\problemdata
{hearts}
{expected value}
{variance}

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%%%% PROBLEM 1
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\begin{problem} %source: spring00 ps10-5
Suppose you are playing the game ``Hearts'' with three of your friends.
In Hearts, all the cards are dealt to the players, in this case the four
of you will each have 13 cards.

\begin{problemparts}

\problempart
What is the expectation and variance of the number of hearts in your hand?

\solution{
Let $H$ be the number of hearts in your hand, 
and let $X_i$ be the indicator random variable 
for the event that the $i$th card in your hand is a heart. 
Then $H = \sum_{i=1}^{13} X_i$.  
So
\[
\expect{H}  = \sum_{i=1}^{13} \expect{X_i}
        = \sum_{i=1}^{13} \Pr(\text{$i$th card is a heart}) 
        = \sum_{i=1}^{13} \frac{1}{4} = \frac{13}{4}.
\]

To compute variance, 
we first compute $\expect{H^2}$, 
following the solution to Problem~4c. 
\[
\expect{H^2} = \expect{\left(\sum_{i=1}^{13} X_i\right)^2} 
        = \sum_i \expect{X_i} + \sum_i \sum_{j \neq i} \expect{X_i X_j} 
\]
Note that $X_i^2 = X_i$, so $\expect{X_i^2} = 1/4$.  
When $i \neq j$,
$X_i X_j = 1$ if cards $i$ and $j$ are both hearts, and 0 otherwise.  The
probability that both cards are hearts is $\binom{13}{2}/\binom{52}{2}$, 
so 
$\expect{X_iX_j} = \binom{13}{2}/\binom{52}{2} = 3/51$.  
In the summation above, there
are 13 $\expect{X_i^2}$ terms, and $13 \cdot 12 = 156$ terms of the
form $\expect{X_i X_j}$ with $i \neq j$.  Thus,
\[
\variance{H} = \expect{H^2} - \expect{H}^2
        = 13 \cdot \frac{1}{4} + 156 \cdot \frac{3}{51} - (\frac{13}{4})^2 
        \approx 1.864. 
\]
}



\end{problemparts}
\fi



\end{problems}
