# A hint for understanding the Monty Hall problem

When I first heard of the Monty Hall problem I didn’t understand why switching was the right idea. You’ve heard of the Monty Hall problem, right? It goes something like this:

You’re on a game show called Let’s Make A Deal. Monty Hall, the host, gestures to three doors. Behind one is a car. Behind the other two are goats. You’d rather have a car than a goat. You choose a door, but don’t open it yet. Monty then opens another door, revealing a goat. He then offers you a chance to switch your choice to the other unopened door. Should you switch?

The correct answer is “you should switch”, as your first guess has a ⅓ chance of having a car behind it whereas, after switching, your new guess has a ⅔ chance of being right.

I didn’t buy this at first.

One of the standard ways to help someone see why the right answer is correct is to reformulate the problem with not just three doors, but, say, 100 doors:

Monty gestures to a corridor of 100 doors. He says “pick a door”. You pick the first door. Monty then opens 98 doors with goats behind them, leaving just two doors: the one you picked but didn’t open yet, and this one other door. Should you switch?

Now it should be more obvious that switching is the better idea, but…it just wasn’t to me. What helped me understand that it’s better to switch was this insight:

Monty is duty-bound to never, ever reveal a car.

After realizing this, I realized that he’s giving you lots of useful information by opening doors with goats behind them. In the 100-door case, he’s reducing the number of doors that *might* have a car behind them from 100 (before you choose a door) to two. Since the odds are only 1 in 100 that your first door had a car behind it, but the odds are *definitely* 1 in 2 that the other unopened door has a car behind it, you should switch.