$\displaystyle \text{Poker Dice is played by simultaneously rolling 5 dice. Show that P(Two Pair) = .2315}\\\text{I'm having a hard time finding the solution}$

Printable View

- Aug 24th 2013, 06:45 AMdownthesun01Combination problem
$\displaystyle \text{Poker Dice is played by simultaneously rolling 5 dice. Show that P(Two Pair) = .2315}\\\text{I'm having a hard time finding the solution}$

- Aug 24th 2013, 07:31 AMMarkFLRe: 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 $\displaystyle 6^5$ different outcomes, hence:

$\displaystyle P(\text{two pair})={6 \choose 2}\frac{6\cdot5\cdot4}{6^5}=\frac{25}{108}$ - Aug 24th 2013, 08:08 AMdownthesun01Re: Combination problem
Please explain the 6 choose 2

Also,

The solution I was given is

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

It gives the same value as your solution, but I was hoping that you could offer some reasoning behind it? - Aug 24th 2013, 08:44 AMMarkFLRe: 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, 08:53 AMPlatoRe: Combination problem
Do you understand that there is no one correct way of doing this?

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

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 $\displaystyle AABBC$ is $\displaystyle \dfrac{5!}{(2!)^2}$.