## Introduction to The Monty Hall Problem - A Game Show Probability Puzzle

What is the Monty Hall Problem?

The Monty Hall problem is a famous brain teaser named after the original host of the American television show "Let's Make a Deal".

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

You choose door A. The host then opens door C, which has nothing behind it. And asks "would you like to switch to door B or stay with door A?"

The question is, is it to your advantage to stay with your choice, or switch to the other door?

Next time I'll explain the answer, until then I'd love to hear your thoughts in the comments.

If you can't wait, I've also provided a written version of this problem along with various ways to think about the solution at the link in my bio.

In my last Racket I introduced the Monty Hall Problem. you should give that a listen first because I'm assuming you already know the problem and I'll be jumping right into the solution and explanation.

A lot of people's intuition is to think that since there are only two doors left, you have a 50/50 chance so it doesn't matter which option you choose.

This would be wrong.

The correct answer is that you should switch, but why?

We know the host will never reveal the winning door, and will never reveal the door you already chose.

There are also only three possible scenarios. The Prize is behind Door A, Door B, or Door C, it can't be anywhere else.

So let's assume you picked door A and try every scenario.

Scenario 1, the prize is behind door A.

The host can reveal either door B or door C. In either case, if you switch you lose, if you stay, you win.

Scenario 2, the prize is behind door B.

The host can only reveal door C. If you switch you win, if you stay you lose.

Scenario 3, the prize is behind door C.

The host can only reveal door B. If you switch you win, if you stay you lose.

So of the three possible scenarios, only one of them results in a win if you staying while two of them result in a win if you switch. So you have a 2 out of 3 chance of winning if you switch and only a 1 in 3 chance of winning if you stay.

If today's explanation didn't convince you keep an eye out for my next racket where I will explain the Monty Hall solution from a slightly different perspective.

If you find these explainers helpful, please like my rackets or tell me in the comments what you'd like me to explain next.

Last time I explained a solution to the Monty Hall problem that made most sense to me.

There are at least two other explanations I know others have found helpful

The first explanation is to think of your options as two groups.

Group 1 is the door you chose which has a 1 in 3 chance of being correct.

Group 2 is all the other doors combined, which have a 2 in 3 chance of being correct.

When the host reveals one of the doors in group 2, the odds of group 2 don't change. Group 2 still has a 2 in 3 chance of having a winning door, but now you know which of the doors in Group 2 is not a winner. Since there is now only 1 remaining door in Group 2, it alone must have a 2 in 3 chance of being a winner, and therefore you should switch.

The last explanation I personally don't find very useful, but I know others do so here we go.

Let's pretend there are 100 doors instead of just 3. Whatever door you choose has a 1 in a 100 chance of being a winner.

After the host reveals 98 of the other 99 doors to be losers, of course you should switch as there is a 99% chance the winner is behind the door you didn't choose.