Re: Probability of 2 PP\'s at 9 handed table?
1 - [(1225-73)/1225]^8 ~ 38.83%
This is a good approximation to solve these types of problems . Notice that I subtract 73 hands from 1225 since there are 73 pairs remaining which may also include the one you're holding . If you're interested in distinct pairs from your own then the answer is 1- [(1225-72)/1225]^8 ~ 38.4%
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