Monty Hall presents you with three doors

Door #1

???

Door #2

???

Door #3

???

Behind two doors are goats. Behind the third is a car.

You are asked to choose one of the three doors. You choose door #1.

Monty knows what is behind each door. He opens one of the doors that you did not choose, which has a goat behind it. Lets say he opens door #2 and it reveals the goat.

You are now given the opportunity to choose door #3 or stay with door #1.

What would you do? Is there a best strategy? What are the odds of winning the car if you choose door #3? What are the odds if stay with door #1?

Argument 1:

Some argue that whether we switch or not, we would have a one in three chance of winning the car. Logic dictates that before Monty opens a door, we have one in three chance of choosing the door with the car. Since Monty is going to open a door that we didn’t pick anyway, nothing has changed. As for switching doors, nothing changes because if we chose that door from the beginning we would be in the exact situation we are now. Switching will have no effect on our odds.

Here are the 3 possibilities of what might be behind the doors.

Doors

Possibilities

#1 #2 #3

a Car Goat Goat

b Goat Car Goat

c Goat Goat Car

If we choose door #1 we have one in three chances of picking the car (Look from top to bottom in column 1). It is the same for #2 and #3. Clearly Monty will always open a door we haven’t chosen (In our case door #2). So that has no effect on our odds of winning.

Doors

Possibilities

#1 #3

a Car Goat

b Goat Goat

c Goat Car

We have one in three chances to win the car if we stay with door #1 or switch to door #3.

Argument 2:

Some argue that after Monty opens the door, we now have one in two chances of winning the car. Logic would dictate that after Monty opens the door we clearly have two doors to choose from. A car is behind one and a goat behind the other. Clearly we have one on two chances to win the car. We either have chosen the car, or we didn’t. Switching will have no effect on our odds.

Here are the 3 possibilities of what might be behind the doors.

Doors

Possibilities

#1 #2 #3

a Car Goat Goat

b Goat Car Goat

c Goat Goat Car

We choose door #1. If Monty opens door #2 and reveals a goat now we know that there is no car at door #2. The only logical possibilities are:

Doors

Possibilities

#1 #2 #3

a Car Goat Goat

c Goat Goat Car

If we stay with door #1 or switch to door #3 we have one in two chances to win the car in both cases

Argument 3:

Some might argue that initially, we have two in three chances of choosing a goat. After Monty opens the door, we most likely have chosen a goat (two in three odds) so since we know where the second goat is, we should switch doors as the new door most likely has a car behind it. Switching increases our odds.

Here are the 3 possibilities of what might be behind the doors.

Doors

Possibilities

#1 #2 #3

a Car Goat Goat

b Goat Car Goat

c Goat Goat Car

If we choose door #1 we have two in three chances of picking a goat (Possibility ‘a’ and ‘b’). It is similar for door #2 and door #3. When Monty opens a door to reveal a goat, we lose if we switch in the case of possibility ‘a’. But if we switch when in possibilities ‘b’ and ‘c’ we win. So if we switch doors we have two chances in three to win the car. If we do not switch doors we have one in three chances to win the car.

All these arguments sound logical. Clearly only one can be correct.

Which argument do you believe?

I have created a JavaScript routine to randomly play the scenario 100 times. (See script link below) It will keep track of how many times a car is won if we stay with our first door or if we choose the other door.

A great explanation at https://www.youtube.com/watch?v=mhlc7peGlGg and https://statisticsbyjim.com/fun/monty-hall-problem/