#1
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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. |
#2
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Re: Probability of making a full house
You aren't accounting for discards. Remember, yiour discards aren't shuffled back in, so you can be sure you won't draw them.
So there are 9 ranks with 4 cards left, and 3 ranks with only 3 cards left. |
#3
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Re: Probability of making a full house
Thank you very much for your help.
I corrected the calculation ... Total of combinations: C(47,3) = 16'215 9 ranks with 4 cards left: 9 * C(4,3) = 9 * 4 = 36 9 * C(4,2) * 2[aces] = 9 * 6 * 2 = 108 3 ranks with 3 cards left: 3 * C(3,3) = 3 * 1 = 3 3 * C(3,2) = 3 * 3 * 2[aces] = 18 36+108+3+18 = 165 combinations of a total of 16'215 combinations make the full house. 16'215/165 = 98.3 = 97.3-to-1 or about 1.02% |
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