Re: How Do I Calculate This?
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I see where the 80% comes from, but why do you cube it. I know on the surface you cube it because there are 3 players, but why?
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You are interested in the probability that each player does not have a top 20% hand. If these are independent events, each person will fold 80% of the time. Probability of all three folding is .8^3. So the probability of getting at least one caller is 1-.8^3.
Now, if you want to think about these as dependent it depends critically on the number of the possible starting hands and what those hands include. This is a much harder problem than it seems because every time someone folds you assume they don't have a good hand, but that then throws out two cards that could be picked. Just assume that the first guy has real trash like 72o. That subtracts two bad cards. Then there are ~2*51 fewer bad hands to pick from. There are a couple of overlaps, but that's not big deal. Let's say those were all bad hands that could have been had. Again, not true (A7 and 77 are probably top 20%), but this will overestimate the calling percentage a little so its fine. The next guy is now picking from 51*50 cards, but he likes more than 20% of those. His percentage is 52/250 (that's .2*52/50). If his hand is crap, then the next guy is missing ~4*51 hands (again similar issues). His percentage is 52/240 (that's .2*52/48). So that gives us a percentage of 49.6% of the time all three fold. So that's pretty close to break even.
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