# Poker Question

• Oct 4th 2010, 07:34 PM
Bushy
Poker Question
Hi all

The question is:

Given a 52 card deck, how many hands of 5 cards have

a) exactly 3 aces?

My guess is 4 x 3 x 2 x 48 x 47. Although this is not correct according to my text book.

b) at least 3 aces?

My guess is, having 3 aces plus having 4 aces = (4 x 3 x 2 x 48 x 47) + (4 x 3 x 2 x 1 x 48) . This is also not correct
• Oct 4th 2010, 08:31 PM
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Quote:

Originally Posted by Bushy
Hi all

The question is:

Given a 52 card deck, how many hands of 5 cards have

a) exactly 3 aces?

My guess is 4 x 3 x 2 x 48 x 47. Although this is not correct according to my text book.

b) at least 3 aces?

My guess is, having 3 aces plus having 4 aces = (4 x 3 x 2 x 48 x 47) + (4 x 3 x 2 x 1 x 48) . This is also not correct

For part (a) you can modify slightly to get the correct answer.

(4*3*2)/3! * (48*47)/2!

This is also C(4,3) * C(48,2) where C(n,k) is binomial coefficient.

4*3*2 means order matters; 4*3*2/3! means order does not matter.

For part (b) you can do as you suggested, but you will need to modify your expression for 4 aces.
• Oct 5th 2010, 07:30 AM
Soroban
Hello, Bushy!

Are you familiar with Combinations?

Quote:

Given a 52-card deck, how many hands of 5 cards have:

(a) exactly 3 Aces?

There are 4 Aces and 48 Others.

We want 3 Aces and 2 Others.

. . There are: . $_4C_3 = 4$ ways to get 3 Aces.

. . There are: . $_{48}C_2 = 1128$ ways to get 1 Other.

Therefore: . $4\cdot1128 \:=\:4512$ hands with exactly 3 Aces.

Quote:

(b) at least 3 Aces?

"At least 3 Aces" means: . $\begin{Bmatrix}\text{3 Aces, 2 Others} \\ \text{or} \\ \text{4 Aces, 1 Other} \end{Bmatrix}$

3 Aces, 2 Others: from part (a), $4512$ ways.

4 Aces, 1 Other:
. . . There is: . $_4C_4 \:=\:1$ way to get 4 Aces.
. . There are: . $_{48}C_1 \:=\: 48$ ways to get 1 Other.

Hence, there are: . $1\cdot48\:=\:48$ ways to get 4 Aces, 1 Other.

Therefore: . $4512 + 48 \:=\:4560$ ways to get at least 3 Aces.