| They are two different problems. Map out all the scenarios exhaustively and you'll find the difference. In both cases we were originally 1/3 chance of being right. That is not in dispute. In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car. In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that: 1. We picked the car and saw Alice 2. We picked the car and saw Bob 3. We picked Alice and saw Bob 4. We picked Bob and saw Alice Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others. This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box. [0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox |