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Explanation of the Monty Hall Dilemma:

You are on a game show, with 3 doors in front of you. You know that behind one door is a new car, and behind each of the other two doors is a goat. You pick one of the doors, and Monty opens one of the other two - a door that he knows has a goat behind it.

Monty asks you whether you would like to change your choice of doors...

Would changing doors affect your chances of winning?

Here are all the possible ways of running the game:

If door 1 wins:

First ChoiceSecond ChoiceWin?Strategy
12 or 3NoChange
11YesStick
22NoStick
21YesChange
33NoStick
31YesChange

If door 2 wins:

First ChoiceSecond ChoiceWin?Strategy
21 or 3NoChange
22YesStick
11NoStick
12YesChange
33NoStick
31YesChange

If door 3 wins:

First ChoiceSecond ChoiceWin?Strategy
31 or 2NoChange
33YesStick
11NoStick
13YesChange
22NoStick
23YesChange

NB: Where I've said 'x or y' in the second choice, I mean whichever of those doors Monty has not opened.

As you can see, there are 18 possible ways of playing the game, 9 of which lead involve changing your choice. The other 9 involve sticking to your choice.

Of the 9 plays where you change, 6 win and 3 lose.
Of the 9 plays where you stick, 3 win and 6 lose.

If you change, you will win 2/3 of the time.
If you stick, you will win only 1/3 of the time.

Changing your choice is therefore the better strategy.