Sometimes you need to give a problem some thought. Tonight, I saw a tweet on the good old Monty Hall problem and decided to do just that. I now understand how it works. I think. Here goes:

There are three doors. Each of the of the doors has 1/3 probability of containing a car (adding up to one). You pick a door. Monty opens one of the other doors, which shows a goat. This leaves two doors: the one you picked, and the ‘other remaining door’. Your options are to stick with your original choice, or to change to the ‘other remaining door’. Frequentist approaches show that you have a better chance of winning the car if you change to the ‘other remaining door’. Why is that?

Explanation from Scientific American podcast:

Your original choice has 1/3 probability of containing a car. The other two doors together have 2/3 probability of containing a car. One of these other two doors is eliminated. That does not change the combined 2/3 probability, but this probability is now transfered onto a single door.

My explanation:

You pick a door, which has either a goat or a car behind it. Either way, one of the other doors will be opened. There are now two possibilities:

(a) if you originally picked a goat, another goat will be eliminated and the other remaining door will show a car;

(b) if you originally picked a car, one of the goats will be eliminated and the other remaining door will show a goat.

Possibility (a) will get you a car; possibility (b) gets you a goat. It is easy to see that possibility (a) (you originally picked a goat) has probability 2/3 and that possibility (b) (you originally picked a car) has a 1/3 probability. This means that “changing” has a 2/3 probability of winning!

Another, more frequentist explanation:

If your strategy is to stay with your original choice:

- You pick goat One; Monty eliminates goat Two; you lose
- You pick goat Two; Monty eliminates goat One; you lose
- You pick the Car; Monty eliminates a goat; you win

–> you win in 1/3 cases

If your strategy is to change doors:

- You pick goat One; Monty eliminates goat Two; you change to the car and win
- You pick goat Two; Monty eliminates goat One; you change to the car and win
- You pick the Car; Monty eliminates a goat; you change to the other goat and lose

–> you win in 2/3 cases

Q.E.D.?

And if you want to have a try, you’re welcome to use my R program:

#

# The Monty Hall problem - "a frequentist approach"

# Jan Verkade

# j.s.verkade@tudelft.nl

# June 2011

#

```
```# Clear all variables from memory

rm(list = objects())

# We store the number of the winning strategy (1 = stick; 2 = change) in a vector

result <- vector("numeric")

# We run the experiment over n cases

n <- 1e3

for (i in 1:n) {

# We create a vector for the three doors

# 0 for a goat

# 1 for a cars

# the car is randomly assigned to either of three doors

doors <- c(0, 0, 0)

doors[sample(1:3, 1)] <- 1

# we choose to open one of three doors, randomly

pick <- sample(1:3, 1)

# Monty has all the information

gts <- which(doors == 0)

car <- which(doors == 1)

# And knows which door to open for you

if ( pick == car ) { opens <- sample(gts, 1) } else { opens <- gts[-which(gts == pick)] }

# This gives you an update on what YOU (not Monty) know; comment out if you're running a lot of cases

# print(doors)

# print(paste("You pick", pick))

# print(paste("Monty opens", opens))

# Strategy 1: you stick with your original choice

# Strategy 2: you change doors

if (pick == car) { result <- c(result, 1) } else { result <- c(result, 2) }

} #i

`print(paste("Strategy 1 (stick with original pick) gave", length(which(result == 1)), "wins"))`

print(paste("Strategy 2 (change to unopened door) gave", length(which(result == 2)), "wins"))

print(paste("Relative frequencies:", sum(result == 1)/length(result), "-", sum(result == 2)/length(result)))

Jan,

You mention there’s a frequentist approach to this… Is there a Bayesian approach that gives another answer?

Tom