# Combination problem

• Aug 24th 2013, 07:45 AM
downthesun01
Combination problem
$\text{Poker Dice is played by simultaneously rolling 5 dice. Show that P(Two Pair) = .2315}\\\text{I'm having a hard time finding the solution}$
• Aug 24th 2013, 08:31 AM
MarkFL
Re: Combination problem
We have 6 choices for the first pair, 5 choices for the second pair, and 4 choices for the odd die. Multiply this by the number of ways to choose 2 from 6, and divide by the $6^5$ different outcomes, hence:

$P(\text{two pair})={6 \choose 2}\frac{6\cdot5\cdot4}{6^5}=\frac{25}{108}$
• Aug 24th 2013, 09:08 AM
downthesun01
Re: Combination problem
Please explain the 6 choose 2

Also,

The solution I was given is

$\frac{4 {6 \choose2} {5 \choose2} {3 \choose2}}{6^5}$

It gives the same value as your solution, but I was hoping that you could offer some reasoning behind it?
• Aug 24th 2013, 09:44 AM
MarkFL
Re: Combination problem
The 6 choose 2 represents the number of ways to choose the two pairs from the 6 different values of a die.
• Aug 24th 2013, 09:53 AM
Plato
Re: Combination problem
Quote:

Originally Posted by downthesun01
The solution I was given is
$\frac{4 {6 \choose2} {5 \choose2} {3 \choose2}}{6^5}$
It gives the same value as your solution, but I was hoping that you could offer some reasoning behind it?

Do you understand that there is no one correct way of doing this?

I would have done as $\frac{\dbinom{6}{2}(4)\dfrac{5!}{(2!)^2}}{6^5}$

Now I can explain reasoning. The six choose 2 is the number of ways to have two pair.
The four is the number of ways to have the odd number out.
The numbers of ways to arrange $AABBC$ is $\dfrac{5!}{(2!)^2}$.