You’re on a game show, and you’re given the choice of three doors: Behind one door is a car and behind the other two doors are goats.

You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?”

**Is it to your advantage to switch your choice?**

This problem has numerous variations and its various forms have stumped many mathematicians.

Source: Monty Hall problem - Wikipedia

Yes, see Solution.

Switching your choice gives you a 2/3 chance of picking the correct door.

Here are the three possible scenarios, all with a probability of 1/3: 1. The player has picked the door with the car. The host has shown one of the two goats. 2. The player originally picked the door hiding Goat A. The game host shows Goat B. 3. The player originally picked the door hiding Goat B. The game shows Goat A.

Not switching your choice means you are giving yourself a 1/3 chance to guess the location of the car. By switching, you effectively ‘open’ 2/3 doors. In other words, there is a 2/3 chance of being wrong initially, and thus a 2/3 chance of being right when changing to the other door.

Alternatively, the Carlton solution:

An intuitive explanation is to reason that a player whose strategy is to switch loses if and only if they initially pick the car; that happens with probability 1/3, so switching must win with probability 2/3 (Carlton 2005).

Simply put, if the contestant picks a goat (to which two of the three doors lead) they will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (to which one door leads) they will not win the car by switching. So, if the contestant switches, they will win the car if they originally picked a goat and they did not win if they originally picked the car. As they have a 2 in 3 chance of originally picking a goat, they have a 2 in 3 chance of winning by switching.

## Comments

Sign uporlog in with Facebookto comment.SurajdubeyYes.. I will.

Jay ElliottThis can definitely be solved using Baye's Rule, but it may be easy to get lost in the math. The Carlton solution in the

Show Solutionbox is definitely the easiest to understand.JangoTask

JangoTask

GlennAnother intuitive way to solve this is to reimagine the problem as having 1000 doors instead of 3... and the host opens 998 doors leaving your door and one other. To assume that you then have a 50/50 chance of sticking with your pick and finding the car on your first pick would also basically assume that you are magic. Your intuition should tell you that the host avoided opening the OTHER door for a reason.

Tarun245yes

GeneThis problem was in Marilyn vos Savant's PARADE Magazine column about 20 or so years ago. She got dozens of abusive letters from math PhD's and other pointy-headed academia telling her how wrong she was. She published some of them, including their names. Hilarious!

Ankit GuptaMore interesting problem and most beautiful solution... Only switching from one door to another gives the high probability to get a success... No other option...

Tarun245It actually, believe it or not, increases your chance of winning from 1/3 to 2/3.