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?

Guest Sep 9, 2021

#1**0 **

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

\[\frac{\binom{13}{2} \binom{4}{2} \binom{4}{3}}{\binom{52}{5}} = \frac{3}{4165}.\]

Guest Sep 9, 2021

#2**0 **

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?

helperid1839321
Sep 9, 2021

#3**0 **

**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

Guest Sep 10, 2021

edited by
Guest
Sep 10, 2021