\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Conditional} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, TU10; S99, TU10} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).

\begin{problem}
Suppose the game {\em 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 {\em
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{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 $1/2$.

Our method for analyzing the ``switch'' strategy is based on (careful,
correct use of) conditional probabilities.  Since we have seen how faulty
conditional reasoning can seem OK in this kind of problem, let's first
illustrate and justify our method to calculate the probability of winning
with the ``switch'' strategy in the \textbf{original game} with 3 doors
and 1 prize.

\begin{figure}[htbp]
\centerline{\includegraphics[height=3.5in]{figures/monty2}}
\caption{This is a tree diagram for the \textbf{original} 3-door Monty
Hall problem.  Each of the 12 leaves of the tree represents an outcome.}
\label{fig:monty}
\end{figure}

Figure~\ref{fig:monty} shows the probability tree we developed for this
game.  We examine the branch in which the prize is located behind door 1.
The winning nodes in this part of the probability tree are $(1,2,3)$ and
$(1,3,2)$ with probabilities that sum up to
\[
\frac{1}{9} +
\frac{1}{9}=\frac{2}{9}.
\]
By symmetry, the same probability applies to the other two branches, and
so the total probability of winning is $3\cdot(2/9)=2/3$.

However we can find a shortcut to calculate these probabilities.
Notice that in the second level of the tree, we can distinguish
between two cases:

(a) The player picks the door that has the prize, \eg $(1,1),(2,2),
(3,3)$.

(b) The player picks the door that does not have the prize, \eg
$(1,2),(1,3),(2,1),(2,3),(3,1),(3,2)$.

These are two disjoint events, and furthermore their union covers the
whole sample space.  Suppose we let $E_1$ be the event in (a).  The event
in (b) is then $\overline{E_1}$.  From the law of total probability
\[
\prob{E}=\prob{E|E_1}+\prob{E|\overline{E_1}},
\]
for every event $E$ in our game.

Let $E_2$ be the event that the player wins---in other words that the door
to which he switches has the prize behind it.  \iffalse Note that it
doesn't matter anymore whether the prize is in door 1,2 or 3.  That
information is not important anymore.\fi
Examining the leaves of the probability tree, we see
\begin{eqnarray*}
\pr{E_2}&=&\prcond{E_2}{E_1}\pr{E_1} + 
    \prcond{E_2}{\overline{E_1}}\pr{\overline{E_1}}\\
   &=& 0\cdot\frac{1}{3}+1\cdot\frac{2}{3} = \frac{2}{3}.
\end{eqnarray*}

So, instead of adding all the $W$ probabilities one by one,  we
group them according to whether $E_1$ occurs or not, and of course
our results match.

\iffalse
So, the fact that we condition on $E_1$ and $\overline{E_1}$ comes
after our observation that we can group all possible outcomes in the
probability tree in just these two cases. 
\fi

Returning to our original problem, for the case with 4 doors and 2
prizes we have:
\iffalse
%To calculate the probability of winning with a ``switch'' strategy, let
%$E_1$ be the event that ``the player chooses the correct door at his first
%chance''. Let $E_2$ be the event that ``the player switches to the correct
%door at his second chance''.
\fi
\begin{eqnarray*}
\pr{\text{winning with ``switch''
strategy}}&=&\prcond{E_2}{E_1}\pr{E_1}+\prcond{E_2}{\overline{E_1}}\pr{\overline{E_1}}\\
&=&\frac{1}{2}\cdot\frac{1}{2}+1\cdot\frac{1}{2} =\frac{3}{4}.
\end{eqnarray*}

This time the two probabilities do not need to sum up to 1 because they
are no longer disjoint events.  It is possible for both the ``switch'' and
the ``stick'' strategy to pick doors that both have a prize behind them,
because there are two prizes.  So the same outcome can be a win for both
strategies.}

\end{problem}