\problemdata                                                             
        {probability-conditional-1} % latex-friendly label for the prob.
        {Conditional Probability, Monty Hall} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S02 PS9; F01 FINAL EXAM} % Usage (list ones you are aware of). 
        {Tina Wang, 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}                                                         
                                                                          
\begin{problemparts}                                                                   
                                                                          
\problempart  What is $\pr{GP}$? \dots $\prcond{OP}{\bar{GP}}$?                 
\solution{1/3,1/2}             

                                                                             
\problempart  What is $\pr{OP}$?                                                    
\solution{                                                                   
\[                                                                           
\prcond{OP}{GP}\pr{GP} + \prcond{OP}{\bar{GP}}\pr{\bar{GP}} = 0\cdot 1/3 +   
1/2 \cdot 2/3 = 1/3.                                                         
\]                                                                           
}                                                                            
\iffalse                                                                     
                                                                             
\problempart  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, $1/\pr{\text{failure}} - 1$.}                       
\fi                                                                         
                                                                            
\problempart  What is the probability the game will continue forever?  Explain.    
                                                                            
\solution{0.  Successive restarts are independent, so the probability of    
an $n$th restart, given there have already been $n-1$ restarts, is 1/2.     
Hence the probability of $\ge n$ restarts is at most $(1/2)^n               
\longrightarrow 0$ as $n \longrightarrow \infty$.}                          
                                                                            
\problempart  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 $\prcond{W}{GP}$.                                                   
\solution{1}                                                              
                                                                          
\item $\prcond{W}{\bar{GP}\intersect OP}$.                                
\solution{$w$}                                                            
                                                                             
\item $\prcond{W}{\bar{GP}\intersect \bar{OP}}$.                             
\solution{0}                                                                 
\end{enumerate}                                                              
                                                                             
\problempart  What is $\pr{W}$?                                                     
                                                                             
\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 $w(1- 1/3) = 1/3$.  Solving for $w$, we get $w=1/2$.}                                                        
                                                                              
\problempart  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                                                                         
                 
\end{problemparts}
\end{problem}                                                         
                                 