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Probability of making a full house
I was wondering how Mike Caro arrives at 97.3 to 1 to a full house in 5draw (no joker) when drawing 3 to 2 aces?
I tried to figure it like that: Since I keep two aces 50 cards are left. Thus there are C(50,3) = 19'600 combinations of 3 cards out of 50. First, I calculated the 3-of-a-kind combinations ... 12 ranks * C(4,3) = 48 Second, I calculated the one pair + aces combinations ... 12 ranks * C(4,2) * 2 aces = 144 Added 144+48 = 192 That is, 192 combinations of a total of 19'600 combinations make my full house. 19'600/192 = 102.08 = 101.08-to-1 or about 1% Am I on the right track or what am I doing wrong? Thanks for any help. |
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