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Message: [QUOTE="swrittenb, post: 1482607, member: 60355"] I had to take a shot at this, as it's good practice for my stats research. :D This is a Bayesian statistics question. Specifically, if you assume that there is some probability associated with Phil Ivey winning a single event (and that the probability is constant across games and tournament sizes, arguably and admittedly not true), you can derive that number using Bayesian inference. Say his probability of winning an event is given as p(win) = x, where x is some value in the range 0 and 1 (0 is dead money, 1 is a lock on the bracelet). You want to consider the probability that he'll win either 0 or 1 bracelets out of the 80 tournaments he plays in. You can look at the opposite side, but then you need to consider the probability that he wins 2 bracelets, 3 bracelets, 4 bracelets ... 79 bracelets, and 80 bracelets. That's more work. Beh! And anyways, say that we want to assume that both sides got money in good on the bet, and it's a coin flip - 50% of the time he hits it, and 50% he doesn't. What are the odds that he wins a given tournament? Bayesian inference states that for the times that he wins k bracelets in 80 tournaments (for some value k), we use the following: 80Ck * x^k * (1 - x)^(80-k) where C is the binomial coefficient, how many combinations of k can you choose from 80. If k is 1, it's like saying "how many ways can you choose one element from 80," and specifically, "how many individual tournaments do we need to consider if he wins one tournament, but we don't know which one." The answer for k = 1 is 80, since he could win any of the tournaments. The last two terms deal with the probability that he won k tournaments (x to the power of k), and the probability that he lost 80-k tournaments (1-x to the power of 80-k). Since we want 0 and 1 wins, solve the following: 80C0 * x^0 * (1 - x)^(80) + 80C1 * x^1 * (1 - x)^(79) = 0.5 This simplifies to: (1 - x)^80 + 80x * (1 - x)^79 = 0.5 If you solve for x there, x works out to be about 0.0209. If you say that the bet is even, Ivey's percentage of winning any tournament is roughly 2.09%. The chance that he'll win zero out of eighty tournaments is about 18.5%, and the chance that he'll win one out of eighty tournaments is about 31.5%. Two or more is just about 50%. If he wins one early, we'd need to revise the numbers. For example, if he won the 15th tournament, and we still believed our original value for the probability that he wins (2.09%), we have 65 more tournaments for him to win a single one. The chance that he doesn't win a tournament in 65 is: 65C0 * x^0 * (1 - x)^65 = 0.2534 So the chance that he doesn't win another tournament after winning the 15th is about 25.3%, and the chance that he does win jumps up to about 74.7%. Gosh, I hope the math is right on that one. [/QUOTE]

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