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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
Jul 23, 2021 at 6:40 history edited Eric CC BY-SA 4.0
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Jul 22, 2021 at 14:41 history edited Eric CC BY-SA 4.0
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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
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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