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Server Time: 2/13/2012 11:06:15 AM PACIFIC |
 a2 in omaha, trogdor, 27. Jan 2003 22:15 | ||
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| what is the probability of being deatl a2 in 4 cards? | ||
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| Re: a2 in omaha, Mano, 28. Jan 2003 09:48 | ||
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| I believe this is right: 4*4*50*49/C(52,4) = .1447 or about 5.9:1 against | ||
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| Above answer is wrong, Mano, 31. Jan 2003 17:01 | ||
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| I redid the math and the above answer is way off. Let C(m,n) = m!/n!(m-n)! (number of ways to choose a subset of n elements from a set with m elements). The number of ways to be dealt a 4 card hand with no aces or 2's is C(44,4) = 135751 The number of ways to be dealt 1 Ace and 3 cards not aces or 2's is 4*C(44,3) = 52976 . The same calculation applies for one 2 no A's. The number of ways to be dealt 2 A's no 2's is C(4,2)*C(44,2) = 6*946 = 5676. Again same for 2's. The number of ways to be dealt 3 's no 2's is C(4,3)*C(44,1) = 4*44= 176. (same for 2's) There is 1 way to be dealt 4 A's and 1 way to be dealt 4 2's. Adding all these up 135751 + 2*(52976 + 5676 + 176 + 1) = 253409. So there are 253409 hands without an ace and a duece in them out of C(52,4) = 270725 possible hands. Dividing, we get that you will not be dealt A2 93.6% of the time, which means you will have a hand with at least one A and one 2 in it about 6.4% of the time, or about once every 15 hands. | ||
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