Odds Vs Number of players |
|
Posted Sun Jul 18, 2004 9:54 am GMT by izb
Does anyone know how the number of players affects the odds of winning?
I can calculate the odds of winning based on what is in my hand and what is on the board, but this is for a two player game.
I was wondering if it's possible to easily calculate the odds for games with the same cards on the table and in my hand, but with a different number of players.
E.g. if my odds of winning are 10:1 with two players, would the odds be 20:1 with three players? I suspect that it's more complex than that, but I'm hoping not.
Did you know that participating in a poker forum can help you improve your own game? Be it by sharing experiences or simply asking for help, participation in a forum helps you focus and keep 'on topic' which will help you improve your game. You can learn from other players feedback and from their experiences. Why the THP poker forums? We offer one of the best managed texas holdem poker forums available, and the community within is far more friendly than those typicaly found on other sites. We've made a 'lurkers edition' of the poker forum available here on Holdem Poker Online, but we encourage all visitors to register and join in on the conversations on TexasHoldem-Poker.com
Posted Sat Aug 21, 2004 4:09 am GMT by vegasholdem
| izb wrote: |
E.g. if my odds of winning are 10:1 with two players, would the odds be 20:1 with three players? I suspect that it's more complex than that, but I'm hoping not. |
I totally agree with that statement.
Posted Mon Aug 23, 2004 2:55 am GMT by infarom
The winning odds for your hand are the same, no matter how many opponents you have (the information - seen cards - is the same).
The parameter n= number of your opponents is involved when calculating your opponents winning odds. For example, for "at least one opponent" type events: the probability formula for at least one opponent to can achieve a specific card formation (or many) is something like that:
P = n*P(one opponent can achieve that formation) - C(n,2)*P(two opponents can achieve that formation) + C(n,3)*P(three opponents...) - ... and so on.
For all odds and all math behind the game, see www.probability.go.ro/page8.html
|
|
