Su, Francis, et. al. have a short description of the paradox here: https://www.math.hmc.edu/funfacts/ffiles/20001.6-8.shtml

I used that link because it concisely sets forth the paradox both in the basic setting but also given the version where the two envelopes contain $( \, \$2^k, \$2^{k+1}) \,$ with probability $\frac{( \,\frac{2}{3}) \,^k}{3}$ for each integer $k \geq 0$.

Where the paradox is formulated by considering one person’s odds when choosing to swap an envelope, my question is whether the paradox might be resolved by considering the paradox from both swapper’s perspective instead of just one (i.e. for one person to swap, there must be another person for the original to swap with).

From a single person’s perspective, the paradoxical odds are traditionally given by the equation: $$0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x$$ To incorporate a two-person perspective, the equation would be [what one person stands to gain] less [what they stand to lose = what their opponent stands to gain]: $$[ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, - [ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, = 0$$ The result is that neither person improves their odds by swapping. Paradox resolved.

Comments, suggestions, agree, disagree…? I’m just fishing here. Thank you!


The result is that neither person improves their odds by swapping.

This isn't what the equation you've written says. To see this, consider the following game: there are two players, player 1 does nothing, and player 2 can choose whether or not to flip a coin. If 2 says no flip, then neither player gets anything; if 2 says flip, then player 1 gets 1 point if the coin comes up heads and player 2 gets one point if the coin comes up tails. Clearly 2 has an incentive to flip, but $$\mbox{(expected gain for 1 from flipping)-(expected gain for 2 from flipping)=0}.$$

All that your equation indicates is that the action is appropriately symmetric; but that doesn't say anything about either player having an incentive to switch.

  • $\begingroup$ Thank you for that comment. Is not the paradox different in the sense that when one person wins the other loses, where in your proposed game, if one person wins a point, the other person loses no points? The equation has to be one person's gain = one person's loss I believe, right? $\endgroup$ – AplanisTophet Apr 22 '17 at 1:49
  • $\begingroup$ @AplanisTophet It depends what you're thinking of as your preference function. If you only care about maximizing your own money, then adding another player doesn't make the envelope paradox any easier: you don't care how much money they make, you're still interested in calculating your expected value. And if your preference function does take into account what your opponent is getting out of this, then you've substantially changed the game. This new game is indeed non-paradoxical, and "obviously" equivalent to the original, but it doesn't shed any light on the original game directly. $\endgroup$ – Noah Schweber Apr 22 '17 at 1:55
  • $\begingroup$ In being concerned only with what you make, you must factor into account what you lose as well because what your opponent stands to make comes at your expense. Where you stand to make just as much as you stand to lose to your opponent, there is no incentive to switch. That is what the equations say to me. If you don't see it that way though, then I completely understand and respect that. This is "fishing" to see if others can come to my concensus so disagreement is important too. Thanks again! $\endgroup$ – AplanisTophet Apr 22 '17 at 1:59
  • $\begingroup$ @AplanisTophet "In being concerned only with what you make, you must factor into account what you lose as well because what your opponent stands to make comes at your expense." Not unless you redefine your preference function. Me failing to make a possible dollar, and me failing to make a possible dollar and my opponent making a dollar, don't look different to me if all I care about is how much money I make. (By the way, +1 - good question!) $\endgroup$ – Noah Schweber Apr 22 '17 at 2:04
  • $\begingroup$ I intend to let this question sit for a while and then I am happy to give Noah a "hard earned" accept on this one after getting more feedback. I've have a very small sample of people going both ways here off-site. Please give Noah a +1 if you accept his logic and please give my question above a +1 if you accept my logic to keep track. Again, thank you! $\endgroup$ – AplanisTophet Apr 22 '17 at 4:14

We can also break it down into the four possible scenarios from a two-person perspective when swapping where, when ambiguously starting with $x$, we have 25% of the time you’ll lose 0.5x to your opponent, 25% of the time you’ll gain 0.5x from your opponent, 25% of the time you’ll lose x to your opponent, and 25% of the time you’ll gain x from your opponent. The same is true for your opponent.

$$x - 0.25( \,0.5x) \, + 0.25( \,0.5x) \, - 0.25( \,x) \, + 0.25( \,x) \, = x - 0.25( \,0.5x) \, + 0.25( \,0.5x) \, - 0.25( \,x) \, + 0.25( \,x) \,$$ $$x = x$$

In addition to what is noticed in the original question, the above shows that there is no incentive to swap.

The same general logic works for all consistent versions of the paradox that prey on the ambiguity of the variable $x$ (see the link in the original question for more insight into that statement).

I assert that the paradox is resolved.


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