Re: ? from ssnl
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It's true that the exact solution isn't much more complicated at all .
It's interesting to note that this method is not bad even when we calculate the probability that at exactly one player holds a pocket pair at a 6 handed table . If you can tolerate being off by about 3% [img]/images/graemlins/smile.gif[/img]
6c1*78/1326*[(1326-78)/1326]^5 ~ 26.06%
Using the other method , we get:
6*78/1326*1147/1225*1050/1128*957/1035*868/946*783/861 ~23.73%
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Actually, for exactly 1 pair, the method of independence is more accurate when done like this:
6*78/1326*[(1225-73)/1225]^5 =~ 25.959%
I computed the exact answer to lie between 25.969% and 25.972%, so this independence approximation is within about 0.01%.
Also, if we use this independence approximation for the AA/KK problem, we get
6*12/1326*(1218/1225)^5 =~ 5.2765%
The exact answer lies between 5.2754326% and 5.2754319%, so this independence approximation is within about 0.001%.
The reason yours is less accurate is actually because you attempt to account for the changing number of cards remaining in the denominator with each term, without a corresponding change in the numerator. It is actually more accurate to keep both of these the same, as the effects then tend to largely cancel out.
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