The Monty Hall problem

[humber]

Yes, really. Actually, that was the way of thinking it out I came to myself, independently. That example demonstrates whether or not the host is deliberately avoiding the prize
In that example, there is a 1 in 100 chance you picked right to start with, a 1 in 100 chance he is opening doors at random and simply missed the prize every time by chance, and a 98 in 100 chance he is deliberately avoiding the prize.
Common sense says you switch, obviously.

This is the way Marilyn first explained it: "Suppose there were 100 doors and Monty opened 98 of them. You'd switch pretty fast then, wouldn't you?"
But it's not about multiple trials, but a rule, a strategy, that does not change with the number of doors.

There are 100 doors. He opens 98 doors. If the contestant has chosen the car, Monty opens 98 doors, and the remaining closed one has a goat behind it. The contestant sees 98 goats.
If the contestant hasn't chosen the car, he opens 98 doors, and the remaining closed door has a car behind it. The contestant sees 98 goats.
OK swap. See one goat in the 3 door case, swap.
But what makes her think that "I would swap pretty fast" if there were 100 doors, but be less inclined to do so if there were 3?

It's not obvious to me that this is wrong. Quite the contrary.
Imagine that happened on multiple iterations. It's the same as I outlined
above.
If he's forgotten and is guessing, then it's the same as if he's selecting either unchosen door at random and not deliberately avoiding the prize. One third of the iterations reveals the car. Of the remaining iterations, where the contestant does get the chance to switch, it's 50/50.

You are on a game show with three doors. A car is behind one; goats are behind the others. You pick door No. 1. Suddenly, a worried look flashes across the host’s usually smiling face. He forgot which door hides the car! So he says a little prayer and opens No. 3. Much to his relief, a goat is revealed. He asks, “Do you want door No. 2?” Is it to your advantage to switch?
—W.R. Neuman, Ann Arbor, Mich.

Nope. If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch.

That is wrong, because the information is the same regardless of whether it is given by intent or by chance.
If he reveals a goat, then swap. 2/3 time, the car is behind the other door.
That is true if he does that by intent or not. The questioner says a goat is revealed.

It does what you programme it to do. You can programme it to open a remaining door with a goat behind it if you like.
If you do that, you will find that switching doubles your chances of winning. As we all know.

The best strategy for the player is; if the door reveals a goat, swap.
If the machine is programmed to open any door with a goat, then the benefit of swapping remains. (edit) The next choice is either a 50/50 chance or 2/3 chance, dependent on which door the machine opens. Always better that 1/3. The programne works to the contestant's advantage.

Or you can programme it to open either of the two remaining doors at random. In that version, one third of the time, the car is revealed. Sorry, you lost. Good game! If that's the version you're playing, then switching (if you haven't already lost, which we know you haven't because that's the scenario we're examining) confers no advantage.

It makes no difference if it's 1/3 of the time or not. The programme could be
random, pseudo-random or deliberate sequence, and the choice based on contents or door number.
If it opens either of the remaining doors, the programme works to the contestant's advantage, or is neutral.

If a goat is revealed, the contestant learns that he should swap to the remaining door.
If a car is revealed, then he learns there is no point in swapping to the remaining door.
It's only of advantage to swap, if a goat is revealed.

The point about Monty's intentions is in effect to demonstrate which version of the game you're playing. If it's the former, switching is an advantage. If it's the latter, switching is neutral. So if you don't know which it is, but you do know it's one or the other, switch anyway, because it can't harm your chance, and might improve it.

Yes, but his intention plays no part. The choice is to swap to a remaining door. (edit)
If Monty is forced to reveal a goat, then the contestant always learns to swap to his advantage.
If Monty is forced to reveal the car, the contestant learns he has won the car, so don't swap, or he has lost, so don't bother to swap.

Therefore, Monty's intent makes no difference to the advantage of swapping, and it is only advantageous to do so, if a goat is revealed.

 

[zooterkin]

I don't really follow you. Are you saying, it wouldn't be good TV if Monty opened his chosen door and revealed the car, so that wouldn't happen and the possibility can be discounted? I dispute that strongly. The possibility that Monty might do exactly that is what would give suspense and drama to that door-opening moment. And Monty reveals the car! Too bad, thanks for playing!

It's the same moment as someone getting kicked out of the Big Brother house. Of course it would be good TV. I cannot see any a priori reason for discounting the possibility that a random choice between the two remaining doors is exactly the game Monty might be playing.

Even closer examples would be Deal or No Deal and the current lottery programme, where the players can lose the big prizes during the game.

 

[humber]

This problem is not just badly described but over-egged. Certainly the Vos Savant version.
I think it's about information, and when it becomes available, but it's only what the contestant knows and when that counts, and Monty and the goats have nothing to do with it.

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.

How different is that from the Monty Hall game?

When the contestant makes the first guess, that is not really relevant.
When Monty shows the goat, then the game begins. Being shown the goat, puts you in the first chosen empty garage, and from there, you chose another.

ETA:
Or perhaps, your first choice is the first garage, but the light is off, so you don't see it, Jeeves shows you an empty garage. If you then turn the light on, 1/3 of the time you will find the car, if you don't, and try the remaining one, you will find it 2/3 of the time.

 

[Vorpal]

humber, I already linked to program and its results for the case that the host does not know, picks a random door that's not the player's pick, and it happens to come out to be a goat. The result is very unambiguous in that switching is not an advantage when a goat is revealed (however, it also does no worse). You can examine it yourself, re-run it, or build your own.

Marylin's statement is correct within that particular interpretation of the problem statement. It's not the only possible interpretation, but it is a very natural one.

Another alternative interpretation is that the host picked a random door, and it could have been the player's (but actually come out different). In that case, things are different despite the outcome being the same: there is an advantage of switching when a goat is revealed in that it gives wins 1/9 of the time more than keeping (see the other link in post 159).

If your point is that we don't actually know how the host picked the door and therefore switching on a goat is a better strategy because it generally does no worse than keeping, then I agree that's true when the host cannot act on a malicious bias. (For example, if the host is not obligated to offer and is biased towards offering when the first pick is a car, then obviously switching suffers as a strategy.)

 

[yy2bggggs]

Full trees starting with a choice of 1.

Ordinary:

Column|A|B|C
You pick|1|1|1
Car behind|1|2|3
Monty shows|2 or 3|3|2
Decline|C|G|G
Accept|G|C|C

Chance of car if accepting gambit is 2/3. Chance if declining 1/3.

Monty guesses:

Column|A|B|C|D|E|F
You pick|1|1|1|1|1|1
Car behind|1|2|3|1|2|3
Monty shows|2|2|2|3|3|3
Game over|-|Y|-|-|-|Y
Decline|C|-|G|C|G|-
Accept|G|-|C|G|C|-

Overall chance of winning a car is 1/3. If you lucked up and didn't immediately lose when Monty revealed the car (columns B and F), then there are only 4 possibilities left (Columns A, C, D, E). You have a 1/2 chance of winning, accept or not.

Monty always picks lowest numbered door, unless it's a car, then highest:

Column|A|B|C
You pick|1|1|1
Car behind|1|2|3
Monty shows|2|3|2
Decline|C|G|G
Accept|G|C|C

Overall chance of winning if you swap is still 2/3. But once you know which door Monty opens, the probability changes. If he opens 3 (column C only), then swap for a sure thing. Probability is outright 1. Otherwise, if he opens a 2 (columns A and B), you have a 1/2 chance of winning, accept or not.

Note that this critically depends on his always choosing 2. If he chose 2 or 3 at random, it becomes the original puzzle again.

ETA: Humber: The specific thing you're missing is that possible scenarios are getting culled out in the cases where you do not believe it should matter. The above tables show exactly which cases. When you're told something happened and that means a portion of your probability space goes away, your probabilities should update.

 

[Rolfe]

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.

 

[Rolfe]

This is the way Marilyn first explained it: "Suppose there were 100 doors and Monty opened 98 of them. You'd switch pretty fast then, wouldn't you?"
But it's not about multiple trials, but a rule, a strategy, that does not change with the number of doors.

There are 100 doors. He opens 98 doors. If the contestant has chosen the car, Monty opens 98 doors, and the remaining closed one has a goat behind it. The contestant sees 98 goats.
If the contestant hasn't chosen the car, he opens 98 doors, and the remaining closed door has a car behind it. The contestant sees 98 goats.
OK swap. See one goat in the 3 door case, swap.
But what makes her think that "I would swap pretty fast" if there were 100 doors, but be less inclined to do so if there were 3?

Because three doors isn't enough for it to be clear to the conststant that it matters whether or not Monty is deliberately avoiding the prize, and if he is, that there is advantage to switching. With 100 doors, it becomes intuitive. (Usually.)

Yes, but his intention plays no part. The choice is to swap to a remaining door. (edit)
If Monty is forced to reveal a goat, then the contestant always learns to swap to his advantage.
If Monty is forced to reveal the car, the contestant learns he has won the car, so don't swap, or he has lost, so don't bother to swap.

Therefore, Monty's intent makes no difference to the advantage of swapping, and it is only advantageous to do so, if a goat is revealed.

Nobody said anything about Monty being obliged to reveal a goat, or alternatively being obliged to reveal the car. If he is obliged to reveal the car, there is no choice offered, and the game is at an end, obviously.

The question is whether he was obliged to reveal a goat, or whether he chose which of the two remaining doors he was going to open, at random. If it was the latter, there is no advantage to switching.

ETA: Humber: The specific thing you're missing is that possible scenarios are getting culled out in the cases where you do not believe it should matter. The above tables show exactly which cases. When you're told something happened and that means a portion of your probability space goes away, your probabilities should update.

Uh, yes, exactly.

Rolfe.

 

[This is The End]

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.

<snip>

ETA:
Or perhaps, your first choice is the first garage, but the light is off, so you don't see it, Jeeves shows you an empty garage. If you then turn the light on, 1/3 of the time you will find the car, if you don't, and try the remaining one, you will find it 2/3 of the time.

Another nice visual explanation.

 

[This is The End]

<snip>

Nice charts. Everything looks correct there for all 3 of those versions of the game.

 

[Rasmus]

I think it's about information, and when it becomes available, but it's only what the contestant knows and when that counts, and Monty and the goats have nothing to do with it.

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.

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. [/quote]

Right.

How different is that from the Monty Hall game?

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.

When the contestant makes the first guess, that is not really relevant.

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.

When Monty shows the goat, then the game begins. Being shown the goat, puts you in the first chosen empty garage, and from there, you chose another.

No.

The first chosen Garage *might not be empty*! But in the Monty game the rules state that it will always be empty. Your choice of a garage is entirely random, Monty's choice is never entirely random and might be entirely forced.

ETA:
Or perhaps, your first choice is the first garage, but the light is off, so you don't see it, Jeeves shows you an empty garage. If you then turn the light on, 1/3 of the time you will find the car, if you don't, and try the remaining one, you will find it 2/3 of the time.

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.

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

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.

Regardless of which door you check first, 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.

 

[Rolfe]

Having watched a number of people work through this, it seems to me there are definded stages that are gone through as the parameters of the problem become clearer. I went through the same stages myself.

1. Immediate "intuitive" belief that the probabilities haven't changed and the prize is no more likely to be behind the remaining closed door than the door originally chosen.
2. Deeper examination of the problem, which is often informed by considering the "Monty Hall card game" version or the one with 100 doors and one prize, reveals that in fact switching doubles the chance of winning, given the assumption which is usually in force at this stage (that is that Monty has deliberately avoided revealing the car, and will always reveal a goat in every iteration of the problem).
3. However, the puzzler then tends to figure out, or have it pointed out to him, that this solution is only valid if indeed Monty was deliberately avoiding the prize, and that if he actually opened his door at random, the probabilities of the two remaining doors are 50/50.
4. Consternation! How can it be that Monty's intention has a bearing on the probabilities? Well, quite easily actually. Strictly, Monty's intention defines the set of possible outcomes you are selecting from, with one set giving the advantage to the switch, while the other remains neutral.
5. But then - if we're considering Monty's intention, surely we have to consider all possible intentions he might have! Maybe he dislikes you, and is trying to steer you away from the prize, so intended to offer you the switch only if you chose right the first time! Maybe he likes you, and is offering you the chance to switch because you got it wrong!
6. Utter exasperation. If these intentions are part of the puzzle, it's unsolvable, end of story. Why did I waste my time with it?

That, in my view is the defect in the way the problem is formulated. There is no explicit declaration that Monty will not use his knowledge of whether or not you already chose correctly to influence his subesquent action. Up till then, it's a fair problem, with two possible solutions depending on the assumptions made. But examining these assumptions raises the question of whether Monty may in fact be being deliberately manipulative, and at that point it is realised that this is not exluded, and if it is not, then the puzzle is completely pointless.

That's why I liked Tim Mann's contribution. If you simply make it clear that whatever Monty does, it is not influenced by the knowledge of whether you picked the prize first time, the problem is valid, and soluble, and interesting.

Put it another way. There are four possible strategies Monty might be employing.

• Consistent Monty will always show you a goat, every time.
• Random Monty chooses his door randomly, without reference to what is behind it. (It's the randomness that's important, not whether or not he knows what's behind the doors - simply saying he knows what's behind the doors doesn't prevent him from revealing the prize.)
• Helpful Monty will only show you a goat if you didn't pick the prize first time.
• Evil Monty will only show you a goat if you already picked the car.

The real-life Monty probably varied between all of these from show to show (leading to interesting possibilities as regards second-guessing him), but in any one example, he has to be following one of these strategies. To figure out what the probabilities are, one then imagines a series of iterations, all with Monty playing to the same rule, and figures out the distribution of outcomes after the point in the game when Monty has revealed a goat.

Stipulating for the purposes of the intellectual puzzle that Monty will not allow his knowledge of whether you already picked right to influence his actions, thus eliminates both Helpful Monty and Evil Monty, which are the possibilities that make the problem unsolvable.

This leaves Consistent Monty and Random Monty. Obviously, when playing Consistent Monty you should switch to maximise your chances, but when playing Random Monty switching is a neutral move.

Thus, if you don't have the information as to whether you're playing Consistent or Random Monty, and there is a possibility you are playing Consistent Monty, then the correct answer is to switch, because you can't harm your chances, and may be increasing them.

And I really can't see what all the fuss is about beyond this. Mathematical papers, indeed.

Rolfe.

 

[Vorpal]

Not in disagreement, but two things to add:
-- The original problem statement excluded Consistent Monty, although what is now known as "The Monty Hall Problem" has this consistency.
-- Random Monty comes in at least two interesting flavors: one that randomly picks a door you didn't pick and one that just randomly picks a door. Whether switching is advantageous depends on flavor.

 

[Rolfe]

Not in disagreement, but two things to add:
-- The original problem statement excluded Consistent Monty, although what is now known as "The Monty Hall Problem" has this consistency.

If there is no possiblilty of Consistent Monty, it's not a very interesting problem. It's the non-intuitive results of the Consistent Monty behaviour that make it worth discussing.

-- Random Monty comes in at least two interesting flavors: one that randomly picks a door you didn't pick and one that just randomly picks a door. Whether switching is advantageous depends on flavor.

As far as I can tell, it doesn't matter which flavour of "Random Monty" is in force for the purposes of the scenario under discussion. Sure, he could have opened the already-chosen door, but that merely equates to ending the game without offering the opportunity to switch. (I can't imagine a game where it was revealed to the contestant what was behind the door they picked, and they were still allowed to switch - once you've "opened the box", that's it.)

In the scenario under consideration, we know Monty didn't do that, so the problem as it stands (for any flavour of Random Monty) is the one where he has in effect randomly chosen between the two remaining doors.

Rolfe.

 

[Rolfe]

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

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.

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

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.

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.

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.

 

[Vorpal]

As far as I can tell, it doesn't matter which flavour of "Random Monty" is in force for the purposes of the scenario under discussion. Sure, he could have opened the already-chosen door, but that merely equates to ending the game without offering the opportunity to switch.

Um... no?

(I can't imagine a game where it was revealed to the contestant what was behind the door they picked, and they were still allowed to switch - once you've "opened the box", that's it.)

Why not? It's like being allowed having two picks: one decided entirely by you and one by a roll of a die (mod 3). If either of them get a car, you win, and you're allowed to change the one you control if you don't like the die's outcome, which is revealed first. Those are reasonable game rules, if a bit biased toward you, but hardly ones that are unimaginable. The only change to a realistic "game show" format would be that the contestant rolls the die rather than Monty, but that's not mathematically relevant.

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.
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.

In the scenario under consideration, we know Monty didn't do that, so the problem as it stands (for any flavour of Random Monty) is the one where he has in effect randomly chosen between the two remaining doors.

Nope. The way he chose the door makes can make a difference of 1/9 to your chances if your strategy is to stay with your choice when you see a goat that's not the door you picked.

 

[Rolfe]

OK, fair enough, I hadn't thought about that one in detail but I will now.

Rolfe.

 

[yy2bggggs]

Nice charts. Everything looks correct there for all 3 of those versions of the game.

Wrong!

In my explanation of the "Monty always picks lowest numbered door, unless it's a car, then highest" case, I have columns B and C swapped. It should read:

Overall chance of winning if you swap is still 2/3. But once you know which door Monty opens, the probability changes. If he opens 3 (column C B only), then swap for a sure thing. Probability is outright 1. Otherwise, if he opens a 2 (columns A and B C), you have a 1/2 chance of winning, accept or not.​

 

[zooterkin]

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.

I don't think Monty opening a door, in the classical version of the problem, actually tells you anything about what's behind the other doors. It's more a case that because he always opens a door with a goat that you can work out that he is not giving you information about your door or the other unopened door, and therefore is not altering the a priori probabilities.

 

[Rasmus]

I don't think Monty opening a door, in the classical version of the problem, actually tells you anything about what's behind the other doors.

A car and a goat.
Behind those two doors, it could have been two goats as well, before Monty opened his other door.

It's more a case that because he always opens a door with a goat that you can work out that he is not giving you information about your door or the other unopened door, and therefore is not altering the a priori probabilities.

I am not sure I follow. Yes, I need to know how he will always act, but because I know this his actual actions give me more information.

 

[Jack by the hedge]

Put it another way. There are four possible strategies Monty might be employing.

• Consistent Monty will always show you a goat, every time.
• Random Monty chooses his door randomly, without reference to what is behind it. (It's the randomness that's important, not whether or not he knows what's behind the doors - simply saying he knows what's behind the doors doesn't prevent him from revealing the prize.)
• Helpful Monty will only show you a goat if you didn't pick the prize first time.
• Evil Monty will only show you a goat if you already picked the car.

I might be tempted to add one variation: Evil Monty who knows everyone knows he doesn't want them to get the car, and that they can learn his tactics.

In the "simple" Evil Monty strategy, if you guess wrong he always opens your door and you lose straight away, 2/3 of the time, and if you guess right, he always opens another door to show you a goat. Everyone learns not to switch, so you win 1/3 of the time.

Maybe he can do better: (In the spirit of making assumptions about how the game can be played, I'm here assuming Monty may either a) open one of the remaining two doors to reveal a goat and offer you a swap or b) let your original choice stand.)

If he modifies his Evil Monty tactic to also offer a swap half the time you initially choose wrongly, then you lose more often. (He flips a coin before the start to decide whether he'll offer you the swap if you're wrong.)

So now;
a) 1/3 of the time you choose wrongly and he doesn't offer a swap
b) 1/3 of the time you choose wrongly and he does offer a swap
c) 1/3 of the time you choose right and he offers a swap

With a) you lose. With b) or c) you have a dilemma. Both scenarios are equally likely to occur, so you have no way of assessing whether you face scenario b) or c).

If your strategy is to swap, in scenario a) you lose anyway, in scenario b) you have a 50% chance of winning and in scenario c) you lose. So swapping gives you an overall chance of winning which is 50% of 1/3. i.e. you only win 1/6 of the time.
If your strategy is to stick, in scenario a) you lose anyway, in scenario b) you lose and in scenario c) you have a 50% chance of winning. So sticking also gives you an overall chance of winning which is 50% of 1/3. i.e. you only win 1/6 of the time

So Evil Monty can be twice as evil if he sometimes offers you a swap when you chose wrong!

<edit> DOH! the above was wrong. In scenario c) you always win if you don't swap. I got a bit carried away with myself. Perhaps there still is a maximising strategy for Evil Monty, if he doesn't make b) vs c) a 50:50 chance... Need to think about that. Sorry.