Timeline for What is known about multiplayer poker with flop?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| S Nov 29, 2014 at 13:58 | history | bounty ended | CommunityBot | ||
| S Nov 29, 2014 at 13:58 | history | notice removed | CommunityBot | ||
| Nov 21, 2014 at 18:44 | comment | added | domotorp | I had independent uniform distribution in mind. The thing is that I could also compute the Nash equilibrium for this particular example, I am more interested in knowing whether this problem has been studied and if there are any interesting results. | |
| Nov 21, 2014 at 18:27 | comment | added | Zack Wolske | You're right, I included the bet in the winnings. How are the cards distributed? | |
| Nov 21, 2014 at 17:15 | comment | added | domotorp | @Zack: Your calculation seems to be incorrect, it should be $(0.4)(21)−(0.8)(10)<0$, so if the others have A and B, you should not call with a B. In fact $(0.3)(21)−(0.7)(10)<0$ shows that you should not even call with an A. | |
| Nov 21, 2014 at 17:11 | comment | added | domotorp | @Zack: My goal is exactly to model situations that can arise in a normal poker game, making it as simple as possible, but containing this phenomenon. | |
| Nov 21, 2014 at 15:36 | comment | added | Zack Wolske | I don't know that this has been studied mathematically by game theorists, but an equivalent scenario comes up often in actual poker, especially limit games. After all cards are dealt, a player can estimate their own chances of winning at showdown against a range of opponents cards, with the perceived range based on previous actions (both actions in the hand, and general playing tendencies). | |
| Nov 21, 2014 at 15:29 | comment | added | Zack Wolske | 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 | |
| S Nov 21, 2014 at 12:49 | history | bounty started | domotorp | ||
| S Nov 21, 2014 at 12:49 | history | notice added | domotorp | Draw attention | |
| Nov 17, 2014 at 12:48 | history | rollback | domotorp |
Rollback to Revision 2
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| Nov 17, 2014 at 12:43 | comment | added | domotorp | @Ilya Yes, just like in poker. | |
| Nov 17, 2014 at 9:09 | comment | added | Ilya Bogdanov | 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? | |
| Nov 16, 2014 at 22:53 | history | edited | Ricardo Andrade |
replaced two new tags (of a less mathematical nature) with relevant existing tags
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| Nov 16, 2014 at 21:45 | comment | added | domotorp | @Sam I am sorry if it was ambiguous, there is no raising in this game, everyone can decide to pay to play or fold. | |
| Nov 16, 2014 at 21:43 | history | edited | domotorp | CC BY-SA 3.0 |
deleted alpha
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| Nov 16, 2014 at 20:37 | comment | added | Sam Hopkins | 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. | |
| Nov 16, 2014 at 18:58 | history | asked | domotorp | CC BY-SA 3.0 |