A Visual Explanation and Simulation of the Monty Hall Problem

Or, trying to understand this nonsense made my head hurt, so I made some pictures.

Not so long ago, my brother introduced me to the Monty Hall Problem, a fairly well-known probability puzzle. The answer, while simple, is maddeningly counterintuitive. What began as a "How can this be?!" Googling quest eventually developed into a feeling of "I guess I get it... but I want to actually see it to believe it."

It Goes Like This...

You're on a gameshow and there are three doors.
There is a prize behind one of them, and nothing behind the others.*

First, choose a door.

Next, the host reveals one of the empty doors and offers you the chance to switch.
Should you switch doors, keep your current choice, or does it not matter?

Switch Doors
Keep Original
Doesn't Matter

Yep, switching gives you a 2/3 chance of winning.

Nope, you'll only win 1/3 of the time keeping your original choice.

I know, seems like it should be fifty-fifty, but it's not.

*Note, the original posing of the question used a car and two goats, but who knows, maybe you really want a goat. They are surprisingly adorable).

When first presented with this problem, my answer was that switching doesn't matter. When one door is revealed, there are two left and they should each have a fifty-fifty chance of holding the prize. This, however, is wrong. If you don't switch, you will only win 1/3 of the time. If you do switch, you will win 2/3 of the time.

A quick look at the Wikipedia page will show you that there are number of explanations to the problem, and people can get a bit crazy with proposed conditions and assumptions. In my search, I was looking for the simplest answer.

A Common, Simple, yet Maddening, Explanation

Initially, all doors have a one in three chance of containing the prize. After you select the first door, there is now a one in three chance that the prize is behind your door, and a two in three chance that the prize is behind one of the other two doors (because they each have their own one in three chance). When the host eliminates a door, most people will come to the conclusion that there are two doors left, each with an equal 1/2 chance. However, this is not the case. The eliminated door's 1/3 chance is added only to the door you do not have selected. The remaining door has a 2/3 chance of containing the prize.

What We Think is Happening

The eliminated door's probability is equally split between the remaining doors.

1/32

1/32

1/3

What is Actually Happening

The eliminated door's probability is added to the door you do not have selected.

1/3

12/3

1/3

While simple, I find it hard to wrap one's head around this transfer of probability. It just feels wrong. Instead, I find it helpful to look at all of the possible outcomes of your initial selection.

Go ahead, choose a door again.

Cool.

Now there are three possible locations for the prize.

First, the host must eliminate a door.

Now you choose to keep your original selection or switch doors.

One out of three times, when you have chosen the correct door on your first guess, the host can choose either of the two remaining doors to reveal.

The other two out of three times, the host cannot reveal the prize, and cannot reveal your door, so he has only one door that can be opened.

Looking closely, you'll see that keeping your original selection only wins one out of three times. If you switch, you'll win two out of three.

Reveal One of Them
Reveal Those Doors
Keep Original
Switch Doors

But, there are two doors left, how are your odds of winning not one in two, damn it?!

I find it's hard to get over the fact that with two doors left, your odds aren't fifty-fifty. But, we can add an additional participant to resolve this.

Let's say a friend enters the game after the host has opened one of the doors, but she does not know which door you have selected. All she knows is that there are two remaining doors, one of which has a prize. She has a fifty-fifty chance at choosing the winning door.

The Simulation

After all that, I still have the desire to just see it happen. I've setup a simulation with three players:

Player 1. Win Rate: 66.6667%
Always switches to remaining door after the host eliminates one

Player 2. Win Rate: 33.3333%
Always keeps original selection

Player 3. Win Rate: 50%
Randomly chooses between the remaining two doors after the host eliminates one

When you run the simulator, the win rates will eventually converge to those listed above. It's possible in some cases it could take quite a few iterations if a random winning streak throws things off. Let it run slowly at first so you can follow what's happening, but then I suggest toggling the faster speed.

The simulator will rapidly run through the following steps:

Step 1: Player 1 & Player 2 will randomly choose the same door

Step 2: The host reveals one of the remaining doors

Step 3: Player 1 switches his selection.

Step 4: Player 3 randomly chooses one of the remaining two doors

Step 5: The winner(s) are revealed and their scores are tallied below

Simulator Running...

OK, Next »
Simulation Speed:
Slower
Faster

Player 1

Wins

0

Rounds

0

Win Rate

0%

Player 2

Wins

0

Rounds

0

Win Rate

0%

Player 3

Wins

0

Rounds

0

Win Rate

0%