\"choose\"- another question from a math illiterate
pretty much any cards mathematics question lists "choose" as an operation... (i.e., 52choose2 or something?)
my question is, how do you do the choose operation? not knowing this seems like a pretty serious mathematical flaw in my card game.
walk me through the steps seen here, taken from a forum about Kings Vs. Aces... what to do on the calculator when you see the "C"?
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There are 6 ways to have AA, so the probability of 1 particular player having it is 6/C(50,2).
Multiply this by 8 players to get 8*6/C(50,2). This alone gets you very close, but it will double count all cases where 2 players have AA, so for the exact answer we have to subtract this off. The probability that 2 specific players have AA is 1/C(50,4) since there is only one way to deal 4 aces to 2 players. Since no more than 2 players can have AA, we can just multiply this by the number of 2 player pairs C(8,2). So the exact answer for the whole thing is:
8*6/C(50,2) - C(8,2)/C(50,4) = 1 in 25.6 = 24.6-to-1.
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all help appreciated.
monte [img]/images/graemlins/club.gif[/img]
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