Earlier quoted context omitted.
> His purposeful behavior could also be random. It doesn't matter. Because the problem explains what happens: he opens a door and reveals a goat. That is crucial information. There's now only one goat and one car left. But the choices have not been shuffled: you're definitely still pointing at your original choice. What were the odds that your original choice is a goat? Two in three. If you picked a goat, what is gua…
It's actually counter-intuitive. I was going to argue on your side, and then I wrote up a quick program that proved me wrong[0]. Let's go through the scenarios, with Goat A, Goat B, and Car C. In the scenario where Monty picks a door purposefully, always selecting a goat, the scenarios are: You picked A, Monty showed you B, and you switch to get C. You picked B, Monty showed you A, and you switch to get C. You picked…
If Monty always shows a goat, it is undeniably a 2/3 chance to win on switch.
The nuance here being discussed is whether you can assume Monty would have shown you the goat had you chosen a different initial door just because he showed you one this time. If you don’t know that, then you don’t learn anything.
Edit: actually I misread. You just chose to ignore the cases where Monty revealed a car. Which is correct although most people chalk that up as a win or loss.