Math - Probability

The Monty Hall Problem Explained Simply

Behind three doors are a car and two goats. You pick one door. The host, who knows where the car is, then opens a different door revealing a goat. Now you may stay with your door - or switch to the last remaining door.

Stay: 33% winning chance  |  Switch: 66% winning chance Switching doubles your chance of winning the car!

đź’¬ Mia asks Grandpa Theo

Mia

But there are only two doors left! Isn't it just 50-50 then, whether I switch or not?

Grandpa Theo

That's what almost everyone thinks at first! But the host doesn't open the door randomly - he always knows where the car is, and is guaranteed to open a goat door. That changes the probabilities.

Mia

Explain that to me in detail!

Grandpa Theo

With your first pick, you have a 1/3 chance of standing on the car and a 2/3 chance of standing on a goat. That 2/3 chance of "standing on a goat" doesn't disappear - it moves entirely to the last remaining door, because the host has already removed the other goat door!

Your pick 1/3 Switch door 2/3

The "vanished" goat-door probability lands entirely on the switch door - which is why it's 2/3 instead of 1/2.

Try it yourself

If you don't believe it: play it 20 times with a friend and three cups (a candy under one). Stay ten times, switch ten times - switching wins noticeably more often!

The problem is named after the US game show host Monty Hall, whose show "Let's Make a Deal" featured exactly this game. It is also widely known as the "goat problem."

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