Associated Topics || Dr. Math Home || Search Dr. Math

### Four of a Kind in a 13-Card Hand

```
Date: 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/
```
Associated Topics:
College Probability
High School Probability

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search