probability of being dealt four of a kind in poker

Question

I have to resolve this exercise:

I have 52 cards. I get 5 cards. Calculate the probability I get a poker hand of four-of-a-kind.

well I applied the formula ${52 \choose 5}=\frac{52!}{5!47!}$ to calculate every possible combination.but after here I don't know how can go ahead. Some help?

Edit

I made mistake.Now it's correct

$\binom{52}{5}=\frac{52!}{5!47!}$! What is a poker? – draks ... Mar 30 '12 at 19:27 — draks ..., Mar 30 '12 at 19:27
Yes, the term "poker" here is non-standard. – Thomas Andrews Mar 30 '12 at 19:32 — Thomas Andrews, Mar 30 '12 at 19:32
I mean for poker: 4 cards with identical value – Mazzy Mar 30 '12 at 19:45 — Mazzy, Mar 30 '12 at 19:45

score 3 · Answer 1 · answered Mar 30 '12 at 19:50

3

Hint: Your formula is the total number of $5$ card hands. How many choices of rank of the four are there? How many choices for the other card?

answered Mar 30 '12 at 19:50

Ross Millikan

374,822

score 3 · Accepted Answer · edited Mar 30 '12 at 21:53

3

According to Wiki:Poker probability, or better this here, you'll have the following:

There are 624 possible hands including four of a kind; the probability of being dealt one in a five-card deal is $\frac {C_{13}^1 C_{4}^4 \cdot C_{12}^1 C_{4}^1} {C_{52}^5} = \frac {13 \cdot 1 \cdot 12 \cdot 4} {2{,}598{,}960} \approx 0.024\% $.

with $C_a^b=\binom{a}{b}$.

edited Mar 30 '12 at 21:53

David Mitra

74,748

answered Mar 30 '12 at 19:51

draks ...

18,449

probability of being dealt four of a kind in poker

2 Answers2

Linked