The Monty Hall Problem

What is the mathematical explanation for the fact that switching is a better strategy than staying with the originally chosen door in the Let's Make a Deal scenario?

Using probability in a situation like this, where there are a finite number of possibilities with equal likelihoods, basically amounts to counting how many different ways there are to do something and in how many of those situations you get the desired outcome. The probability of winning is then the number of ways to get the desired outcome divided by the total number of possible scenarios!

Let's consider two people playing the game: Mr. Steed (who always stays with his first choice) and Mr. Switzer (who always switches after the goat is revealed). Neither of them knows that the car is really behind door number one. So, let's compute the probabilities that they will will by looking at how many possibilities there are and in how many of those they will win.

Mr. Steed has three choices; he can pick door number one, two or three. So, the number of possibilities (and therefore the denominator of the probability that he will win) is 3. To get the numerator we need to work out when he will win. If he picks door number one, he will win because the car is behind it and he is going to stay with his first choice. (He's a very confident guy!) However, if he picks two or three as his door he will go home with a goat. So, his probability of winning is 1/3. This means that if he plays with his strategy for a long time, he'll win just about 33 percent of the time.

On the other hand, Mr. Switzer has a different probability of winning! Like Mr. Steed, he has three choices: door one, two or three. And so, again, the denominator of his likelihood of winning is going to be the same number 3 as before. But, if Mr. Switzer picks door number one he will not get the car. After he picks door number one, Monty will reveal a goat behind one of the two other doors (it doesn't matter which one) and Mr. Switzer (true to his name) will switch to another door...he will switch away from the car. (Sounds bad, right?) But wait, if Mr. Switzer picks door number two, then Monty will show him the goat behind door three and Mr. Switzer will switch to door number one and get the car! So, he will win if he picks door number two. In fact, he'll also win the car if he picks door number three for the same reason (Monty will open door two and Switzer will switch to door one). So, Mr. Switzer's chances of winning are 2/3. He will win about 66 percent of the time...twice as often as Mr. Steed.

Yes, this might seem counter-intuitive. But, it makes sense because Monty is revealing information to you about the two doors that you do not pick, but gives you no information about the door you picked. Essentially, when the door with the goat is revealed, the 1/3 probability that the car was behind that door all gets transfered to the door that you can switch to since Monty only had the choice of opening one of those two and never opens the door that the contestant picked.

Still don't believe me? Try playing the game yourself. Try each of the strategies at least 10 times and see what happens! Or, look at the statistics so far for all of the times the game has been played on this webpage.


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Page written and maintained by Alex Kasman (2005)