Yes, I am making the assumption that,before anyone chooses a door, you flip a coin for each and every door to decide if a goat or a car is behind that door.
I agree my wording is awkward, but it was my way of trying to show some scenarios will show up twice as often as others. If you scratch scenarios 3 and 5, then scenarios 4 and 6 are twice as likely to occur as 1 and 2, and you get the same result.
We don’t know the odds of cars and goats. My earlier post also partially ignored this. We do know that if player 2 is shown to have a goat, you are more likely to not have a goat than box 3. So we don’t know what the winning percentage is, just that in this scenario staying is better than switching.
If we knew the distribution, say goats show up 9 out of 10 times, we could calculate the exact odds (by we I mean someone else, because lazy). But it doesn’t change that staying is better than switching, because of the extra info of p2’s goat, and the rules of its reveal.