\problemdata
{expectation}
{S99, pset11}
{Eric Ho S02}

\begin{problem} 
\begin{problemparts}

\problempart Suppose that I roll a 4-sided die, a 6-sided die, an
8-sided die, a 10-sided die, a 12-sided die, and a 20-sided die.  What
is the expected number of 6's that come up?  (Assume that all of the
dice are fair.)

\solution{
\begin{eqnarray*}
\Ex(\text{6's on all dice})
        & = &   \Ex(\text{6's on 4-sided die}) + \Ex(\text{6's on 6-sided die}\
) + \ldots \\
        &   &   \mbox{} \quad + \Ex(\text{6's on 20-sided die}) \\
        & = &   \frac{0}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10} +
        \frac{1}{12} + \frac{1}{20} \\
        & = &   \frac{21}{40}
\end{eqnarray*}

The first step uses linearity of expectation, and the second step is
simplification.  
}

\problempart Suppose that I roll $n$ dice that are 6-sided, fair, and
mutually independent.  What is the expected value of the largest
number that comes up?

\solution{Let the random variable $M$ be the largest number that comes
up.  We will use the identity

\begin{eqnarray*}
\Ex(M)  & = &   \sum_{i=0}^{\infty} \Pr(M > i)
\end{eqnarray*}

We must compute the probability of the event $M > i$; that is, the
event that some die shows a value greater than $i$.

\begin{eqnarray*}
\Ex(M > i)
        & = &   \sum_{i=0}^5 \Pr(M > i) \\
        & = &   \sum_{i=0}^5 1 - \Pr(\text{every die $\leq i$}) \\
        & = &   \sum_{i=0}^5 1 - (\Pr(\text{one die $\leq i$}))^n \\
        & = &   \sum_{i=0}^5 1 - \left(\frac{i}{6}\right)^n \\
        & = &   6 - \frac{1^n + 2^n + 3^n + 4^n + 5^n}{6^n}
\end{eqnarray*}
}
\end{problemparts}
\end{problem}
