Poker Hand with At Least One Ace
Date: 04/07/2003 at 17:55:35 From: Matti Subject: A poker hand that contains at least 1 Ace Dr. Math, There is an exercise in my textbook that says, "What is the probability that a five-card poker hand contains at least 1 ace?" The textbook gives the answear to this probability: 1-C(48,5)/C(52,5) I do not agree; I feel that this would be a more accurate answer: (C(4,1)*C(51,4))/C(52,5) I first choose 1 ace out of 4 and then I have "at least one ace." Then I can choose the next 4 out of the 51 remaining cards. Then I've got C(4,1) * C(51,4) possible correct outcomes. I divide this by all possible poker hands or: C(52,5) Could anyone tell me which is correct? Kind regards, Matti
Date: 04/07/2003 at 18:13:13 From: Doctor Roy Subject: Re: A poker hand that contains at least 1 Ace Hi, Thanks for writing to Dr. Math. That's not correct. Since it is "at least" one ace, you have to account for drawing 1, 2, 3, or 4 aces. It is easier to calculate the following: Probability of drawing 0 aces. The probability of drawing "at least" 1 ace is: Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces] The probability of drawing 0 aces is: C(48,5)/C(52,5) The numerator is the ways we can draw 5-card hands out of 48 cards (all the cards except for the aces). The denominator is the ways we can draw any 5-card hands. So the ratio is the number of ways of drawing no aces. If we subtract it from 1, we get the probability of drawing "at least" 1 ace. Does this help? Please feel free to write back with any questions you may have. - Doctor Roy, The Math Forum http://mathforum.org/dr.math/
Date: 04/08/2003 at 08:57:37 From: Matti Subject: Thank you (A poker hand that contains at least 1 Ace) Your explanation was different from what we had and I can see now that this approach is correct. Thank you.
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