If it says he opens a door, and it has a goat, then you’ve learned nothing about the remaining two.
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 guaranteed to be behind the other door? A car.
So what are the odds that behind the other door is a car? Two in three.
You should switch.
If that were the case then switching is a guaranteed loss. Simply knowing that he opened a door and showed a goat does not mean he would have done this regardless of your choice.
Not a probability problem without this info.
What do you mean?
It's specified in the scenario. He shows you a goat. It's right there. This isn't a variable. It's a fact.
Your only job is to work out whether, given the scenario described, it makes sense to switch. Given all your possible choices.
That’s why this problem is kind of bad. It does not describe the behavior of the host. It describes the perspective of a contestant halfway through the game.
Dumb example. Host flips a coin 9 times in a row and lands heads each time. If you believe this to have happened by chance then it’s probably rigged because that’s insane and it’ll likely be heads again.
If it’s ALWAYS heads then you haven’t learned anything.
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 C, Monty showed you either A or B, and you switch to get the other goat.
If Monty is choosing randomly, we have the following scenarios:
Initial Choice | Monty's choice | Remaining Door
A | B | C
A | C | B
B | A | C
B | C | A
C | A | B
C | B | A
Initial Choice | Monty's choice | Remaining Door
A | B | C
B | A | C
C | A | B
C | B | A
I'm not great with probabilities, but the major difference I can see is that in the first scenario, if you pick the car, Monty will either show you the goat A or B with equal probability as a part of the same 1/3 scenario. So you have really:
1/3: You picked A, Monty showed you B, and you switch to get C.
1/3: You picked B, Monty showed you A, and you switch to get C.
1/6: You picked C, Monty showed you A, and you switch to get B.
1/6: You picked C, Monty showed you B, and you switch to get A.
[0]: https://gist.github.com/Taywee/2ba202b1bf7af40293ecffb01c2ab...
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.
It's always "You choose a door, the host reveals a goat behind another door. Do you switch?" The scenario does not include him revealing the car.
If he intentionally chooses a goat, switching gets the car 2 times out of 3.
If he chose randomly and just happened to choose a goat, switching doesn't matter. It's 50/50.
Of course, these aren't the only scenarios. As others have mentioned, if Monty can decide whether to reveal what's behind a door, he can act maliciously and only reveal a goat if you've already selected a car, and switching will always make you lose. Without knowing these constraints, a single answer isn't knowable, as you mentioned before.