\problemdata
{s96-ps9}
{expected value}
{variance}

\begin{problem}
In bridge, the 52 cards of a standard deck 
are dealt to four players.  A hand is the set of 13 cards received by
a player.  
\begin{problemparts}
\problempart
What is the expected value and variance of the number of
spades in a hand? 

\bigskip
\noindent
{\bf Answer: }First observe that the sample space in this problem is
the set of all possible hands of one of the players. Let
$N_{\spadesuit}$ denote the number spades in a hand. Let
$X_{\spadesuit,i}$ be the indicator variable which is $1$ if card $i$
is a spade and $0$ otherwise. Then
$N_\spadesuit=\sum_{i=1}^{13}{X_{\spadesuit,i}}$. The expectation of
$N_\spadesuit$ can be computed by the linearity of expectation. Since
$X_{\spadesuit,i}$ is an indicator variable, $E(X_{\spadesuit,i})$ is
the probability that card $i$ is a spade which is $1/4$. Thus
\[
E(N_\spadesuit)=E(\sum_{i=1}^{13}{X_{\spadesuit,i}}) = \sum_{i=1}^{13}{E(X_{\spadesuit,i})} = 13/4{.}
\]
Not a surprising result! To compute the variance, recall the formula
\[
\var(\sum_i{X_i}) = \sum_i{\var(X_i)} + \sum_{i\not=j}{\cov(X_i,X_j)}{,}
\]
where the second sum is taken over all possible {\em ordered} pairs of
distinct $i$'s and $j$'s. Variance of each of the indicator variables is
\[
\var(X_{i,\spadesuit}) = E(X_{i,\spadesuit}^2) - E^2(X_{i,\spadesuit})
= \frac{1}{4} - \frac{1}{16} = \frac{3}{8}{.}
\]
The covariance of any pair of the indicator variables is
\[
\cov(X_{i,\spadesuit},X_{j,\spadesuit}) = E(X_{i,\spadesuit}X_{j,\spadesuit}) - E(X_{i,\spadesuit})\times E(X_{j,\spadesuit}) = \frac{1}{4}\times \frac{12}{51} - \frac{1}{16} = -\frac{1}{272}{.}
\]
Putting this all together we get that
\begin{eqnarray*}
\var(N_\spadesuit) &=& \sum_{i=1}^{13}{\var(X_i)} + 2\sum_{1\leq i<j\leq 13}{\cov(X_i,X_j)}\\
 &=& 13\times \frac{3}{8} + 2 {13 \choose 2} (-\frac{1}{272}) = \frac{585}{136} = 4.30147
\end{eqnarray*}


\problempart
What is the expected value and variance of the
number of different suits in a hand?

\bigskip
\noindent
{\bf Answer: }We do a similar computation as in the first part. Let
$N$ denote the number suits in a hand. Let $X_{\spadesuit}$ be the
indicator variable which is $1$ if there is a spade in a hand and $0$
otherwise. Define $X_\heartsuit$, $X_\clubsuit$, and $X_\diamondsuit$
analoguously. Then
$N=\sum_{s\in\{\spadesuit,\heartsuit,\clubsuit,\diamondsuit\}}{X_s}$.
The expectation of each of the indicator variables is the probability
that a hand contains the corresponding suit. Thus
\[
E(X_s) = 1-{{52-13\choose 13}\over{52\choose 13}} = 0.987209 {,}
\]
since ${52 - 13 \choose 13}$ is the number of hands
that are missing the particular suit, and ${52 \choose 13}$ is the
total number of hands.
\[
E(N)=E\left(\sum_{s\in\{\spadesuit,\heartsuit,\clubsuit,\diamondsuit\}}{X_s}\right)
= \sum_{s\in\{\spadesuit,\heartsuit,\clubsuit,\diamondsuit\}}{E(X_s)}
= 4 \times 0.987209 = 3.94884{.}
\]
To compute the variance, first compute the variance of each of the
indicator variables:
\[
\var(X_s) = E(X_s^2) - E^2(X_s)
= 0.987209 - 0.987209^2 = 0.0126273{.}
\]
(Note: $E(X_s^2) = E(X_s)$ since $X_s$ is an indicator variable.)  To
compute the covariance of any pair of the indicator variables we need
to compute $E(X_{s_1}X_{s_2})$, i.e. the probability that a hand
contains both suit $s_1$ and suit $s_2$. Let $S_1$ be the event that a
hand contains suit $s_1$ and let $S_2$ be the event that a hand
contains suit $s_2$. Then,
\begin{eqnarray*}
P(S_1\cap S_2) &=& 1 - P(\overline{S_1}\cup \overline{S_2})\\
 &=&  1 - P(\overline{S_1}) - P(\overline{S_2}) + P(\overline{S_1}\cap \overline{S_2})\\
 &=& 1 - 2 {{52-13\choose 13}\over{52\choose 13}} + {{52-13-13\choose 13}\over{52\choose 13}} = 0.974434
\end{eqnarray*}
Then the covariance is
\begin{eqnarray*}
\cov(X_{s_1},X_{s_2}) &=& E(X_{s_1}X_{s_2}) - E(X_{s_1})\times E(X_{s_2})\\
 &=& 0.974434 - 0.987209^2 = -0.00014723{.}
\end{eqnarray*}
Putting this all together we get that
\begin{eqnarray*}
\var(N) &=& \sum_{s\in\{\spadesuit,\heartsuit,\clubsuit,\diamondsuit\}}{\var(X_s)} + \sum_{s_1\not=s_2\in\{\spadesuit,\heartsuit,\clubsuit,\diamondsuit\}}{\cov(X_{s_1},X_{s_2})}\\
 &=& 4\times 0.0126273 + 2 {4 \choose 2} (-0.00014723) = 0.487424
\end{eqnarray*}

\end{problemparts}
\end{problem}
