Monty Hall Problem: Win by Switching

[zhentil]

There is no such thing as a probability of an event that has already occurred. Care to rephrase?

Schrodinger's cat does not appreciate your jest.

 

[MisterX]

Dave, I have modified your list to include probabilities. Some of these are conditional probabilities; for "1.1 Monty opens B (1)", (1) means this would always happen given that "1 You choose A"
There is an equation for the probability of each outcome. Winning outcomes are in bold.

1 You choose A (1/3)
1.1 Monty opens B (1)
1.1.1 You stay with A (1), You lose, probability = 1 * 1 * (1/3) = 1/3
1.1.2 You switch to C (1), You win, probability = 1 * 1 * (1/3) = 1/3

2 You choose B (1/3)
2.1 Monty opens A (1)
2.1.1 You stay with B (1), You lose, probability = 1 * 1 * (1/3) = 1/3
2.1.2 You switch to C (1), You win, probability = 1 * 1 * (1/3) = 1/3

3 You choose C (1/3)
3.1 Monty opens A (1/2)
3.1.1 You stay with C (1), You win, probability = 1 * (1/2) * (1/3) = 1/6
3.1.2 You switch to B (1), You lose, win probability = 1 * (1/2) * (1/3) = 1/6
or
3.2 Monty opens B (1/2)
3.2.1 You stay with C (1), You win, probability = 1 * (1/2) * (1/3) = 1/6
3.2.2 You switch to A (1), You lose, probability = 1 * (1/2) * (1/3) = 1/6

So, the total probability of winning with the strategy of staying is 1/6 + 1/6 = 1/3.
The total probability of winning with the switch strategy is 1/3 + 1/3 = 2/3.

 

[Dr. Seafood]

^ Read the second post at the top of this page and was going to post exactly what you did. Thank you.