Four of a Kind in a 13-Card HandDate: 03/24/2001 at 15:42:20 From: Stanford Tran Subject: Compound Probability What is the probability of receiving four-of-a-kind when dealt 13 cards from a regular 52-card deck? Since you can get four cards in any order, I don't know how to set up an equation. Date: 03/26/2001 at 15:17:51 From: Doctor Rob Subject: Re: Compound Probability Thanks for writing to Ask Dr. Math, Stanford. Let C(n,k) = n!/(k!*[n-k]!). There are C(52,13) ways to deal a 13-card hand from the deck. There are C(13,1) ways to choose the denomination for the four of a kind, and C(48,9) ways to deal the remaining 9 cards from the 48 other cards in the pack. Some of these hands are being counted twice, however, because you can have two or three four-of-a-kind sets in 13 cards. The actual number comes out to be: C(13,1)*C(48,9) - C(13,2)*C(44,5) + C(13,3)*C(40,1) Divide this by C(52,13) to get the probability of one or more fours-of-a-kind in 13 cards. I get between 3% and 4%. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/