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you are close but not correct

The problem is your proof 3a.

"The winning door will be the selected door 1/3 of the time. So there is only a 2/3 chance that the winning door is behind the two non-selected doors."

Absolutely correct.

"In the random selection the odds are 1/2 that the winning door is revealed so the odds of randomly selecting the winning door is 1/2 * 1/3 = 1/6"

Sorry you are off by a factor of 2.

Probability calculations are really just counting.

There are 2 different doors that your 1/6 probability applies to. You need to count both possibilities.

The situation is very clean.

If random selection is made, the odds are 1/3 that you've picked the right door, 1/3 that the prize will be revealed in the 2nd round, and 1/3 that you will pass the 2nd round and it will be in the other door than which you have selected.

I did not want to do this but here is the complete list of possible outcomes:

Given doors X, Y, Z- you pick X

1. prize behind X- Monty reveals Y- you win by sticking
2. prize behind X- Monty reveals Z- you win by sticking
3. prize behind Y- Monty reveals Y- game ends you lose
4. prize behind Y- Monty reveals Z- you win by switching
5. prize behind Z- Monty reveals Y- you win by switching
6. prize behind Z- Monty reveals Z- game ends you lose

This is the complete set and they are equally likely given a random pick. If you count them up you will see the odds are as I have said above.

There is a similar but different count if Monty specifically chooses to reveal the losing door. Then, as you know, the 1 of 3 cases where you lose in the 2nd round become 1 of 3 cases where you also win by switching, giving 2/3 in this category.


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