\problemdata     
{S01-PS8-6}           
{probability-intro}              
{} 		
{S98-PS9, S01-PS8}
{S01 Karger}


\begin{problem}



Suppose the game \emph{Let's Make A Deal} is changed slightly. Instead
of having 3 doors with 1 grand prize, in the new game there are 4
doors with 2 grand prizes. The rule of the game remains the
same. Therefore the game now becomes:
\begin{quote}
  At one stage of the game, a contestant is shown 4 doors. The
  contestant knows that there are 2 grand prizes behind 2 of the
  doors and that there are goats behind the 2 other doors. At the
  beginning, the contestant picks a door. To build suspense, the
  assistant of the game then opens a \emph{different} door, revealing
  a goat. The contestant can then stick with his original door or
  switch to another unopened door. He wins a prize only if he now
  picks the door with the prize.
\end{quote}
In the new game, what is the probability of winning with the ``stick''
strategy?  How about the ``switch'' strategy? Can you explain why the
two probabilities do not sum up to 1?

\solution{
{\bf Solution}

The player wins with a ``stick'' strategy iff he chooses the ``correct
door'' in the first place. Therefore the probability of winning with
the ``stick'' strategy is $\frac{1}{2}$.

To calculate the probability of winning with a ``switch'' strategy,
we use a tree diagram.  The prize location can initially be any one of the pairs ${(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}$ each with probability $\frac{1}{6}$.  Because of the symmetry in the tree, we consider the following sub-tree for the pair $(1,4)$.

\graphicspath{{/toc/a/class/6.042/spring02/freshRepository/figures/}}

\bigskip
\centerline{\resizebox{!}{7.0in}{\includegraphics{ps8-monty}}}
\bigskip

Therefore we get that:

\begin{eqnarray*}
  P(\mbox{winning with ``switch'' strategy})
  &=& \mbox{\# of prize combinations} \cdot P(\mbox{winning with ``switch'' strategy for (1,4)})\\
  &=& 6 \cdot (4 \cdot \frac{1}{48} + 4 \cdot \frac{1}{96})\\
  &=&\frac{3}{4}
\end{eqnarray*}

The two probabilities do not need to sum up to 1. This is because the
two strategies are not mutually exclusive, that is there are some
strategies for which the player would win whether or not he ``sticks''
or ``switches''.  For example, if the prizes are behind doors 1 and 4, and the player chooses 1 first then switches to 4, then he would have won
by either strategy.  Therefore, the same winning event can be counted
in both cases at the same time.  }

\end{problem}
