Re: odds of opponent holding pocket Aces given you holdpocket Kings
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with 50 unknown cards you can have a posiblity of 1225 hands 6 are AA. Against 1 opponent then (6/1225).
Against X number of opponents then aprox:
1 - (1219/1225)^x
If x approx.
2 then .00977
3 then .01462
4 then .01945
6 then .02903
9 then .04323
Matt
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x*6/1225 is more accurate for any number of players. The error is C(x,2)/C(50,4).
Your independence approximation works better for some problems where several players can have a hand, such as the probability that any pair is out.
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