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Comment by bemusedthrow75 | Hacker News Reader

> 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 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.

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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 C, Monty showed you either A or B, and you switch to get the other goat.

So a 2 / 3 chance of getting the car if you always switch.

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

But we know in the problem statement that Monty hall showed us a goat, so we can eliminate possibilities 2 and 4 to get:

    Initial Choice | Monty's choice | Remaining Door
    A | B | C
    B | A | C
    C | A | B
    C | B | A

Whether you switch or not, you have a 50/50 chance.

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.

But in the second scenario, each of those options is actually 1/4, because he was choosing randomly. Most importantly, each option was a 1/6, but two options where you selected a goat were eliminated because those were ones where Monty selected the car.

[0]: https://gist.github.com/Taywee/2ba202b1bf7af40293ecffb01c2ab...

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.

I'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.

Yes 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.

Incorrect. You have no way to know if he would only show you a goat if you picked the car.

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.

> You have no way to know if he would only show you a goat if you picked the car.

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.

You cannot assume that your context would apply from all starting conditions!

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.

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