1.3.6 problem 34

There are $\left(\frac{52}{5}\right)$ equally likely $5$-card hands.

(a) Choose a suit in $\left(\frac{4}{1}\right)$ ways, then choose $5$ of its $13$ ranks in $\left(\frac{13}{5}\right)$ ways. This includes the royal flush for that suit, so we subtract the $1$ way to get the royal flush. Hence, the probability of a flush (excluding the royal flush) is

(b) Choose the $2$ ranks that form the pairs in $\left(\frac{13}{2}\right)$ ways, then choose $2$ suits out of $4$ for each of those ranks in ${\left(\frac{4}{2}\right)}^2$ ways. Then choose the rank of the remaining card from the $11$ leftover ranks in $\left(\frac{11}{1}\right)$ ways, and its suit in $\left(\frac{4}{1}\right)$ ways. Hence, the probability of two pair is