Consider the following problem

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, 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?

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(Formulation by Marilyn vos Savant at Whitaker, Craig F. (1990).)

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So what is the answer? Let’s see a simple solution. The following table shows the three possible arrangements of two goats and one car behind three closed doors and the result of switching or staying after initially picking Door 1

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Door 1 |
Door 2 |
Door 3 |
Outcome if switching |
Outcome if staying |

Car |
Goat |
Goat |
Goat |
Car |

Goat |
Car |
Goat |
Car |
Goat |

Goat |
Goat |
Car |
Car |
Goat |

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Which means that in 2 out of 3 cases we would win the car by switching doors. Hence it would make perfect sense to always switch. To understand why this is happening do take into consideration that the event “staying” is composed by only one element while its complement by two (note that all elements are equiprobable).

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Frequency of winning when staying/switching doors in a sequence of "Monty-Hall" games

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Something’s not right with the table? It refers to “If switching” or ” if staying”, which means that the table applies to an outcome prediction after door number 3 has revealed the goat. The third line of the table, in which door no. 3 would have a car, no longer applies, therefore the chances at this point remain 50-50 and only the two first lines of the table should still be active. Switching at this point or not switching, after one of the three doors no longer holds a mystery, the chances are just 50-50? What did I miss?

In the third case the presenter will open the door number 2. That is because he knows where the goat is! If he opens door number 3 he would prematurely end the game…

The following sentence implies that he always opens the door with the goat:

“opens another door, say No. 3, which has a goat”

(I now realize that this was probably not too clear)

You’re right! I misinterpreted the word “say” as referring to this one case, and the table as referring to chances after one door was opened.

Now, it makes sense!

I am glad it now makes sense

My apologies for confusing you…

This is counter intuitive at first thought but it makes perfect sense if you think about it. Another way to think about this is to say the following:

When you select a door the first time you have 2/3 chances of being wrong.

By eliminating the “bad” of the other doors and offering you the possibility to swap your selection, I am essentially offering you those 2/3 chances at that point. This is the “mathematically challenged” person’s proof I recon 😉

You are right manblogg. This problem is known to be counter-intuitive. Try it on your friends and you will certainly confirm that.

“I am essentially offering you those 2/3 chances at that point”

Yeah. That is what I meant by saying that the event “staying” is composed by only one element while its complement by two. It is quite tricky to realize what is really at stake

There is also a way to derive a formal answer by using the Bayes’ theorem but I was too lazy to do it… 😀

Thanks for the question AND the explanation 😉 I am now going to use this as a trick up my sleeve.. 😀

Well, I was planning to just guess the answer based on my gut feeling rather than use graphs! Ouch!

Hi. I am glad you liked the problem. I am sure you can trick many people with this 😉 But bear in mind that this problem is so counter-intuitive that some people won’t want to accept your explanation…

Can you please explain to me how to make the graph that you have made above?

Hello Nelao. There are many ways to make this graph such as the following (in R):

n <- 1000 #sample size

plot(cumsum(apply(replicate(n,sample(c(0,0,1),2)),2,max))/(1:n), type="l")

# frequencies when you switch