\problemdata      
        {?} % latex-friendly label for the prob.
        {Probability-Conditional} % The topic(s) of the problem content.
        {?} % Source (if known)
        {Fall 2001, final} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).



\begin{problem}
We consider a variation of Monty Hall's game.  The contestant
still picks one of three doors, with a prize randomly placed behind one
door and goats behind the other two.  But now, instead of always opening a
door to reveal a goat, Monty instructs Carol to \emph{randomly} open one
of the two doors that the contestant hasn't picked.  This means she may
reveal a goat, or she may reveal the prize.  If she reveals the prize,
then the entire game is \emph{restarted}, that is, the prize is again
randomly placed behind some door, the contestant again picks a door, and
so on until Carol finally picks a door with a goat behind it.  Then the
contestant can choose to \emph{stick} with his original choice of door or
\emph{switch} to the other unopened door.  He wins if the prize is behind
the door he finally chooses.

To analyze this setup, we define two events:
\begin{description}
\item[$GP$:] The event that the contestant \textbf{g}uesses the door with
the \textbf{p}rize behind it on his first guess.

\item[$OP$:] The event that the game is restarted at least once.  Another
way to describe this is as the event that the door Carol first
\textbf{o}pens has a \textbf{p}rize behind it.
\end{description}

\bparts

\ppart  What is $\pr{GP}$? \brule{0.5in}.  $\prcond{OP}{\bar{GP}}$?\brule{0.5in}.
\solution{1/3,1/2}

\ppart What is $\pr{OP}$? \brule{0.5in}
\solution{\[
\prcond{OP}{GP}\pr{GP} + \prcond{OP}{\bar{GP}}\pr{\bar{GP}} = 0\cdot 1/3 +
1/2 \cdot 2/3 = 1/3.
\]
}

\ppart Let $R$ be the number of times the game is restarted before Carol
picks a goat.

What is $\expect{R}$?\brule{0.8in}\\
You may express the answer as a simple closed form in terms of $p \eqdef
\pr{OP}$.

\solution{
$1/(1-p) - 1 = 3/2 -1 = 1/2$.

Think of \emph{not} having to restart as a failure.  So
$\pr{\text{failure}} = 1-p$, and $\expect{R}$ is mean time to failure
minus one---because we are only counting the number of
``successes''---namely, $1/\pr{\text{failure}} - 1$.}

\ppart What is the probability the game will continue forever?\brule{0.5in}

\solution{0.  In terms of the previous problem part, this is the
probability of \emph{never} failing.}

\ppart When Carol finally picks the goat, the contestant has the choice of
sticking or switching.  Let's say that the contestant adopts the strategy
of sticking.  Let $W$ be the event that the contestant wins with this
strategy, and let $w \eqdef \pr{W}$.  Express the following conditional
probabilities as simple closed forms in terms of $w$.
\begin{enumerate}
\renewcommand{\labelenumi}{\roman{enumi})}

\item $\prcond{W}{GP} = $ \brule{0.5in}
\solution{1}

\item $\prcond{W}{\bar{GP}\intersect OP} = $ \brule{0.5in}
\solution{$w$}

\item $\prcond{W}{\bar{GP}\intersect \bar{OP}} = $ \brule{0.5in}
\solution{0}
\end{enumerate}

\ppart What is $\pr{W}$? \brule{0.5in}

\solution{1/2, because
\[\begin{array}{rcll}
w & = & \prcond{W}{GP}\pr{GP} 
        & +\ \prcond{W}{\bar{GP}\intersect OP}\pr{\bar{GP}\intersect OP}\\
  &   & & +\ \prcond{W}{\bar{GP}\intersect \bar{OP}}\pr{\bar{GP}\intersect \bar{OP}}\\
  & = & 1 \cdot 1/3 & +\ w \cdot 2/3 \cdot 1/2 + 0 \cdot 2/3 \cdot 1/2\\
  & = & 1/3 + w/3.
\end{array}\]
So $w(1- 1/3) = 1/3$.}


\ppart For any final outcome where the contestant wins with a ``stick''
strategy, he would lose if he had used a ``switch'' strategy, and vice
versa.  In the original Monty Hall game, we concluded immediately that the
probability that he would win with a ``switch'' strategy was $1-\pr{W}$.
Why isn't this conclusion quite as obvious for this new, restartable game?
Is this conclusion still sound?  Briefly explain.


\solution{Switching strategies turns wins to losses for \emph{terminated}
games.  So the probability of win with switch is $t-\pr{W}$ where $t$ is
the probability of termination.  In original Monty Hall, the game obviously
terminated after one door-opening, so $t$ was 1.

The extra complication here is that it is possible for the game to run
forever, but this event has probability 0, so $t$ is still 1, and the
conclusion is still sound.}

\iffalse

\ppart Ben Bitdiddle claims that we didn't need all this analysis to
determine a good strategy.  He says that sticking will be a losing
strategy for the contestant, just as in the original Monty Hall game.  His
reasoning is that, because it is more likely than not that the prize is
behind one of the doors the contestant didn't pick, it's to the
contestant's advantage to switch whenever he has the opportunity.  How
would you explain to Ben, briefly, what the fallacy is in his reasoning?
\fi

\eparts
\end{problem}