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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 Choice	Second Choice	Win?	Strategy
1	2 or 3	No	Change
1	1	Yes	Stick
2	2	No	Stick
2	1	Yes	Change
3	3	No	Stick
3	1	Yes	Change

If door 2 wins:

First Choice	Second Choice	Win?	Strategy
2	1 or 3	No	Change
2	2	Yes	Stick
1	1	No	Stick
1	2	Yes	Change
3	3	No	Stick
3	1	Yes	Change

If door 3 wins:

First Choice	Second Choice	Win?	Strategy
3	1 or 2	No	Change
3	3	Yes	Stick
1	1	No	Stick
1	3	Yes	Change
2	2	No	Stick
2	3	Yes	Change

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.