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3

Frisbee's answer is correct. But it only answers for eight players (although the other numbers are hidden in that table). Since the OP asked for 8-10, I will add the answers for nine and ten players. The odds of an ace not occurring in the next x cards dealt (after you're dealt two aces) is ( 48!/(48-x)! )/( 50!/(50-x)! ). That's the probability that no one ...


2

Pick is it 8 or 10 At 8 I think it is 0.4114 Or 1.4306 : 1 This is for exactly one ace out - (not two) Using combination (2/1) * (48/13) / (50/14) (2 aces need 1) (48 non ace need 13) / (50 cards need 14) Both other aces out would be 0.0743 Add them for 1 or 2 aces out = 0.4857 You can get that same number with 1 - 48 / 50 * = ...


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I have been struggling over the same example for the same reason. Let P(A) = Prob A has nuts = 0.2, P(A') = Prob A has dead hand = 0.8, P(B) = Prob A bets, P(A'|B) = Prob A has dead hand given A bets P(A|B) = Prob A has nuts given A bets Then P(A|B) = [P(A)P(B|A)]/[P(A)P(B|A)+P(A')P(B|A')] = (0.2*1)/(0.2*1 + 0.8x) and P(A'|B) = ...


1

Just outs to improve on the next card is fairly straight forward. Probability to win is much more complex. If it is just you against a single player on the river then you can calculate. A starting hand against 3 random hands is very complex. I assume you mean chances (not change). You have poker calculators but there is no formula A straight can get ...


1

You can use this this calculator, but basically you would need to know your opponents hand in order to actually calculate the odds. The probability you have of hitting your outs however, can be calculated. Take a look at my other answer to learn how. If you have a solid read you could try to include the probability of him hitting his hand after you have ...



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