 and 0 as probabilities. If a coin is biased (unfair) but you do not know in which direction, your probability estimates of heads and tails are each .5. Probability is in the mind. http://lesswrong.com/lw/oj/probability_is_in_the_mind/
Omega may have filled the envelopes with $50 and $100 because his son was born in the galaxy his people call "50zba100". Or he may have filled them with $100 and $200 for any other whismsical reason. We do not have access to his algorithm. But since he has not given any hints as to which he picked, why should you not switch? Why should you not act under P(50) = .5 and P(100) = .5, which most closely approximates your state of knowledge? Furthermore, NOT switching is implying that $50&$100 is LESS likely than $100&$200 (otherwise you're a fool). That is your implied conclusion when you do not switch, having seen $100. How did you arrive there with such confidence?
