View Full Version : Top Pair Flop

05-12-2003, 04:29 AM
Hey. I was trying to work out the probability the top pair coming out (assuming no one beats it with pocket pairs) with n amount of players in the game.

I ran into problem after problem with this and really had no idea how to think about it math wise.

SIde question, are there any poker programs just for the use of probability?

Thanks to everyone and thanks for this board.

05-12-2003, 09:41 AM
I'm not quite sure what you're asking - are you trying to figure the probability of *someone* having top pair when n players see the flop? The problem with this sort of question is that it is much more likely that someone has top pair on a board like A45 than on a board like 743... you have to assume *something* about your opponents' play, i.e. that there are many more hands containing an A that they would play than hands containing a 7 that they would play.

If you had infinite time and a personal computer to use during the hand, you would figure this by putting each opponent on a range of hands, each with some % likelihood, and then you would add up all the hands that make top pair with the board, and divide by the total... unfortunately, you won't have this available to you, so you need to "ballpark" it.

so, if you have a hard time ever throwing away QQ to a Kxx flop with some betting action, you will probably give away money to top pair. on the other hand, if you auto-muck A6 to an 863 flop, you will probably also be giving away money, since your opponents will have top pair less often here... the answer, as always, is "it depends" - on the flop, the texture of the game, and what you know about your opponents' play.


05-12-2003, 01:11 PM
Assume 100% of people saw the flop, so those lower cards are still in play.

05-12-2003, 01:39 PM
so, if the flop is xyz, where x > y >= z (x is the TOP card on the board), what are the chances, given n opponents with random cards, that at least one of them holds an x?

assuming that you don't have an x, there are 47 unseen cards, and 3 of them are x's. if you have n opponents, there are 2*n cards in their hands combined. the ways that there could be 1, 2, or 3 x's in those hands are:

(C[3,1]*C[44,2*n-1] + C[3,2]*C[44,2*n-2] + C[3,3]*C[44,2*n-3])

divided by the number of possible distributions for those hands...

/ C[47,2*n]

which works out to the following:

N ... %

1 ... 12.488%
2 ... 23.891%
3 ... 34.258%
4 ... 43.639%
5 ... 52.081%
6 ... 59.636%
7 ... 66.352%
8 ... 72.279%
9 ... 77.465%

so, if you are against 5 players, and you don't have top pair, odds are that at least one of your opponents does, assuming that they all played independent of the quality of their starting hand...


05-12-2003, 02:35 PM
Then *1/2 that their 2nd hard is higher than their pair. Yeah, i'll force those early folds /forums/images/icons/smile.gif.

Thanks for that, i'm in my last year of A level maths, i didn't do stats though so i'm pretty bad at combination probability, not too bad at conditional though. Thanks again /forums/images/icons/smile.gif