The Monty Hall problem

The Monty Hall problem is a famous game which was played in the television show "Let's make a deal".

The game goes like this:

There are three doors, behind each door there is either a goat or an amazing sportcar. The contestant wins if they guess where the car is. There are in total 2 goats and the car. Therefore the initial chance of choosing the car is 1/3.

The host asks the contestant to pick a door. Once a door has been chose, the host, who knows where the goats and the car are, opens a door behind which there is a goat and asks the contestant if they want to switch door.

Now the question is: should the contestant change door or should they stay? Is there any statistical reason which could justify either choice?

It might not be that immediate to understand, however the optimal strategy is to change door. In this case, the probability of winning the car increases from 1/3 to 2/3. You can check this by analysing the favourable scenarios over the possible scenarios. However, should we stick with this or should we make a simulation to test this statement out? Let's go for the simulation with R.

	#----------------------------------------------------------------------
	# Monty Hall game SIMULATION
	# This function generates 3 random doors where 2 are set equal to 0 (the goats) and
	# 1 is set equal to 1 (the car)
	generate_random_3_door <- function()
	{
	sampl = c(0,1)
	probvec = c(.5,.5)
	a <- sample(sampl,1,prob=probvec)
	b <- sample(sampl,1,prob=probvec)
	c <-sample(sampl,1,prob=probvec)
	if((a|b|c==1) &amp; (a&amp;b)!=1 &amp; (b&amp;c)!=1 &amp; (c&amp;a)!=1)
	{
	vect <- c(a,b,c)
	return(vect)
	}else
	{
	return(generate_random_3_doors())
	}
	}
	# Library needed to plot the pie chart
	library(plotrix)
	# This function simulates n games both changing and not changing door
	trialswsw <- function(n)
	{
	# We generate our sample matrix
	first <- c(1,0,0)
	doors <- matrix(first,ncol=3)
	for(j in 1:n){
	doors <- rbind(doors,generate_random_3_doors())
	}
	#View(doors)
	# Choices matrix, choice 1: no change, choice 2: change
	choices <- matrix(ncol=2)
	for(i in 1:nrow(doors))
	{
	# Since there's no reason to pick a particular door, we always choose the first
	choice1 <- doors[i,1]
	# Now Monty gets rid of one of the remaining wrong doors (either number 2 or 3)
	if((doors[i,2] == 0) &amp; (doors[i,3] == 0))
	{
	doors[i,2] <- NA
	}
	else if(doors[i,2] == 0)
	{
	doors[i,2] <- NA
	}else if(doors[i,3] == 0)
	{
	doors[i,3] <- NA
	}
	# We switch door, while being careful not to choose the open one! (NA) ;)
	if(is.na(doors[i,2])){
	choice2 <- doors[i,3]
	}else{
	choice2 <- doors[i,2]
	}
	# This vector contains the choice
	choicevec = c(choice1,choice2)
	# we add each vector (game) to the result matrix
	choices <- rbind(choices,choicevec)
	}
	# A small adjustment, not relevant if n is big enough
	choices[1,]= c(1,1)
	# We calculate relative frequencies
	freqrelchoice1 = sum(choices[,1])/nrow(choices)
	freqrelchoice2 = sum(choices[,2])/nrow(choices)
	data = c(freqrelchoice1,freqrelchoice2)
	labels = c("% win without changing","% win changing")
	names(data) = labels
	# View(scelte)
	pie3D(data,labels=labels,explode=0.1,main="Monty Hall % chance of winning")
	return(data)
	}
	# Let's try our function and go for the simulation
	generate_random_3_doors()
	# we test our random doors generator
	trials(10000) # Please ignore this function which I have integrated in the trialswsw()
	trialswsw(10000) # We simulate 10000 games

view raw monty_hall_simulation.R hosted with ❤ by GitHub

Here are the results:

The simulation confirms that, on the long run, the odds are more favourable if the contestant decides to switch door. Hope this was useful and interesting.
More on the Monty Hall problem: http://en.wikipedia.org/wiki/Monty_Hall_problem