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You confused yourself. There are four possible outcomes at the end, but they are not equally likely. So not 50/50. If Monty always shows a goat, it is undeniab
by jncfhnb 3y ago
You confused yourself. There are four possible outcomes at the end, but they are not equally likely. So not 50/50.
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.
- LegibleCrimson 3y agoI'm covering the problem statement. I'm ignoring the cases where Monty revealed a car because it didn't happen. You can't chalk it up as a win or a loss, because it didn't happen. 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.
- jncfhnb 3y agoYes you’re correct. You’re just the first person I’ve ever seen cover the true random Monty variety who doesn’t force the situation where he shows a car in as a win or loss.