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I am interested in the following simplified version of poker. Each player gets a card (for example, either A or B). Then they bet knowing their own cards (for example, the pot initially has 1 euro, then they go in order and decide to fold or add 10 euros to the pot, then after the last player the betting is over). After the betting, either those distribute the money who have A or those who have B (for example, with 60% chance A wins and with 40% chance B wins). What are the optimal strategies if the players want to maximize their expected income?

Note that I am most interested if the game has at least 3 players and it is crucial that the outcome can depend on the flop, unlike most theoretically studied versions, like Kuhn poker or when the inputs are from the [0,1] interval. Has this ever been studied?

I wonder if paradoxical situations can happen, like the last player should bet with a higher probability when his card is B than with A (when A has a higher chance of winning).

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  • $\begingroup$ It's a little unclear to me what the exact set-up is. You make it sound like players can raise the bet, but in this case it seems like you need to assume players have fixed chip stacks or else re-raising forever (or raising arbitrarily high) could be correct. $\endgroup$ – Sam Hopkins Nov 16 '14 at 20:37
  • $\begingroup$ @Sam I am sorry if it was ambiguous, there is no raising in this game, everyone can decide to pay to play or fold. $\endgroup$ – domotorp Nov 16 '14 at 21:45
  • $\begingroup$ Am I right that the players do not know each other's cards, but the next players see the bets of all the previous ones? $\endgroup$ – Ilya Bogdanov Nov 17 '14 at 9:09
  • $\begingroup$ @Ilya Yes, just like in poker. $\endgroup$ – domotorp Nov 17 '14 at 12:43
  • $\begingroup$ Are all players equally likely to have A or B? If so, the example game won't be much fun. If the third player faces two callers, then the worst scenario is to have card B while one other player has B and one has A, giving a $40%$ chance to win half the pot by calling, and a $60%$ chance to lose the bet. $(0.2)(33)-(.6)(10)=0.6>0$, so the third player should always call. The second player knows this, so facing one call, she should always call. The first player knows both will call if he does, so he should call. The probability of A winning needs to be $>\frac{3x+3}{5x+3}$ for bet $x$ and ante 1 $\endgroup$ – Zack Wolske Nov 21 '14 at 15:29

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