Monty Hall Explanation

The Monty Hall Problem is Very Counterintuitive

• If you still don’t believe it, you are in good company:

Simplest Explanation

• Your first choice has a 1/3 chance of having the car
• So there’s a 2/3 it is behind one of the other doors

• Monty will never open the door with the car behind it
• After he opens a door, the entire 2/3 probability shifts to the other closed door

Monty is Key!

• Imagine you chose door 1, and you are asked whether you want the prize behind door 1 or the better of the prizes behind doors 2 and 3

• This is exactly what you get when you switch (because Monty always removes the worse prize of the two when he opens a door)!

Another Appeal to Intuition

• Imagine there are 100 doors, 99 goats, and 1 car
• You pick a door, so you have a 1/100 chance of winning the car
• Then Monty opens 98 of the doors revealing 98 of the goats

• There is still one other unopened door: Should you switch?

• If you pick a goat on your first choice (99/100), you will always get the car if you switch

• By switching, you only get a goat if you had the car at first, and that’s super rare (1/100)

More Involved (and More Satisfying) Explanation

Bayes’ Rule

• OK, so you chose door 1 and then monty opened door 3

• What are the chances that the car is behind door 2 (i.e., should you switch)?

• We are interested in \(p(car2 \mid open3)\)

• From above, we already have the flip of it, but this is not the same: \(p(open3 \mid car2)\)

• We can turn it around using Bayes’ rule, which states that

\[ p(A\mid B)=\frac{p(A)p(B\mid A)}{p(B)} \]

In our case,

\[ p(car2 \mid open3)=\frac{p(car2)p(open3 \mid car2)}{p(open3)} \]

• We already have the two pieces in the numerator (from above):

\[ p(car2 \mid open3)=\frac{(\frac 13 )\left(1\right)}{p(open3)} \]

• We just need the denominator (the probability Monty opens door 3)
• We can get this by adding together the probabilities of all the different situations in which Monty opens door 3 (this is the Law of Total Probability):

\[ \begin{aligned} p(open3)&=p(open3\ \& \ car1)+p(open3\ \& \ car2)+p(open3\ \& \ car3)\\ &=p(car1)p(open3 \mid car1)+p(car2)p(open3 \mid car2)+p(car3)p(open3 \mid car3)\\ &=(\frac13)(\frac 12)+(\frac13)(1)+(\frac13)(0)\\ &=(\frac16)+(\frac13)+(0)\\ &=\frac 12 \end{aligned} \]

• Now we have the denominator, so

\[ p(car2 \mid open3)=\frac{(\frac 13 )\left(1\right)}{(\frac12)}=2/3 \]

• This tells us that, if we picked door 1 and Monty opened door 3, then there is a \(\frac23\) chance that the car is behind door 2

• That means there is only a \(\frac13\) chance the car is behind door 1 (let’s use Bayes’ rule again)

\[ p(car1 \mid open3)=\frac{(\frac 13 )(\frac 12)}{(\frac12)}=1/3 \]

• Thus, we would be much better off switching!

• This will be true no matter which door you pick initially!

# The probability the car is behind door 1, door 2, and door 3 is initially 1/3

car1 <- 1/3
car2 <- 1/3
car3 <- 1/3

# Say you choose door 1

# Monty will never open the door with the car behind it (!)
# What is the probability Monty opens door 3, given your choice (door 1) and the actual location of the car?

open3_car1 <- 1/2 #if the car was behind 1 (your door), he could open either goat door 2 or 3 
open3_car2 <- 1 #if you chose 1, and the car's behind 2, he has to open 3
open3_car3 <- 0 #if you chose 1, and the car is behind 3, he cannot open 3

## Monty is CERTAIN to open door 3 if the car is behind door 2
## Monty is only 50:50 to open door 3 if the car is behind door 1
## Monty will NEVER open door 3 if the car is behind it

## Say Monty opens door 3, revealing a goat.

## What is the probability that the car is behind door 1, given door 3 was opened? 
## (What is the probability your initial choice was correct)?

car1 * open3_car1 / (car1*open3_car1 + car2*open3_car2 + car3*open3_car3)

## [1] 0.3333333

## What is the probability that the car is behind door 2, given door 3 was opened?
## (What is the probability that the other unopened door has the car?)

car2 * open3_car2 / (car1*open3_car1+car2*open3_car2+car3*open3_car3)

## [1] 0.6666667

We can simulate this in R too (you can edit/run the code below):

#randomly assign cars and goats

doors <- sample(c("goat","goat","car"))

#you pick door 1 (below you can change it to whatever you want)!

yours <- 1

others <- doors[-yours]
  
if(others[1]=="goat" & others[2]=="goat"){
  open <- sample(c(1,2),size=1)
} else if(others[1]=="goat" & others[2]!="goat"){
  open <- 1
} else {
  open <- 2
}

open=ifelse(yours==1,open+1,ifelse(yours==3,open,ifelse(yours==2,floor(open*3/2))))

print(paste("Monty opened door",open,"revealing a",doors[open]))

## [1] "Monty opened door 3 revealing a goat"

print(paste("If you kept your original door:", doors[1]))

## [1] "If you kept your original door: goat"

print(paste("If you switched:", doors[!c(1,2,3)%in%c(yours,open)]))

## [1] "If you switched: car"

######################################

##Simulate the game 5000 times

results <- replicate(n = 5000, {
  
  doors <- sample(c("goat","goat","car"))
  yours <- 2
  others <- doors[-yours]
  
  if(others[1]=="goat" & others[2]=="goat"){
    open <- sample(c(1,2),size=1)
  } else if(others[1]=="goat" & others[2]!="goat"){
    open <- 1
  } else open <- 2

open <- ifelse(yours==1,open+1,ifelse(yours==3,open,ifelse(yours==2,floor(open*3/2))))

return(c(stay=doors[yours], switch=doors[!c(1,2,3)%in%c(yours,open)]))
})

#full results
head(t(results))

##      stay   switch
## [1,] "goat" "car" 
## [2,] "car"  "goat"
## [3,] "goat" "car" 
## [4,] "car"  "goat"
## [5,] "goat" "car" 
## [6,] "goat" "car"

#proportions if you stay
table(t(results)[,1])/5000

## 
##    car   goat 
## 0.3322 0.6678

#proportions if you switch
table(t(results)[,2])/5000

## 
##    car   goat 
## 0.6678 0.3322