Now that it is morning and I have had some coffee instead of evening beverages.
This is what I meant by Monty can't take a win away.
(2) Monty always opens a door concealing a goat
I mixed that rule up a bit in some of my posts. My apologies for that confusion.
For example you pick door #1
1. Door #1 has the car, monty shows you a goat in either #2 or #3.
Stay = win (outcome 1,2)
Switch =lose (outcome 3,4)
2. Door #2 has the car, monty shows you a goat in #3
Stay = lose (outcome 5)
Switch = win (outcome 6)
3. Door #3 has the car, Monty shows you a goat in #2
Stay = lose (outcome 7)
Switch = win (outcome 8)
You list 6 outcomes. If we separate your 1. into "monty shows you a goat in #2" and "monty shows you a goat in #3", then we would have 8 outcomes, 4 winners and 4 losers evenly split on stay and switch. You can't account for all possible outcomes by ignoring some.
Here is another representation you gave which again falsely combines the 2 possibilities and that is why it does not come away with the right result.
Assuming pick of door 1. Line 1 in that image: car, goat, goat. Monty eliminates door 2:goat, stay wins, change loses. Monty eliminates door 3:goat, stay wins, change loses. Line 2, Monty can only eliminate door 3, no changes to this line, stay loses and change wins. Line 3, Monty can only eliminate door 2 no changes to this line, stay loses and change wins. Now count: winners on stay 2, losers on stay 2, winners on change 2, losers on change 2.
I've looked at your matrix, and there are too many possibilities listed.
The first sheet leaves out your rule 2 (above) so that it can clearly be seen to cover all possible combinations. The second sheet applies that rule by removing lines that would violate the rule.
If you would like I can write out the math to show that 3=4, and I will tell you it is wrong and it uses the same method of falsely ignoring some outcomes to reach that conclusion.