The Monty Hall problem

[69dodge]

My brain will fry if I tr< to understand this any longer .... but thank you for the effort. Maybe I'll have tome over the weekend.

Yes, it's a bit of a mess. I don't see any easy way of understanding why that's the answer. I just went through the calculation, and that's what came out.

But the upshot is that you should switch again. It's better to switch to one of the remaining doors, but if you can't, you should even switch back to your original door.

That's if Monty was allowed to open your original door as the second door he opened, but just happened not to. If he wasn't allowed to, you should switch to one of the remaining doors but not back to your original door.

 

[Rasmus]

Yes, it's a bit of a mess. I don't see any easy way of understanding why that's the answer. I just went through the calculation, and that's what came out.

But the upshot is that you should switch again. It's better to switch to one of the remaining doors, but if you can't, you should even switch back to your original door.

That's if Monty was allowed to open your original door as the second door he opened, but just happened not to. If he wasn't allowed to, you should switch to one of the remaining doors but not back to your original door.

Cool, thanks.

 

[Jack by the hedge]

Yes, it's a bit of a mess. I don't see any easy way of understanding why that's the answer. I just went through the calculation, and that's what came out.

But the upshot is that you should switch again. It's better to switch to one of the remaining doors, but if you can't, you should even switch back to your original door.

That's if Monty was allowed to open your original door as the second door he opened, but just happened not to. If he wasn't allowed to, you should switch to one of the remaining doors but not back to your original door.

This is a multi-door game, where you get to swap every time Monty opens a door?

If so, then I do see that you should swap every time if Monty can't choose your own door, but it doesn't seem to me to make any difference if you swap when Monty is able to choose from all the doors including your own. If your choice doesn't affect Monty's choice, then his choice tells you nothing useful about your current choice. Might as well stick.

Also, if Monty can't choose your door you should always swap but it doesn't matter if you swap back to a previous choice. You benefit equally from swapping to any of the doors from which Monty chose, including your previous choices.

With N doors and after Monty has opened X doors, your choice has 1/(N-X) chance of winning. When he opens another, all the remaining doors except yours have 1/(N-(X+1)) chance of winning. Previous selections don't retain their previous probability when they return to the pool from which Monty can choose.

 

[Rolfe]

In the classic Monty Hall problem the contestant also has the information that Monty only ever opens one of the two doors the contestant hasn't picked with a goat behind it.

At the start of the game the probability the door the contestant picks has the prize behind it is 1/3 and the probability is 2/3 that it is behind one of the other two doors. Monty cannot open the door the contestant picked and cannot open the door with the prize behind it. This means the probability of the one unopened door of the two doors not picked by the contestant having the prize behind it becomes 2/3 after Monty excludes one of the two unpicked and unopened doors by showing the contestant one of them that has a goat behind it.

If the wording of the puzzle is sloppy or the rules of the game changed then the correct answer may be different or there may be no correct answer.

It's still interesting if you allow Monty to open either of the doors the contestant didn't choose, at random. That's what supplies the curious paradox, that the probabilities change according to Monty's intention.

Once you start considering that Monty may be biassed and trying to steer the contestant one way or another based on whether or not he originally selected the car, it's nothing but a sterile game of second-guessing.

Does anyone but Humber really think it's necessary to specify that all doors have equal probability of having the car behind them? I've never seen anyone else get into "well if the car is always behind door 1 and you picked door 1 you'd be nuts to switch" territory.

Rolfe.

 

[Dr.Sid]

I think the problem is clear to everyone. From some posts I actually think it's clear to humber too .. but then from another posts ..

 

[Rolfe]

If he really does get it, but he's trolling for fun, I have to say it's a class act.

Rolfe.

 

[humber]

If he really does get it, but he's trolling for fun, I have to say it's a class act.

Rolfe.

OK. I will make the effort to put the whole thing down in one post, and address each point. It's useless this way. But thanks for being curious enough to "lurk".

 

[69dodge]

This is a multi-door game, where you get to swap every time Monty opens a door?

There are many doors, yes.

I only analyzed the case where Monty opens two doors.

It starts off as usual. You pick a door, but don't open it. Monty opens a door, which is definitely not yours and definitely is a goat. You switch to another door, because it's to your advantage, but you don't open it yet. Then, Monty opens a second door, which is definitely a goat and is definitely not the door you switched to.

Here there are two possibilities. Either way, the second door that Monty opens isn't your original door; the difference between the two is whether Monty deliberately avoided your original door, or whether he just happened not to open it.

Now you can switch again if you want to.

See post #354 for the first possibility and post #356 for the second.

Also, if Monty can't choose your door you should always swap but it doesn't matter if you swap back to a previous choice. You benefit equally from swapping to any of the doors from which Monty chose, including your previous choices.

With N doors and after Monty has opened X doors, your choice has 1/(N-X) chance of winning. When he opens another, all the remaining doors except yours have 1/(N-(X+1)) chance of winning. Previous selections don't retain their previous probability when they return to the pool from which Monty can choose.

I disagree.

It is not true that previous selections retain their previous probability in the sense of keeping that probability entirely unchanged, but it is necessary to take the prior probability into account when calculating a new probability.

Bayes's Theorem for the probability of A, given new information B, is

P(A | B) = P(A) [P(B | A) / P(B)].​

That 'P(A)' in front of the right side is the prior probability of A.

 

[Jack by the hedge]

I disagree.

It is not true that previous selections retain their previous probability in the sense of keeping that probability entirely unchanged, but it is necessary to take the prior probability into account when calculating a new probability.

Bayes's Theorem for the probability of A, given new information B, is

P(A | B) = P(A) [P(B | A) / P(B)].​

That 'P(A)' in front of the right side is the prior probability of A.

I'm not quite grasping what exactly the A and B events are.

How about considering a 4 door version for a moment: If you choose door 1 and Monty opens door 2, then door 1 has probability 1/4 of winning and doors 3 & 4 have p = 3/4 between them. So 3 and 4 have p = 3/8 each.

If you switch to, say, 3 and Monty opens either 1 or 4, then door 3 still has p = 3/8 so whichever remains between 1 and 4 must have p = 5/8. Mustn't it? So it doesn't matter that 1 and 4 have different histories. Or did I miss something?

PS Post #356 doesn't seem to give sensible answers for n = 4.

 

[69dodge]

I'm not quite grasping what exactly the A and B events are.

I didn't have anything particular in mind when I wrote that. I just wanted to show that the prior probability always matters.

For the Monty Hall problem, A will be something like "the car is behind the door you originally picked" or "the car is behind the door you're thinking about switching to." B will generally be "Monty does what he did."

PS Post #356 doesn't seem to give sensible answers for n = 4.

#356 does work for n = 4.

But #354 doesn't, which I neglected to mention. In that case, the probabilities are 2/5 for the door you originally picked, and 3/5 for the door you switched to. The general formula doesn't work when n = 4 because Monty might not be able to open any door the second time. There's only one door left and it might hide the car. So, the problem is different.

How about considering a 4 door version for a moment: If you choose door 1 and Monty opens door 2, then door 1 has probability 1/4 of winning and doors 3 & 4 have p = 3/4 between them. So 3 and 4 have p = 3/8 each.

Yes.

If you switch to, say, 3 and Monty opens either 1 or 4, then door 3 still has p = 3/8 [...]

Why?

Door 3 has whatever probability Bayes's Theorem says it has. So let's figure it out.

Suppose Monty opens door 1.

Let c1 mean "the car is behind door 1", and similarly for c2, c3, and c4. Let m1 mean "Monty chooses door 1 as his second door to open". We know that before Monty opens his second door, we have P(c1) = 1/4 and P(c3) = P(c4) = 3/8. What we want to figure out is the probability of c3 after Monty opens door 1, that is, we're looking for P(c3 | m1).

P(c3 | m1) = P(c3) P(m1 | c3) / P(m1)
= (3/8) (1/2) / [P(m1 | c1) P(c1) + P(m1 | c2) P(c2) + P(m1 | c3) P(c3) + P(m1 | c4) P(c4)]
= (3/8)(1/2) / [(0)(...) + (...)(0) + (1/2)(3/8) + (1)(3/8)]
= 1/3.

Some explanation: If the car is behind door 3, Monty has a choice between doors 1 and 4. I assume that he chooses randomly in that case, so P(m1 | c3) = 1/2. If the car is behind door 4, Monty has no choice---the only door left is door 1---so P(m1 | c4) = 1. Monty won't show the car, so P(m1 | c1) = 0. And, finally, P(c2) = 0 because Monty already opened door 2 to show a goat.

 

[Vorticity]

If he really does get it, but he's trolling for fun, I have to say it's a class act.

Rolfe.

I dare you to read this response of humber's to one of my posts and still say he isn't trolling.

 

[Rolfe]

Yeah, and I saw the tortoise and Axchilles thing too.

If he's trolling, I'm overwhelmed with admiration.

Rolfe.

 

[Jack by the hedge]

If you switch to, say, 3 and Monty opens either 1 or 4, then door 3 still has p = 3/8...

Why?

Door 3 has whatever probability Bayes's Theorem says it has. So let's figure it out.

Suppose Monty opens door 1.

Let c1 mean "the car is behind door 1", and similarly for c2, c3, and c4. Let m1 mean "Monty chooses door 1 as his second door to open". We know that before Monty opens his second door, we have P(c1) = 1/4 and P(c3) = P(c4) = 3/8. What we want to figure out is the probability of c3 after Monty opens door 1, that is, we're looking for P(c3 | m1).

P(c3 | m1) = P(c3) P(m1 | c3) / P(m1)
= (3/8) (1/2) / [P(m1 | c1) P(c1) + P(m1 | c2) P(c2) + P(m1 | c3) P(c3) + P(m1 | c4) P(c4)]
= (3/8)(1/2) / [(0)(...) + (...)(0) + (1/2)(3/8) + (1)(3/8)]
= 1/3.

Some explanation: If the car is behind door 3, Monty has a choice between doors 1 and 4. I assume that he chooses randomly in that case, so P(m1 | c3) = 1/2. If the car is behind door 4, Monty has no choice---the only door left is door 1---so P(m1 | c4) = 1. Monty won't show the car, so P(m1 | c1) = 0. And, finally, P(c2) = 0 because Monty already opened door 2 to show a goat.

That looks entirely convincing, but it's taking me a while to get my head around why Monty's choice of second door should influence the probability of your second choice. It seemed counterintuitive to me.

Once again the 100 door variant came to the rescue. If those doors 1, 3 and 4 were the last three remaining from a vast number of doors, and the player had stuck with door 1 throughout, then the situation would be that door 1 would have almost no chance of winning, doors 3 and 4 would each be almost 50% likely to win. So when the player swaps to 3, Monty can open 1 or 4. He can almost always choose 1, and he can choose 4 about half the time. If he opens door 4, then the choice is between door 3 and door 1. But door 1 had virtually no chance of being right, so it must increase the chances of door 3 being the winner, rather than making door 1 most likely to win. That seems intuitive. For 100 doors, I get P(c3 | m4) = 94.9%.

Suddenly this problem seems interesting again. Thanks.

 

[Dr.Sid]

My understanding is this: in case you choose goat as your first choice, Monty will show you the other goat, and thus the third door must be car. Of course you don't know if you have chosen car or goat .. but goat has 2/3 change, while car just 1/3.
It's all about the case you choose goat in first choice, because in such case Monty's choice is dependent on your choice. If you choose car in the first choice, Monty can show any goat he likes, his choice is independent, and it brings no new information.

 

[69dodge]

For 100 doors, I get P(c3 | m4) = 94.9%.

How'd you get that?

I got 99/103 = 96.1%.

But I'm not sure I'm solving the same problem as you.

 

[bruto]

My understanding is this: in case you choose goat as your first choice, Monty will show you the other goat, and thus the third door must be car. Of course you don't know if you have chosen car or goat .. but goat has 2/3 change, while car just 1/3.
It's all about the case you choose goat in first choice, because in such case Monty's choice is dependent on your choice. If you choose car in the first choice, Monty can show any goat he likes, his choice is independent, and it brings no new information.

I don't see it that way. You make an initial choice, which has a 1/3 chance of being right. No matter whether it is right or wrong, it is dead certain, since there is only one car and two unchosen doors remaining, that one of the unchosen doors contains a goat. Knowing this certainty does not change the odds of your original choice. Therefore, whether you were right or wrong in your first choice, Monty can and will open a goat door, and if you have no way of knowing whether there is a trick or a system to the opening, you have no more information after it's open than you did before. If you have no new information, you are stuck with the original odds.

 

[CurtC]

My understanding is this: in case you choose goat as your first choice, Monty will show you the other goat, and thus the third door must be car. Of course you don't know if you have chosen car or goat .. but goat has 2/3 change, while car just 1/3.

That's the classical solution, which is the right answer if the puzzle is worded in a way that Monty always has to show you a non-prize door and then offer the chance to switch. However, most of the time when I've seen the puzzle, it has not included that constraint and therefore cannot be solved without making some assumptions about why Monty did what he did.

For example, if we assume that Monty's goal is to avoid giving away the prize car, then he may offer the switch only to those players who guessed the right door with their first choice. As you can see here, if this is the game he's playing, you will always lose if you switch.

 

[Jack by the hedge]

How'd you get that?

I got 99/103 = 96.1%.

But I'm not sure I'm solving the same problem as you.

P(c3 | m4) = P(c3) P(m4 | c3) / P(m4)
= (0.495) (0.5) / [P(m4 | c1) P(c1) + P(m4 | c2) P(c2) + P(m4 | c3) P(c3) + P(m4 | c4) P(c4)]
= (0.495)(0.5) / [(1)(0.01) + (...)(0) + (0.5)(0.495) + (0)(...)]
= 96.1%

I cannot brain today. I have the dumb.

 

[Dr.Sid]

That's the classical solution, which is the right answer if the puzzle is worded in a way that Monty always has to show you a non-prize door and then offer the chance to switch. However, most of the time when I've seen the puzzle, it has not included that constraint and therefore cannot be solved without making some assumptions about why Monty did what he did.

For example, if we assume that Monty's goal is to avoid giving away the prize car, then he may offer the switch only to those players who guessed the right door with their first choice. As you can see here, if this is the game he's playing, you will always lose if you switch.

Wait .. so there is not even agreement on the problem definition ?

 

[bruto]

Wait .. so there is not even agreement on the problem definition ?

Apparently not, and this is part of the problem, since until you get an agreed definition you cannot know what information is conveyed by Monty's actions. The problem as I have always seen it defined is that Monty knows what's behind the doors, opens a goat door and then offers you the chance to switch, and that we are not privy to any other information about what he might do in another session, or whether there is any system or rationale to which door he opens. In other words the only information conveyed by Monty's door opening is the confirmation of what we already knew before he opened it: that there's always at least one goat door left unchosen. Since we have no new information, we must assume the odds remain the same as before, and switch.