The Monty Hall problem

[Reno]

Ahh..I thought it was going to be something more interesting and perhaps even mildly scandalous Aw well...

 

[bruto]

The problem is complicated even if Monty always shows a goat and offers the switch, because the probability also depends on how he chooses which door to open when he has a choice. (If you pick a goat, he has no choice, as there's only one goat left to show you, but if you pick the car, he can show you either of the two goats.)

You picked door 1. Monty opened door 2. The probability that door 1 now hides the car is p / (p + 1), where p is the probability that Monty opens door 2 supposing he has a choice. The usual answer of 1/3 is obtained only if we assume that p = 1/2.

I don't see why this is complicated, or why it matters whether or not Monty had a choice. What we know is that he opened a door with a goat, and that there will always be at least one door with a goat whatever was behind the first door you chose. The result would be the same if, instead of opening that door, Monty simply invited to you to swap your choice from the one door you first chose to all the doors you did not. Your chance of being right when you switch is still the same as your chance of being wrong on your first choice.

 

[yy2bggggs]

Or use Bayes's Theorem, which is how I did it:
...
= p / (p + 1).

So if p=1, then the probability would be 1/2.

This would correspond, in my problem, to something like Monty always wanting to show me the billy, unless I picked it; in that case he shows me the nanny. So if I see the nanny, I know I picked the billy. But if I see the billy, either I picked the car, or the nanny. Those being equally likely, the probability is 1/2.

I buy that.

 

[Ziggurat]

I don't see why this is complicated, or why it matters whether or not Monty had a choice.

Probability is about information. When Monty has no choice, the information you have is easy to analyze, and the solution is indeed simple. But if Monty has a choice, then the information is NOT simple to analyze. When Monty has a choice, then his motives will make a difference to his actions. The information contained within his choice is concealed from you because you don't know his motives.

The result would be the same if, instead of opening that door, Monty simply invited to you to swap your choice from the one door you first chose to all the doors you did not.

True, that is equivalent. But if Monty has a choice, then he might only offer you this option if you picked correctly the first time. Or he might only offer you this option if you picked wrong the first time. If you don't know his motives, and furthermore cannot even assign a probability to that, then you cannot determine your odds for switching vs. not switching. You could guess at the probability of his motives being one way or the other and calculate the odds based on that, but that's still a more complicated problem than the original.

 

[ZirconBlue]

Reno: I don't think it does. In DoND nobody involved in the game knows any of the box contents.

One similarity is that many contestants do seem irrationally tied to "their" case.

3. The situation is incredibly contrived. Where else, outside a game show, you have to make a decision based on a partial information, from a knowledgeable source, who is deliberately withholding part of information?

Dating?

 

[Ziggurat]

3. The situation is incredibly contrived. Where else, outside a game show, you have to make a decision based on a partial information, from a knowledgeable source, who is deliberately withholding part of information? It is just not a common occurrence.

Dating?

Good point. But there are also plenty of directly economic examples, like buying a used car. Or (closer to your dating example) hiring an employee.

 

[bruto]

Probability is about information. When Monty has no choice, the information you have is easy to analyze, and the solution is indeed simple. But if Monty has a choice, then the information is NOT simple to analyze. When Monty has a choice, then his motives will make a difference to his actions. The information contained within his choice is concealed from you because you don't know his motives.



True, that is equivalent. But if Monty has a choice, then he might only offer you this option if you picked correctly the first time. Or he might only offer you this option if you picked wrong the first time. If you don't know his motives, and furthermore cannot even assign a probability to that, then you cannot determine your odds for switching vs. not switching. You could guess at the probability of his motives being one way or the other and calculate the odds based on that, but that's still a more complicated problem than the original.

But wouldn't the possession of that information amount to beating the odds? Knowledge of how Monty arrives at his choice changes the solution because it changes the problem. Of course there are assumptions here however you look at it, but I have always figured that the problem as presented is based on the assumption that Monty always opens a goat door, and always offers the choice, and that his action could be reproduced mechanically by the instruction "open any goat door that remains and offer the choice." If that is the only thing you can know, then other complications are inapplicable... or so my simple mind would think.

 

[69dodge]

I don't see why this is complicated, or why it matters whether or not Monty had a choice. What we know is that he opened a door with a goat, and that there will always be at least one door with a goat whatever was behind the first door you chose.

We don't only know that there is a goat behind door 2. We also know that Monty chose to open door 2, even though he might have been able to open door 3 instead. This affects the answer.

For example, suppose he dislikes door 2, and only opens it when he has to. Then, if we see him open it, we know that he had no choice. That is, we may conclude that door 3 definitely hides the car, rather than hiding it with probability 2/3.

(In this setup, the probability is p = 0 that Monty chooses door 2 when the car is behind door 1, and the probability is p / (p + 1) = 0 / (0 + 1) = 0 that the car is behind door 1.)

The result would be the same if, instead of opening that door, Monty simply invited to you to swap your choice from the one door you first chose to all the doors you did not. Your chance of being right when you switch is still the same as your chance of being wrong on your first choice.

Here, Monty doesn't choose to open a door at all. So we can't infer anything from his choice of door.

 

[Ziggurat]

But wouldn't the possession of that information amount to beating the odds?

In that case, knowing that information amounts to knowing what Monty knows, and Monty knows where the car is, so there ARE no odds anymore, it's a certainty.

Knowledge of how Monty arrives at his choice changes the solution because it changes the problem.

Indeed it does.

Of course there are assumptions here however you look at it, but I have always figured that the problem as presented is based on the assumption that Monty always opens a goat door, and always offers the choice

That is the standard presentation of the problem, yes. And it's a neat enough problem even in this simple form. Since you're supposed to calculate odds, you'd need more information if you wanted to play it any other way, information which is generally not presented.

But it's not how the real game show is played. Monty can do whatever he wants in the real game. So aside from tracking his history, there's no way to calculate the odds in real life. But that doesn't make for a good thought problem because it's too cumbersome at that point.

 

[69dodge]

[...] I have always figured that the problem as presented is based on the assumption that Monty always opens a goat door, and always offers the choice,

Yes, I am making the same assumption.

and that his action could be reproduced mechanically by the instruction "open any goat door that remains and offer the choice."

That is not mechanical enough. Two goat doors might remain. Which one will the mechanism open?

 

[lomiller]

The 'proper' description of the Monty Hall puzzle is that Monty *must* show you a goat door, no matter what (as indicated in the o/p). He only tosses a coin, figuratively speaking, if he has two goats to choose from, in which case it makes no difference anyway.

Also, if he is allowed to choose Monty can actually control the outcome. I.E. he could decide to only offer a choice to switch when you already had the door with the car, in which case the decision to switch would never be correct.

 

[Ziggurat]

We don't only know that there is a goat behind door 2. We also know that Monty chose to open door 2, even though he might have been able to open door 3 instead. This affects the answer.

For example, suppose he dislikes door 2, and only opens it when he has to. Then, if we see him open it, we know that he had no choice. That is, we may conclude that door 3 definitely hides the car, rather than hiding it with probability 2/3.

You have introduced information into the problem that the original formulation did not have, and information changes probabilities. In the absence of this information (meaning that Monty might still dislike door 2, but we don't know that), the odds remain what they were. If we initially pick door #1 and he chooses door 2, that could be because the car is behind door #3, or it could be because he doesn't hate door #2 but rather door #3. Since (a priori) he's as likely to hate door #3 as door #2 (if he hates any door at all), if you know nothing about his hatred of doors, the odds work out the same as him not having any preference between doors.

 

[69dodge]

Since (a priori) he's as likely to hate door #3 as door #2 (if he hates any door at all), if you know nothing about his hatred of doors, the odds work out the same as him not having any preference between doors.

Interestingly, putting a uniform distribution on p from 0 to 1 doesn't give the same answer here as assuming that p = 1/2. The integral from 0 to 1 of p / (p + 1) dp is 1 - ln 2 = 0.30685..., whereas (1/2) / (1/2 + 1) = 1/3 = 0.33333... . Not sure what that means, exactly. Perhaps 1 - ln 2 is really correct, or perhaps a uniform distribution is "wrong" for some reason.

 

[Ziggurat]

Interestingly, putting a uniform distribution on p from 0 to 1 doesn't give the same answer here as assuming that p = 1/2. The integral from 0 to 1 of p / (p + 1) dp is 1 - ln 2 = 0.30685..., whereas (1/2) / (1/2 + 1) = 1/3 = 0.33333... . Not sure what that means, exactly. Perhaps 1 - ln 2 is really correct, or perhaps a uniform distribution is "wrong" for some reason.

What is your p? The probability that he hates door #2?

Because if you want a uniform probability distribution, it should be uniform across all three doors, not just between door 2 and 3.

 

[69dodge]

What is your p? The probability that he hates door #2?

The probability that Monty opens door 2, given that the contestant picked door 1 and the car is behind door 1.

Because if you want a uniform probability distribution, it should be uniform across all three doors, not just between door 2 and 3.

If I pick door 1, it doesn't matter whether Monty likes door 1, because he can't open it anyway. All that matters is his preference between doors 2 and 3, because he might have a choice between those.

But, the more I think about it, the weirder it seems (that 1 - ln 2 is right, rather than 1/3). Maybe I've made some other mistake.

 

[Ziggurat]

The probability that Monty opens door 2, given that the contestant picked door 1 and the car is behind door 1.

Then what is p/(1+p) ?

If I pick door 1, it doesn't matter whether Monty likes door 1, because he can't open it anyway. All that matters is his preference between doors 2 and 3, because he might have a choice between those.

But if he hates door #1, then he has no preference between the other doors. That makes a 50/50 choice more probable than a 90/10 or 10/90 choice. A uniform probability distribution between all three options (a 2D triangle) doesn't produce a uniform probability distribution when projected onto one axis (the edge of the triangle).

 

[BenBurch]

http://en.wikipedia.org/wiki/Monty_Hall_problem

 

[69dodge]

Then what is p/(1+p) ?

The probability that the car is behind door 1, given that I picked door 1 and then Monty opened door 2.

But if he hates door #1, then he has no preference between the other doors. That makes a 50/50 choice more probable than a 90/10 or 10/90 choice. A uniform probability distribution between all three options (a 2D triangle) doesn't produce a uniform probability distribution when projected onto one axis (the edge of the triangle).

I don't think that's it.

I forgot that seeing Monty open door 2 makes higher values of p become more probable, even though p started out uniformly distributed.

 

[bruto]

Yes, I am making the same assumption.



That is not mechanical enough. Two goat doors might remain. Which one will the mechanism open?

Of course if, mechanically, Monty or the Monty-bot always opened the first available goat door in a given row, then we'd gain some insight if he opened the second available one, but none if he opened the first. However, if we have no knowledge of the process used to choose the door, or no time to establish a pattern, then we're stuck with the odds. Granted, the whole problem relies on some unwritten assumptions, but I'm working on the assumption that knowledge not stated in the problem is knowledge we cannot otherwise obtain.

 

[69dodge]

Of course if, mechanically, Monty or the Monty-bot always opened the first available goat door in a given row, then we'd gain some insight if he opened the second available one, but none if he opened the first.

I'm not sure what you mean by "none if he opened the first". In both cases you mention, the probability of winning if you switch is different than it is in the case where Monty opens the same door but we know that he chooses randomly when he has a choice.

However, if we have no knowledge of the process used to choose the door, or no time to establish a pattern, then we're stuck with the odds. Granted, the whole problem relies on some unwritten assumptions, but I'm working on the assumption that knowledge not stated in the problem is knowledge we cannot otherwise obtain.

That's reasonable, but it's not the same as saying that, even if we had such knowledge, it wouldn't make any difference anyway, which is what I thought you meant.