Re: Top Pair Flop
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...
-switters
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