
\problemdata     
{S01-PS10-3}           
{probability-expectation}              
{} 		
{S01-PS10}
{S01 Karger}





\begin{problem}



\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{
  
  Let $X_i$~be the number of 6's on the $i$-sided die for $i \in
  \{4,6,8,10,12,20\}$. Then the number of 6's on all dice is $X_4 +
  X_6 + X_8 + X_{10} + X_{12} + X_{20}$ and by linearity of expectation
  \begin{displaymath}
    \expect{X_4 + X_6 + X_8 + X_{10} + X_{12} + X_{20}} = 
    \expect{X_4} + \expect{X_6} + \expect{X_8}
    + \expect{X_{10}} + \expect{X_{12}} + \expect{X_{20}}.
  \end{displaymath}
  Since every die has exactly one 6, $\expect{X_i} = 1 / i$, which gives
  \begin{displaymath}
    \expect{X_4 + X_6 + X_8 + X_{10} + X_{12} + X_{20}} = 
    \frac{0}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10} +
    \frac{1}{12} + \frac{1}{20}
    = \frac{21}{40}\,.
  \end{displaymath}
}

\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?

\hint You may want to use the ``alternative" formula for expectation:
$\expect{M} = \sum_{i=0}^{\infty} \Pr(M > i)$. (This formula is valid
since the random variables involved are non-negative.)

\solution{  
  Let the random variable $M$ be the largest number that comes up. To
  use the identity $\expect{M} = \sum_{i=0}^{\infty} \Pr(M > i)$ we
  must compute the probability of the event $M > i$; that is, the
  event that some die shows a value greater than $i$.
  \begin{displaymath}
    \Pr(M > i)
    = 1 - \Pr(\text{every die $\leq i$})
    = 1 - \prod_{k=1}^n \Pr(\text{$k$th die $\leq i$})
    = 1 - (i/6)^n,
  \end{displaymath}
  where the second equality holds since the dice are independent. Therefore,
  \begin{displaymath}
    \expect{M}
    = \sum_{i=0}^5 \Pr(M > i)
    = \sum_{i=0}^5 1 - (i/6)^n
    = 6 - \frac{1^n + 2^n + 3^n + 4^n + 5^n}{6^n}\,.
  \end{displaymath}}


\end{problem}