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by miloshh
5788 days ago
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OK, now I see there are two slightly different possible formalizations of the problem: 1. You are told that the two envelopes contain amounts A and 2A, but you aren't told what A is. After you pick one envelope, you are allowed to open it, and then you're given the choice to switch. Here the optimal move depends on the distribution of A, and if you don't know it, you can't do much other than pick randomly. After some googling, this is the more common formalization, and it is analyzed in several math papers and blogs. 2. (The version I was assuming.) You are told that the envelopes have, say, $100 and $200. You pick one and you aren't allowed to open it yet. Now you're given the option to switch one last time. There is no problem with undefined priors and weird conditional probabilities in this version. However, the freaking paradox still holds! The expected value you get by switching is $150, no question about that. But the expected relative gain you get by switching is 1.25, there's also no question about that! This is the real paradox to me. Taking an expectation of a relative quantity is intuitively wrong, but why exactly? |
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If you have problem with that imagine stack of cards. You and your opponent pick one card. Whoever picks higher card wins. Until you see your card you have 1/2 probability of winning. But after you see you just picked 3 you see your probability of winning sharply changes (depending on what cards are left in the stack).
Ad.2 When you don't open your envelope you must treat amount in it (A) as random variable. Then all I wrote in post above applies.
It doesn't matter that you did not looked into an envelope. Because whether you got higher or lower amount is significant and you don't know which occurred you must split your reasoning and consider both cases. But upon splitting you must remember that condition you assume for each given branch of your reasoning must be taken into account. You can do this by pruning your random variable A with the appropriate condition and that influences its expected value. So by all means you can sum up expected values but not before appropriately adjusting them with the conditions you assumed for branches of your reasoning.
Consider solving equation x^2 + yx +1 = 0 While solving your may split you reasoning to three cases depending on whether y^2-4 is larger, smaller or equal to zero (even though you don't know what is the value of y). But when summing up solution you must remember that you assumed something about y in your branches so for the x you've found y can no longer be any real number.