# Simulating the Monty Hall Problem

Back in 1990, the following question was posed toÂ PARADE magazine columnist,Â Marilyn Vos Savant (once holder of theÂ GuinnessÂ World Record for highest IQ):

Suppose youâ€™re on a game show, and youâ€™re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows whatâ€™s behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?

Marilyn’s response: “Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance.” And thus began a firestorm ofÂ controversy. Over a thousand people with PhDs (many of them statisticians and mathematicians) wrote in to her column to tell her she was wrong. Wikipedia has a nice summary of how it went down. AndÂ Marylin’s own site has a nice sample of some of the thousands of letters she received. In the end, though, she was right.

If you want to know why she was right, visit the links above. You won’t believe how many ways there are to analyze this seemingly simple problem in probability. What I want to do in this post is simulate it using R. If what she said was right, I should win 2/3 of the time.

So here’s what we do. First define a variable to store the number of simulations and a vector to store the results Â of each simulation.

# number of simulations
n <- 10000
# allocate memory to store results
x <- vector(mode = "integer", length = n)

Now we do the simulation. Picking the first door can be thought of as a single binomial trial with p = 1/3. We either pick the car (1) or we don't (0). We have a 2/3 chance of not picking the car. After we pick our door, we know another door will be opened to reveal a goat. That happens every time. That means we'll either switch to a goat or to a car. Hence the "if" statement: if we picked the car (1), switch to goat (0); else switch to car(1).

# simulation
for (j in 1:n){
# pick a door; either goat=0 (p=2/3) or car=1 (p=1/3)
x[j] <- rbinom(1,1,prob=(1/3))
# switch; if 1, switch to 0, and vice versa
if (x[j] == 1) {
x[j] <- 0
} else x[j] <- 1
}

When the simulation is done, calculate the percentage of times you switched to a car (1):

sum(x)/n # percentage of time you get car

When I ran this I got 0.6667, which is in excellent agreement with the theoretical answer of 2/3. Yay winning cars on game shows!

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