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Note: I was brought here by the 'wonderful' community of Philosophy.

I realise the title 'By Changing My Option, How Does This Double My Chance of Winning?' was a little vague so let me go into detail.

Problem:

I'm on a gameshow. The gameshow host presents me with three doors. Door A, B and C. Behind two of the doors are angry goats and behind one of the doors is a new car. I choose door A. The gameshow host eliminates door C. Behind it is an angry goat. I am left with two doors: One has a new car and the other has an angry goat.

Question:

By changing my option to door B instead of A, how does my chance of winning increase by 50%?

Take note that I'm just wondering why this is. Thanks.

J. M. ain't a mathematician
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3 Answers3

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This is the Monty Hall problem: https://en.wikipedia.org/wiki/Monty_Hall_problem.

The solution that changing increases your chance of winning is more obvious if you consider the same problem, but with more doors:

You have 1000 doors behind which are 999 goats and 1 car. You choose a door, and then the host shows 998 doors of his choice which have a goat behind it.

Now he gives you the choice of either selecting the door which is left, or keeping your original door.

The chances of selecting the car when you selected the first door is $\frac{1}{1000}$.

What the host is actually doing is giving you the chance of changing your probability of winning into $1-\frac{1}{1000}$ if you select the other door as the other 999 doors had a much better chance of hiding the car than the one you originally selected*.

Now consider the original problem and see how this relates to it.

*Remember that when the host selects the 998 doors (or 1 door in the original problem), he is not selecting them completely randomly - your choice that you originally made limits his choice of doors too, and he is not allowed to select the door with the car behind it.

Shuri2060
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You may see that sticking to your original choice does not increase your winning probability: Because the host can and does always exhibit a bad door, no matter what is behind your first choice, his action gives you no information about your first choice door. As the winning probability of your first choice thus remains at $\frac13$, the remaining $\frac 23$ must somehow distribute among the other doors. As one of the doors is obviously not a winning door, the full $\frac 23$ must to the other door.

Hagen von Eitzen
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If you changed your option with the door still closed then nothing would change for your chance of winning and it would still be possible for you to choose everyone of the three door. If instead the door has been open and it happens to be a door with the goat then you don't have all the choice you had before (3), you have less choices (only 2) and one bad choice is now impossible therefore it is clear that you are in a better position.

Marco Disce
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