(Almost) every introductory course in probability introduces conditional probability using the famous Monte Hall problem. In a nutshell, the problem is one of deciding on a best strategy in a simple game. In the game, the contestant is asked to select one of three doors. Behind one of the doors is a great prize (free attendance to an R workshop, lets say), and there is a bum prize behind each of the other two doors. The bum prize is usually depicted to be a goat, but I don’t think that would be such a bad prize, so let’s say that behind two of the doors is a bunch of poorly collected data for you to analyse. Once the contestant makes her first selection, the host then opens one of the other two doors to reveal one of the bum prizes.

At this point, the contestant is given the choice to either stay with her original selection, or switch to the other remaining unopened door. What should she do?

If you’re reading this blog, you no doubt gleefully shouted “Switch! Switch! Daddy needs a ticket to that R workshop!”. And you also, no doubt, can prove this to be the best strategy using the logic of probability. You reason that the contestant’s chance of selecting the winning door from the onset was 1/3, giving her a 2/3 probability of being wrong. Once one door with a bum prize has been opened, the contestant is now choosing between two doors. Knowing that there was only a 1/3 chance that original selection was correct, there is now a 2/3 chance that the alternate door is the winner.

As I have mentioned in previous posts, I have found that engaging students to think through logical reasoning problems can be greatly enhanced by appealing to their desire to see it in action. To that end, I whipped up this little script to simulate repeatedly playing the Monte Hall game.

##################################################### # Simulation of the Monty Hall Problem # Demonstrates that switching gives you better odds # than staying with your initial guess # # Corey Chivers, 2012 ##################################################### rm(list=ls()) monty<-function(strat='stay',N=1000,print_games=TRUE) { doors<-1:3 #initialize the doors behind one of which is a good prize win<-0 #to keep track of number of wins for(i in 1:N) { prize<-floor(runif(1,1,4)) #randomize which door has the good prize guess<-floor(runif(1,1,4)) #guess a door at random ## Reveal one of the doors you didn't pick which has a bum prize if(prize!=guess) reveal<-doors[-c(prize,guess)] else reveal<-sample(doors[-c(prize,guess)],1) ## Stay with your initial guess or switch if(strat=='switch') select<-doors[-c(reveal,guess)] if(strat=='stay') select<-guess if(strat=='random') select<-sample(doors[-reveal],1) ## Count up your wins if(select==prize) { win<-win+1 outcome<-'Winner!' }else outcome<-'Losser!' if(print_games) cat(paste('Guess: ',guess, '

Revealed: ',reveal, '

Selection: ',select, '

Prize door: ',prize, '

',outcome,'



',sep='')) } cat(paste('Using the ',strat,' strategy, your win percentage was ',win/N*100,'%

',sep='')) #Print the win percentage of your strategy }

Students can then use the function to simulate N games under the strategies ‘stay’, ‘switch’, and ‘random’. Invoking the function using:

monty(strat="stay") monty(strat="switch") monty(strat="random")

Hey, look at that – it is better to switch! Feel free to use this in your own class, and let me know if you use it or adapt it in an interesting way!

Edit: Thanks to Ken for pointing out that saying that ‘switching is always better’ is not quite right (you will loose 1/3 of the time, after all), but rather that switching is a rational best strategy, given your state of knowledge.