Timeline for Is there an equilibrium for this non-zero-sum game?
Current License: CC BY-SA 4.0
12 events
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Jul 25, 2021 at 9:48 | vote | accept | Eric | ||
Jul 23, 2021 at 6:52 | history | edited | Eric | CC BY-SA 4.0 |
formally define strategy
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Jul 23, 2021 at 6:40 | history | edited | Eric | CC BY-SA 4.0 |
edited body
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Jul 22, 2021 at 14:41 | history | edited | Eric | CC BY-SA 4.0 |
added 53 characters in body; edited tags
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Jul 22, 2021 at 5:29 | comment | added | Eilon | I would like to comment on the part "by symmetry the best strategy is the same for everyone". It is true that this is a symmetric game, hence there is a symmetric equilibrium. But a player's equilibrium strategy depends on the current number of chips in the pot, and it is not clear that for K (where K is the number of chips in the pot) the symmetric equilibrium strategy will dictate the same behavior (same is whatever meaning you would want). | |
Jul 22, 2021 at 5:24 | history | edited | Eric | CC BY-SA 4.0 |
added 357 characters in body
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Jul 22, 2021 at 4:59 | answer | added | Eilon | timeline score: 1 | |
Jul 21, 2021 at 15:57 | comment | added | Eric | @PeterTaylor That sounds reasonable. But if $M\gg N$, will it be better to bet $M-K$ chips if the current number of chips $K$ is very close to $M$? | |
Jul 21, 2021 at 15:38 | comment | added | Peter Taylor | The obvious play is to bet all but one of the chips if there's more than one. You can't lose immediately, and if you throw tails you have 50% odds that the player after you loses. If everyone does this (and by symmetry the best strategy is the same for everyone) then with large $N$ it's increasingly likely that someone will lose before it comes back to you. | |
Jul 21, 2021 at 15:27 | comment | added | Eric | @PietroMajer Yes, they do. | |
Jul 21, 2021 at 15:26 | comment | added | Pietro Majer | What if after the move of player N there are 0<x<M chips left in the pot? Do they start another round from player 1? | |
Jul 21, 2021 at 14:32 | history | asked | Eric | CC BY-SA 4.0 |