## What is the Monty Hall Problem?

A game show host presents you with three closed doors. Behind one of the doors is a prize, behind the other two there is nothing.

You get to choose one of the doors, if you guess the right door, you win the prize. If you guess incorrectly, you lose and win nothing.

You guess door A.

Before telling you whether you've won, the host reveals one of the other doors to show that it's a loser.

The host now gives you the option to keep your selection or change it. What do you do?

## The Answer

I will explain things below, but first let's jump straight to the answer.

Most people's intuition is to say "well, there's a 50/50 chance between the remaining two doors, so there's no difference if I keep my selection or switch it. This would be false.

**The right answer is to switch your choice. **

It sounds counterintuitive, but it's true and I'll show you why below.

## The Explanation

The simplest way I've found to show why switching is the right approach is by breaking down the options.

Before you start there are three potential scenarios. The prize can only be behind one of the three doors.

We also know that the host will never reveal the winning door and will never reveal your door.

Let's assume you picked door A, the the scenarios would play out as follows:

### Scenario 1: The prize is behind door A

You picked door A, so the host can reveal either door B or door C to be losers. In either case the options play out as:

- Switch = Lose
- Stay = Win

### Scenario 2: The prize is behind door B

You picked door A, so the host can only reveal door C to be a loser. So your choices are:

- Switch = Win
- Stay = Lose

### Scenario 3: The prize is behind door C

You picked door A, so the host can only reveal door B to be a loser. So your choices are:

- Switch = Win
- Stay = Lose

As you can see, of the three scenarios, staying loses twice and wins once. Switching wins twice and loses once. So your best bet is to switch.

## Another way of looking at it

Some folks like to explain the Monty Hall solution with probability.

When you begin, there is a 1 in 3 chance (or roughly 33%) that the prize is behind the door that you choose. And therefore a 2 in 3 chance (roughly 66%) that the prize is behind one of the other two doors.

Don't think of them anymore as 3 individual doors, rather two options.

- Option 1: your selection with 33% chance
- Option 2: the other doors with 66% chance.

So when the host reveals that one of the other doors is a losing door, that doesn't change the probability of Option 2 having 66% chance. The probability doesn't reset just because you have new information. What it does is tell you which of the other doors it is not. So now you know all of that 66% is with the one remaining door.

## One further perspective

Some folks find it easier to understand the probability example by increasing the number of doors. Instead of 3 doors, imagine there are 100.

Before you start, each door has a 1% chance of being the winning door.

You pick door 1, which has a 1% chance.

The rest of the doors as a whole have a 99% chance.

If the host revealed 98 of the 99 doors as losers, the last remaining door has a 99% chance of being the right choice compared to your selection that always has a 1% chance.