主题：【文摘】一道有趣的概率题 -- 天下第一银杏树

My only assumption is that the door you picked at your first try is random. It doesn't matter how the hostess picked her door (as long as it revealed a goat). Even in the second case of your understanding, my derivation would still hold.

Some webpages have discussed this problem, and have mentioned that if the hostess had chosen her door randomly, which happened to reveal a goat, then switching door wouldn't enhance your chance of getting the car. I suspect you might have been influenced by those webpages, and I don't think that idea is correct.

In my Monte Carlo simulation, actually the hostess also chose the door randomly, but the events in which the door revealed a car were thrown away, as they violate the constraint that the door has to reveal a goat.

Intuitively, we can understand this problem as follows. Let event Ai = {you picked door #i}, and event Gj = {door #j, j!=i, reveals a goat}. We can verify Ai and Gj are not independent, i.e., they contain nonzero mutual information. So it's understandable that given this new piece of information Gj, your strategy can improve.

Anyhow, if my previous analysis doesn't convince you, you're welcome to write a small program to verify it--your case 2 specifically.