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Su, Francis, et. al. have a short description of the paradox here:

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 [itex]( \,\$2^k, \$2^{k+1}) \,[/itex] with probability [itex]\frac{( \,\frac{2}{3}) \,^k}{3}) \,[/itex] for each integer [itex]k \geq 0[/itex].

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 perspectives 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:

[itex] 0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x[/itex]

To incorporate a two-person perspective, the equation would be one person’s odds for gain less “their opponent’s” odds for gain because their opponent’s gain comes at their expense. In other words, if you stand a 50/50 shot of losing exactly as much as you stand to gain, there is no longer incentive to swap envelopes:

[itex][ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, - [ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, = 0[/itex]

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

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

__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 [itex]( \,\$2^k, \$2^{k+1}) \,[/itex] with probability [itex]\frac{( \,\frac{2}{3}) \,^k}{3}) \,[/itex] for each integer [itex]k \geq 0[/itex].

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 perspectives 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:

[itex] 0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x[/itex]

To incorporate a two-person perspective, the equation would be one person’s odds for gain less “their opponent’s” odds for gain because their opponent’s gain comes at their expense. In other words, if you stand a 50/50 shot of losing exactly as much as you stand to gain, there is no longer incentive to swap envelopes:

[itex][ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, - [ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, = 0[/itex]

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

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

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