
\problemdata     
{S01-PS8-5}           
{probability - intro}              
{} 		
{S99-PS10, S01-PS8}
{S01 Karger}

\begin{problem}


 What is the probability that a random poker
hand contains cards from at most two suits?

\solution{The sample space consists of the $\binom{52}{5}$ possible
hands.  Each hand is equally likely and comes up with probability $1 /
\binom{52}{5}$.  Therefore, the probability that a hand contains at
most two suits is equal to the number of different hands of this type
multiplied by $1 / \binom{52}{5}$.  All that remains is to count the
number of poker hands with at most two suits.

There are $\binom{4}{2}$ ways to choose two suits and $\binom{26}{5}$
ways to choose five cards from these two suits.  However, this
triple-counts hands containing cards from a single suit.  (For
example, a hand with all spades is counted once as a spades-hearts
hand, a second time as a spades-clubs hand, and a third time as a
spades-diamonds hand.)  The number of hands with a single suit is $4
\cdot \binom{13}{5}$.  Therefore, the number of hands with at most two
suits is

\[
\binom{4}{2} \cdot \binom{26}{5} - 2 \cdot 4 \cdot \binom{13}{5}.
\]

The probability that a random poker hand contains cards from at most
two suits is:

\begin{eqnarray*}
\Pr(\text{hand has at most two suits})
        & = & \frac{\binom{4}{2} \cdot \binom{26}{5} -
                        2 \cdot 4 \cdot \binom{13}{5}}
                {\binom{52}{5}} \\
        & = & \frac{88}{595} \approx 0.15
\end{eqnarray*}
}

\end{problem}
