We have a standard deck of 52 cards, with 4 cards in each of 13 ranks. We call a 5-card poker hand a full house if the hand has 3 cards of one rank and 2 cards of another rank (such as 33355 or AAAKK). What is the probability that five cards chosen at random form a full house?
You choose a rank for the three cards, then choose a rank for the two cards. There are C(52,5) ways of choosing 5 cards, so the probability is
\[\frac{\binom{13}{2} \binom{4}{2} \binom{4}{3}}{\binom{52}{5}} = \frac{3}{4165}.\]
+633 I'm actually not to familiar with statistics but would be very happy to learn where you learned this, do you think you'd be able to point me to a resource?
+2490 Guest’s answer is wrong. So it appears he didn’t learn it.
There are numerous examples of online poker-hand probability texts and examples. Google is a good starting point ...Bing is too.... so is Yahoo.... and Dogpile....
Solution for this question:
Choose the rank for the triple nCr(13,1)
Choose three suits for the triple nCr(4,3)
Choose a different rank for the pair nCr(12,1)
Choose two suits for the pair nCr(4,2)
Then
Divide this by the total number of five-card hands. nCr(52,5)
\(\hspace {7cm} \large \text{ } \dfrac{\dbinom {13}{1}\dbinom {4}{3}\dbinom {12}{1}\dbinom {4}{2}}{\dbinom {52}{5}} = \dfrac{6}{4165} \approx 0.144\% \)
GA
