The Monty Hall problem

[Rolfe]

Oh yes! Once Monty is allowed to vary his behaviour according to whether or not you already picked the prize, the problem simply becomes insoluble with the information given. You can go on adding convolutions and second-guesses to it almost indefinitely though!

ETA: I mean, oh yes you can go on formulating second-guess scenarios till the cows come home, not that Monty can get your chances below 1/3rd.

Rolfe.

 

[JoelKatz]

The maximizing strategy for evil monty is to never allow you to swap when you have chosen a losing door. It matters not one bit what he does when you have a winning door because the maximizing strategy for you against evil monty is to never swap.

 

[Vorpal]

He's mistaken. No matter how evil Monty gets, he can't drop your chances below 1/3. If you stick with your first choice no matter what, that's that.

 

[69dodge]

Put that way, it's straightforward that the following two flavors are different:
(1) He picked a random door out of the set that excludes the player's choice. It had a goat.
(2) He picked a random door out of all of them, and it happened to be one that was not the player's choice. It had a goat.
Given the rules are described above, the overall probability of winning for each goat-case is, based on what your strategy in those cases:
(1) 2/3, regardless of what you do when a goat appears
(2) 5/9 if you stay with your choice, 2/3 if you switch
Because if you resolve to stay on a goat that's not yours, then your possible winning conditions are (Your Pick,Random Pick): (<Any>,Car) + (Car,Goat) = (1/3) + (1/3)(2/3) = 5/9.

What about (Goat, Your Goat)?

To make the two cases comparable, shouldn't you do the same thing here as in the following case, namely, switch?

But if you resolve to switch, then your winning conditions are: (<Any>,Car) + (Goat,Other goat) + (1/2)(Goat,Your Goat) = 1/3 + (2/3)(1/3) + (1/2)(2/3)(1/3) = 2/3
The extra (1/2) because if you're switching but have two choices in that subcase.

 

[Vorpal]

What about (Goat, Your Goat)?

To make the two cases comparable, shouldn't you do the same thing here as in the following case, namely, switch?

You're right; I'm a dummkopf. Staying would be silly in those circumstances.
It would work if the random pick is not revealed before you choose to keep or switch, but then the problem becomes very dull.

 

[Jack by the hedge]

He's mistaken. No matter how evil Monty gets, he can't drop your chances below 1/3. If you stick with your first choice no matter what, that's that.

Yeah. I let my keyboard run ahead of my brain.

 

[Rolfe]

Vorpal, maybe I'm being dim here, but I'm still not getting it.

If Random Monty chooses from any of the doors, and lights on your original choice, then there is no ambiguity. Either it's "game over" if that's the rules, or your subsequent course of action is clear. However, that possibility is off the table before we get to the scenario in the problem, where he has picked one of the other two doors, and picked it at random.

How does it make any difference to the probabilities whether he might have picked your original choice at the earlier stage, but didn't? I'm honestly not seeing it.

Rolfe.

 

[Vorpal]

That's because I'm being much dimmer. See 69dodge's comment above.

The 5/9 follows only if you elect to stay when you see a goat regardless whether the one revealed was your first pick or not. On this strategy, what Monty could have picked (regardless of whether he actually did) really does make a difference. The problem is that it's a very dumb strategy.

 

[humber]

Oh God, I thought Humber understood it and had additional insights, but it looks as if I was wrong about that. That's what I get for assuming someone using maths-techie language must have "got" it. The paper he linked to was almost unreadable after the first few paragraphs, but was massively over-complicating a fairly simple situation and sort of re-defines the word "pretentious".

I do agree the problem is massively over-egged though.

Rolfe.

We can get back to that paper and the application of mimimax game theory, but I don't think I need Monty of the goats.
How is my butler game any different other than it may not contain so many variants. I can have "evil jeeves" too.

 

[Rolfe]

That's because I'm being much dimmer. See 69dodge's comment above.

The 5/9 follows only if you elect to stay when you see a goat regardless whether the one revealed was your first pick or not. On this strategy, what Monty could have picked (regardless of whether he actually did) really does make a difference. The problem is that it's a very dumb strategy.

So, do you think my post 191 above actually covers all the bases? Apart, of course, for all the pointless ramifications and second-guessing that go on when Helpful Monty and Evil Monty are allowed as possibilities?

My point is that I think the problem as stated becomes both finitely soluble and interesting if you simply add the information that Monty is not allowed to let his knowledge of whether or not you already picked the prize influence his choice of action.

Rolfe.

 

[humber]

Humber, do you still not understand that if Monty has chosen his pick at random, switching is neutral (that is, confers no advantage)?

We can get back to the game theory paper and the three propositions, but the probabilistic approach is irrelevant.

In the simplest case, where Monty is not evil and opens one of the remaining doors to reveal a goat, that generates the "surprising" 2/3 gain by swapping.
AFAIK, conditional probability does not do that. I did think that, and told myself that where the information went explained it, but in fact, it is something else, and I suspect that Erdos was right, but mugged by simulation.

The problem is a turkey, and relies on a great deal of verbal ambiguity. For example, the contestant "chooses" a door, but Monty "opens" a door. If you apply the same action to each, the problem loses its flavor.

It makes no difference at all if he opens a remaining door at random, or not. The simplest one-line algorithm that covers most situations is: "If Monty is forced to reveal a goat: swap"

There is some ambiguity about what happens should he reveal the car.
If the car is behind the contestant's chosen door and he reveals that, then the contestant can win or lose. If is behind one of the remaining doors, the contestant can lose, or win if he is allowed to swap to that door.

If he reveals a goat, be it the contestant's door or a remaining door; swap.
Intent changes nothing about that. That can be seen from game theory. The contestant can not improve on 2/3 chance of gaining the car, and Monty can't improve on 1/2 of not giving the car away.

You can run a simulation in which Monty always opens a door with a goat, and this will demonstrate that you win twice as often if you switch than if you don't. You've got that. Fine.

Agreed, but the simulation models buy into the errors of the question.
There is no difference in numerical outcome between the probabilistic approach, and the game/information approach, but it is not chance.

You need to run a different simulation, where Monty opens a door at random.

There is no information passed between, or retained from, a previous game and a current game, so randomizing makes no difference.
There are cases where conditional probability produces an "apparent" change in probability, but this game is not one of them.

If he opens the door you already picked, the game is over and you have either lost or won, but that isn't the situation the problem is examining, so we know that possibility is off the table.
If he opens one of the two remaining doors and reveals the car, again the game is over and you have lost. But again that isn't the situation we're examining, and that possibility is again off the table.

OK. We can exclude "very evil Monty". If the contestant chooses (ambiguity again) the car, he opens that door, and the contestant loses. If it is behind one of the remaining doors, he opens that one, and the contestant loses.
The contestant can never win.

The situation the problem describes is those iterations where he opened one of the two doors you didn't choose, and revealed a goat. In this case, that isn't invariable, but the other possibilities have been discarded because we know they didn't happen in the case under consideration.

Good, that simplifies the matter. The intent of the question as commonly posed, is the engender the 2/3 win case, and to associate that with "intuition", whereas the solution is in the question. It is it's own algorithm.

Examining the relevant iterations, from this formulation of the problem, will demonstrate to you that you win 50% of the time regardless of whether or not you switch.
Rolfe.

Why 50% of the time? Not at all.
That is Monty's best strategy, other than forcing a loss to the contestant by revealing the car. He can't do better than prevent you from having a 50/50 chance if there is the opportunity to swap.
That is when the contestant has chosen a goat. He opens that door, and the contestant then has a 50/50 chance with the remaining doors.

 

[humber]

He's mistaken. No matter how evil Monty gets, he can't drop your chances below 1/3. If you stick with your first choice no matter what, that's that.

Yes. Then Monty does nothing but to leave you with the usual 1/3 chance of gaining the car.
If Monty opens a door, he can't improve on 1/2 of losing the car.

In the first case, the contestant strategy is negated. The contestant can't act to improve on 1/3, but when Monty acts, the contestant can improve to 2/3.

 

[humber]

The trick is to impute statistical chance on the contestant, but not on Monty.
The canon is that you have a 1/3 chance of picking the car off the bat, and that is true if each subsequent trial is independent. But, the question is not about that, but the chance of a contestant in a given circumstance. Should he swap?
That is to say at the start " I have 1/3 chance", but in reality, you simply don't know at all, but that 1/3 is "kept in mind" so, hey presto, it magically becomes 2/3
when the goat is revealed.

 

[GlennB]

There is a wrinkle (probably mentioned already, but the thread is getting long) concerning a stickler-for-rules-and-straight-down-the middle-Monty. It relates to his not giving you any more information than is strictly necessary - as opposed to setting out to deceive - and applies where the game is repeated many times for the public to watch and learn (stress that last bit, which seems reasonable as we're talking about a TV show situation).

Stickler Monty lets you choose a door.
Then SM *will* show you a goat from the remaining 2 doors.
Then SM *will* offer you the choice to swap.

The wrinkle (alluded to in my card-game parallel) is that when SM has both goats to offer you (you chose the car) he must take care to offer doors at random.

If, in this situation, he always shows the left goat door then you are in the 'classical' Monty Hall situation and should swap.

But if SM ever then shows a goat in the right door it can only because the car is behind the left door.

Thus, just as the bridge player holding QJ in the trump suit declarer has led must take care to play them at random, SM, when faced with 2 goats to offer you, must do the same.

 

[marplots]

You can win 100% of the time if you get Monty to commit before you choose. So, Monty might say, for example: "If you choose door one, I will reveal a goat behind door 2. If you choose door 3, I will do the same. If you choose door 2, I will reveal a goat behind door 1."

"Thank you Monty."

 

[humber]

But that's just the thing: In the classical problem, Monty does not act entirely randomly, so his decision tells you something beyond the fact that there is X behind door #.
If Monty acts randomly, then you lose that extra bit of information.

Then you can't hold the contestant to the vagaries of random choice applied to multiple trials. Monty acts with knowledge. The contestant knows only that a car is there. Monty knows where the car is, but the rules limit how he can use that against you. The worst case for the contestant is "very evil Monty", but that is excluded. The next is strategy is where he may open the contestant's door to reveal a goat, so leaving the contestant with a 50/50 chance on the swap. The other leaves the contestant with a 2/3 chance when swapping to the remaining door. His actions are limited to that, and intention plays no part. He can't improve on 50/50 of losing the car.
If there can be "very evil Monty" who never lets you win, then there can be
"very benevolent Monty" who opens the appropriate door, and gives you the car. Both are silly.

[/quote=humber]
Instead of a game show, make the task to find the car.
You are wealthy, and have 3 garages, a car and a chauffeur, Jeeves, who normally drives the car, and randomly parks it in one of the 3 garages.
You sometimes drive it, so open one of the garages. 1/3 of the time the car will be there but, if it's not, you say to yourself "Jeeves must have parked in one of the other two garages, so I will try one of those", you then have a 50/50 chance of being right, so 2/3 of the time, you will find the car.

Right.
It is not like the Monty Hall game in any way whatsoever.

In your example, you pick a door and immediately check what is behind it. AT THIS POINT, YOU MIGHT ALREADY HAVE FOUND THE CAR!
You then randomly pick one of the other remaining doors.

There is the difference between choosing and revealing.
The garages are large. I open the door, but the light is off, so I don't see the contents. IF I turn the light on, and the car is there, I stop. Job done.
If I turn the light on, and the car isn't there, I try one of the other garages with a 50/50 chance.

Add Jeeves. The light is off, so I don't see the car in my selected garage, but Jeeves has not just chosen the other garage, but opened it and turned the light on. By ignoring that difference, I apparently gain to 2/3, by not turning the light on in my garage, and moving to the remaining one.
Well, of course. I had 1/3 chance of being right the first time, Jeeves informs me that that his garage does not have car, so it must be that 2/3 of the time, it will be in the remaining garage. It looks like magic, because the 1/3 chance of the correct choice initial choice remains. Erdos is right.

The simulations are corrupted. There is a statistically, 1/3 chance that the car may be selected, so a random generator assigns the car to one of the doors, but all that can actually be said of any single case, is that one of the doors has a car. To assign the car to a specific door, is to confuse chance and position.

But it is: It determines if Monty can freely chose which if the remaining doors he should open or if he will be forced to take the one doesn't have the car behind it.

If he merely chooses, that leaves the contestant with the option to select that door, but he opens the door.

The first chosen Garage *might not be empty*! But in the Monty game the rules state that it will always be empty.

Monty has no control over the first selection, and the puzzle starts from a random choice by the contestant. A car is there. That is all the information that is needed to form the remaining conclusions.

Your choice of a garage is entirely random, Monty's choice is never entirely random and might be entirely forced.

Monty/Jeeves acts after I have selected the garage, but have not seen the contents.

Yes, but that works only iff we assume that Jeeves deliberately picks an empty door rather than pick one at random or just turn on the lights on the garage you chose.

It is the same. Like Monty, Jeeves selects one of the remaining doors and shows the contents. If the car is there, well I find that out. Job done.
If the car is not in Jeeves' garage, then it must be in the garage I have chosen, or the other one. Do I stick or try again?
Calling that "winning or losing" makes no sense.

Should Jeeves open doors randomly, then 1/3 of the time, he will reveal the car.

If he includes my garage and turns the light on, then yes. No change from him not acting at all. If I turn the light on, I will learn just like the contestant who opens the door, if the car is there or not. If so, job done. If not, 50/50 chance with the other two, just like Monty opening the contestant's door to reveal a goat. See goat, swap. See empty garage, swap.

Or, put differently: You want to find the car, so both you and Jeeves walk to one garage each and open it. Each of you will have the car 1/3 of the time, the remaining 1/3 it will be in the garage that neither of you picked.

Monty and the contestant select the same door...

Regardless of which door you check first,

"Check". If you do a random trial of three doors, statistically, there is a 1/3 chance of success, but you need to open the doors to find that out.

or if Jeeves picks his door before you open the first chosen door, each door will have the car 1/3 of the time. If the first door opened doesn't have the car, it's 50/50 for the two remaining doors.

It's not about "each time" but one particular game.
If Jeeves "picks" the door but does not show the contents, I learn nothing.
Flat out 1/3 if all doors are available to me.
If he opens it, and reveals the contents, then I learn if it is there or not.
If it is, the job is done, if not, then 50/50 with the other two. The question concerns only the benefit of swapping.

 

[humber]

There is a wrinkle (probably mentioned already, but the thread is getting long) concerning a stickler-for-rules-and-straight-down-the middle-Monty. It relates to his not giving you any more information than is strictly necessary - as opposed to setting out to deceive - and applies where the game is repeated many times for the public to watch and learn (stress that last bit, which seems reasonable as we're talking about a TV show situation).

Except it isn't. The question is specific. Do you swap or not when faced with some given circumstance? Game theory gives exactly the same result in each case as the multiple iteration version.
It's not a difference between "sticklers" and "deceivers" but between "intuitionists" and "rule followers". The former unnecessarily insist on multiple trials, so they get the "counter-intuitive" buzz, that is just a forgone conclusion of the later.

 

[69dodge]

Stickler Monty lets you choose a door.
Then SM *will* show you a goat from the remaining 2 doors.
Then SM *will* offer you the choice to swap.

The wrinkle (alluded to in my card-game parallel) is that when SM has both goats to offer you (you chose the car) he must take care to offer doors at random.

If, in this situation, he always shows the left goat door then you are in the 'classical' Monty Hall situation and should swap.

Are you sure?

Say the doors are numbered 1, 2, 3, from left to right. You initially pick door 1. Monty always opens the left goat door when he has a choice, and here he opens door 2 and shows a goat.

According to my calculations, the probability is 1/2 that your original door (door 1) hides the car, so it doesn't matter whether you switch to door 3.

 

[GlennB]

Are you sure?

Say the doors are numbered 1, 2, 3, from left to right. You initially pick door 1. Monty always opens the left goat door when he has a choice, and here he opens door 2 and shows a goat.

According to my calculations, the probability is 1/2 that your original door (door 1) hides the car, so it doesn't matter whether you switch to door 3.

Yep. That was nagging me during dinner and in retrospect I have no idea why I wrote it. In fact it contradicts the whole gist of what I was trying to say. Ah well

 

[Rolfe]

If anyone else understands what Humber's argument is, then I'd be grateful if they would explain it to me in comprehensible terms. What he is typing now reads like so much gibberish to me. I'm open to the possibility that this is because his reasoning is so rarefied that he has left plebs like me floundering in the gutter, but at the moment I'm certainly not seeing it.

He quotes a section of my post where I specifically discuss the situation where Monty is selecting the door he opens completely at random, and describes this as "Monty's best strategy". He sees no distinction between choosing a door (selecting it without opening it, thus remaining ignorant about what is behind it), and opening a door, which of course involves revealing what is behind it. He talks about what to do if Monty is "forced" to reveal a goat, when the whole discussion, more or less, centres around the point that we don't know whether Monty is obliged to do (or not do) anything at all.

He claims several times that Erdos was right, while apparently maintaining that the chance of winning is always doubled by switching, irrespectove of what Monty's intention is. But according to the poster who brought up Erdos, he in fact believed that there was never any advantage to switching, even if Monty was obliged always to reveal a goat.

It's possible I'm simply not understanding his erudition. Can anyone else oblige?

Rolfe.