# Combination and Permutations

• Sep 14th 2010, 08:30 AM
algorith
Combination and Permutations
What is the probability that a five-card poker hand has the following?
(a) Two Pairs (not four of a kind or a full house).

Choose a kind - 13
Choose 2 of a kind - C(4,2)
Choose another kind - 12
Choose 2 of a kind - C(4,2)
Choose last card - 44
Multiply all number above for answer.

Is that right?
Another thing that came to mind is to replace the 13 and the 12 by C(13,2), I'm having trouble choosing one over the other. Can somebody confirm the right answer and explain why. Thank you.
• Sep 14th 2010, 09:33 AM
Soroban
Hello, algorith!

Your afterthought is correct . . .

Quote:

What is the probability that a five-card poker hand has the following?
(a) Two Pairs (not four of a kind or a full house).

Choose a kind - 13
Choose 2 of a kind - C(4,2)
Choose another kind - 12
Choose 2 of a kind - C(4,2)
Choose last card - 44
Multiply all number above for answer.

Is that right?

Another thing that came to mind is to replace the 13 and the 12 by C(13,2),
I'm having trouble choosing one over the other. .
This one is correct.
Can somebody confirm the right answer and explain why?

We choose 2 values from the possible 13 values: .$\displaystyle _{13}C_2\,=\,78$ ways.
. . (The order does not matter.)

Choose a pair: .$\displaystyle _4C_2\,=\,6$ ways.

Choose another pair: .$\displaystyle _4C_2\,=\,6$ ways.

Choose the fifth card: .$\displaystyle 44$ ways.

So there are: .$\displaystyle 78\cdot6\cdot6\cdot44\:=\:123,\!552\,\text{ possible Two Pairs.}$

There are: .$\displaystyle _{52}C_5 \:=\:2,598,960$ possible 5-card hands.

Therefore: .$\displaystyle P(\text{Two Pairs}) \;=\;\dfrac{123,\!552}{2,\!598,\!960} \;=\;\dfrac{198}{4165} \;\approx\;4\frac{3}{4}\%$
• Sep 14th 2010, 03:42 PM
algorith
Thanks a lot, that was brilliant.