Hi

I have the feeling I am wrong somewhere, I try to calculate the number of 5-card hands that make full house.

$\displaystyle 13\cdot \binom{4}{3}\cdot 12 \cdot \binom{4}{2} $

thanks

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- Jun 4th 2009, 11:01 AMTwigHow many 5-card hands make full house
Hi

I have the feeling I am wrong somewhere, I try to calculate the number of 5-card hands that make full house.

$\displaystyle 13\cdot \binom{4}{3}\cdot 12 \cdot \binom{4}{2} $

thanks - Jun 4th 2009, 12:58 PMSoroban
Hello, Twig!

Quote:

I try to calculate the number of 5-card hands that make Full House.

. . $\displaystyle 13\cdot \binom{4}{3}\cdot 12 \cdot \binom{4}{2} $

. . $\displaystyle \begin{array}{cc}\text{Number of values for the Triple:} & 13 \\

\text{Number of ways to get the Triple:} & {4\choose3} \\

\text{Number of values for the Pair:} & 12 \\

\text{Number of ways to get the Pair:} & {4\choose2} \end{array}$

*Absolutely correct!*