Poker With a Pinochle DeckDate: 7/1/96 at 9:29:41 From: Anonymous Subject: Some what confused Imagine that we are playing poker with a pinochle deck of 48 cards: two each of 9,10, J, Q, K, A in each of the four suits. If we are dealt a 5 card poker hand, what is the probability that it will be a full house? Can you explain, please? Thank you. Date: 7/1/96 at 19:17:43 From: Doctor Anthony Subject: Re: Some what confused Full house = (for example) 2 aces + 3 jacks. We could pick the face value of the pair of cards (aces in this case) in 6 ways, and then the face value of the threesome (jacks in this case) in 5 ways. So face values could be chosen in 6 * 5 = 30 ways. Now we must choose 2 aces from 8 and 3 jacks from 8 while there are a total of 48 cards from which we must select 5. So the probability of 2 aces and 3 jacks is (8_C_2)*(8_C_3)/(48_C_5). This result must be multiplied by 30 to cover other possibilities for the face values of the cards making up the full house. Required probability = 30*28*56/1712304 = 0.02747. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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