\problemdata                                                             
        {probability-intro-2} % latex-friendly label for the prob.
        {Basic Probability, Poker Hands} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S02 CP9M} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\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 divided
by $\binom{52}{5}$.  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
\[
\binom{4}{2} \cdot \binom{26}{5} - 2 \cdot 4 \cdot \binom{13}{5},
\]
and the probability that a random poker hand contains cards from at most
two suits is:
\begin{equation*}
\frac{\dbinom{4}{2} \cdot \dbinom{26}{5} - 2 \cdot 4 \cdot \dbinom{13}{5}}
                {\dbinom{52}{5}}
\approx 0.15
\end{equation*}
}
\end{problem}
