
\problemdata     
{S01-PS10-7}           
{expectation}              
{} 		
{S01-PS10}
{S01 Karger}



\begin{problem}


 %{\bf (15 points)} 
In the board game Monopoly, the number of
squares that a player advances in a single turn is determined by up to
three rolls of dice as follows.

\begin{itemize}
\item Initially, the player rolls two dice, sums the results, and
advances that many squares.

\item If the same number comes up on both dice---this is called
\emph{rolling a double}---then the player rolls the dice a second time,
sums the results, and advances that many additional squares.

\item If the player rolled on double on the second roll, then the player
rolls the dice a third time, sums the results, and advances that many
additional squares.

\item However, if third roll was a double, then the player ``goes to
jail''.  We will treat this as though he had to reset his position to
where he was just before the first roll, so the total number of squares
advanced by the player in this case is zero.
\end{itemize}

What is the expected number of squares that a player advances in a single
turn?  Assume that the dice are fair and six-sided and that rolls are
mutually independent.

\solution{
Here is one approach to the problem.

It's convenient to assume the dice are \emph{always} rolled three times,
with the player simply ignoring results of any roll which was preceded by
a non-double roll.

Let the random variable $R_i$ be the sum of the two dice on the $i$th
roll, and let $I_i$ be an indicator for the event that the $i$th roll was
a double.

We can express the random variable $R$, the number of squares advanced by
the player, as follows:
\[
R = R_1 + I_1 R_2 + I_1 I_2 R_3 - I_1 I_2 I_3 (R_1 + R_2 + R_3)
\]

Since the dice rolls are independent, we know that $R_i$'s and $I_j$'s at
different rolls are independent.  We can take expectations of both sides
and reason as follows:
\begin{eqnarray*}
\expect{R}      & = &   \expect{R_1}
                 + \expect{I_1 R_2}
                 + \expect{I_1 I_2 R_3} \\
        & & \mbox{}
                 - \expect{I_1 I_2 I_3 R_1}
                 - \expect{I_1 I_2 I_3 R_2}
                 - \expect{I_1 I_2 I_3 R_3} \\
        & = &   \expect{R_1}
                 + \expect{I_1} \cdot \expect{R_2}
                 + \expect{I_1} \cdot \expect{I_2} \cdot \expect{R_3} \\
&&\mbox{}        - \expect{I_2} \cdot \expect{I_3} \cdot \expect{I_1 R_1} \\
&&\mbox{}        - \expect{I_1} \cdot \expect{I_3} \cdot \expect{I_2 R_2} \\
&&\mbox{}        - \expect{I_1} \cdot \expect{I_2} \cdot \expect{I_3 R_3}
\end{eqnarray*}

Note that the expectation of $R_i$ is $7$, and the expectation of $I_i$ is
$1/6$.  Finally, note that $\expect{I_i R_i} = 7/6$ by direct calculation.
Substituting into the formula above gives:
\begin{eqnarray*}
\expect{R}      & = &   \frac{595}{72}.
\end{eqnarray*}

}
\end{problem}