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The Monty Hall problem

You are on a game show, and Monty shows you three closed doors. Behind one there is the prize, a car, and behind the other two there are booby-prizes, in this case goats. You are invited to choose a door, and do so.

Monty has two basic options now. He can open a door so as to end the game then and there - that is, either by opening the door you chose and reveal whether you won or not, or opening one of the doors you didn't choose, to reveal the car (too bad, you lost). Alternatively, he can open one of the doors you didn't choose, and reveal a goat, and if he does this, he will offer you the opportunity to switch your choice to the other unopened door.

Monty is operating under only one constraint. He has to make up his mind whether or not he will offer you the chance to switch, before you choose your door.

Should you switch?


Now there's a possible variant on that, which I didn't include. But how does it play out if I alter that third paragraph to read,

"Monty is operating under only one constraint. He has to make up his mind whether or not he will offer you the chance to switch, or alternatively whether he will open a door completely at random, before you choose your door."

Again, discuss.

Rolfe.
 
To remove ambiguity, I assume you mean Monty has to make up his mind before you choose your door whether or not he will offer you the chance to switch, or alternatively whether he will open a door at random.

My first thought is that even if he consistently acts with the same probability of making each choice, however much you study his behaviour you can't work out that probability because you can't tell the random acts from the deliberate ones. For any particular pattern of wins, losses and switches (with and without having chosen correctly) there might have been zero random acts, 100% random acts or any value in between.
 
I was trying for a more structured presentation of the original problem. I think that the second version is like the original problem, but removes the possibility of Biassed Monty.

The answer then being that you should switch. Because switching cannot harm your chance of winning (neutral if in fact he made a random choice), and will double it if he made a deliberate decision to offer you the choice.

And no you can't tell which he was doing on any one occasion.

Rolfe.
 
Well, maybe not that bored.
I was just thinking about a different formulation of the problem.
You are on a game show, and Monty shows you three closed doors. Behind one there is the prize, a car, and behind the other two there are booby-prizes, in this case goats. You are invited to choose a door, and do so.

Monty has two basic options now. He can open a door so as to end the game then and there - that is, either by opening the door you chose and reveal whether you won or not, or opening one of the doors you didn't choose, to reveal the car (too bad, you lost).

I have already covered that. "very evil Monty"/"very benevolent Monty".
Monty ends the game as decribed, and allows the contestant to take the prize or not. The distinction between the two is trivial.

Alternatively, he can open one of the doors you didn't choose, and reveal a goat, and if he does this, he will offer you the opportunity to switch your choice to the other unopened door.
Standard format.

Monty is operating under only one constraint. He has to make up his mind whether or not he will offer you the chance to switch, before you choose your door.
Should you switch?

Phenomenology is not an influence. What Monty thinks, is hidden from all but himself. He can only make that mental content explicit by action.
The answer is in my reply about him knowing or not.

In the standard game, he expresses that knowledge by showing the contestant a goat that is behind one of the remaining doors.
He may know where he car is or not, but as far as the contestant is concerned, he sees a goat. If that happens, before or after the choice is made, the rule is simple: "see a goat, swap." That is the contestant's best strategy. The worst pay off is 1/2, and the best 2/3.

The following stands, if the offer to swap is made or not.

1) If the goat is behind the chosen door and revealed, the odds are then 1/2 for each of the other 2 doors.

2) If the goat is behind one of the remaining doors and revealed, the odds are 1/3 and 2/3 for the remaining doors.

3) If a goat is revealed before the contestant's choice, the odds are 1/2 for each of the remaining doors.

It is a mistake to see Monty has "knowing where the car is". It is just as valid to say he knows where the goats are.

Very Evil Monty and Very benevolent Monty have counterparts in the contestant. The contestant who will not take the prize, or refuses to guess etc. They can be excluded on the basis of rational operation.

If the above oddities are excluded, and Monty and the contestant are seen as opponents; the contestant wants to win the car, Monty wants to keep it, then it is then a zero-sum game.

If each opponent plays his optimum strategy, they can tell each other what they plan to do, and neither can gain from that. The minimum pay off for the contestant is 1/2, and that is the best case for Monty. Monty's participation in the game raises the contestant's chances from 1/3 to 1/2, not matter what.

ETA: To make the format less ambiguous, don't use "winning" or "swap", but the probability of any particular door containing the car. "knowing" is dodgy too.
 
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For any particular pattern of wins, losses and switches (with and without having chosen correctly) there might have been zero random acts, 100% random acts or any value in between.


I think you could work it out in retrospect. Figure out how often contestants who were offered the switch would have won the prize if they had switched. If he's being completely random, then it should be half of them. If it's more often than that, some of the offers were deliberate. If it's as much as 2/3rds, all of them were deliberate.

Rolfe.
 
I think you could work it out in retrospect. Figure out how often contestants who were offered the switch would have won the prize if they had switched. If he's being completely random, then it should be half of them. If it's more often than that, some of the offers were deliberate. If it's as much as 2/3rds, all of them were deliberate.

Rolfe.

The game is contentious because of the use of word. It takes the meaning of "probability" "intent" and "winning" to breaking point, and then dances on the consequences.
If "winning" is replaced by "the car is revealed" then a great deal of ambiguity is removed.

The matters of intent and random selection, are a straw men. The game may be played such that the contestant can select one of three doors. If that door is opened to reveal the car, the contestant "wins".

The game may be played like that for many iterations, but occasionally, and after the contestant has selected the door ( but the contents not revealed), a bell rings. At that point, Monty offers to the contestant that he will open one of the remaining doors that does not contain the car. "should you let him do so?" It's a no-brainer.

ETA:
The same can be said of Monty offering to open a door that reveals the car such that the contestant wins or loses, and before or after the initial choice. The contestants reaction depends not on how much he knows of probability, but what may be learned from former iterations of the game.

The Bayesian approach can't be denied, but it seems to me that the dispute as to which version is right - if conditional or unconditional probability should be applied - depends solely on the individual's interpretation of some key words.
This game may well be the silliest storm in a tea cup every to hit the mathematical world.
Mathematicians should learn from game theorists, and perhaps both from Gilbert Ryle.
 
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OK, here's my amended wording.

You are on that infernal game show, and Monty shows you three closed doors. Behind one there is the prize, a car, and behind the other two there are booby-prizes, in this case goats. You are invited to choose a door, and do so.


Monty, who knows what is behind the doors, has three basic options.
  • He can decide to open any door competely at random.
  • He can decide to open a door so as to end the game then and there - that is, either by opening the door you chose and reveal whether you won or not, or opening one of the doors you didn't choose, to reveal the car (too bad, you lost).
  • He can decide to open one of the doors you didn't choose, and reveal a goat.
If he opens one of the doors you didn't choose, and reveals a goat, he will then offer you the opportunity to switch your pick to the other unopened door.

Monty is operating under only one constraint. He has to make up his mind which of these three options he is going to take, before you pick your door.

If you get the opportunity to switch doors, should you switch?

The trouble is, it gets wordy. One really wants to make the ground rules as clear as possible, and yet the wording succinct. Sigh.

Rolfe.
 
I think you could work it out in retrospect. Figure out how often contestants who were offered the switch would have won the prize if they had switched. If he's being completely random, then it should be half of them. If it's more often than that, some of the offers were deliberate. If it's as much as 2/3rds, all of them were deliberate.

Rolfe.

That's true up to a point. If the proportion who should have switched is 2/3 (or zero) then the probability that any of Monty's acts were random declines toward zero. If it's any other value, you can't tell what proportion of the choices were random (steering the result towards 1/2) and what proportion were deliberate (steering the result toward some other, undefined value).
 
Are you sure? If the proportion who would have won by switching is 2/3rds, then he was almost certainly choosing to offer the switch every time. If the proportion was 1/2, then surely he was picking doors at random every time. For values between 1/2 and 2/3rds, then you're on a sliding scale of how often he's making the deliberate offer.

Rolfe.
 
If you get the opportunity to switch doors, should you switch?

Yes.

If Monty made the decision in advance, then there are three possibilities.

(1) He chooses to end the game after your choice, in which case there's no opportunity to switch.
(2) He chooses in advance to open a door at random.
(2a) If he opens the door with the car, you win or lose immediately, so there's no opportunity to switch.
(2b) He opens a door with a goat. We can argue all day over this one, but there's no rational argument that gives a worse than 50% chance if you switch.
(3) This is the classic Monty Hall formulation, in which he decides in advance to open a door with a goat behind it. In this case, your chance in switching is 67%.

Overall, your chance on switching is at least 0.5*A/(A+B) + 0.67*B/(A+B), where A and B are the probabilities of cases 2b and 3. This cannot be less than 0.5, so the winning strategy is always to switch if you get the chance.

Dave
 
With the amended wording, Monty loses the capacity for Evil. So you ought to switch if offered the choice as your chances will then be either 1/2 or 2/3 compared to your original chance of 1/3.
 
Are you sure? If the proportion who would have won by switching is 2/3rds, then he was almost certainly choosing to offer the switch every time. If the proportion was 1/2, then surely he was picking doors at random every time. For values between 1/2 and 2/3rds, then you're on a sliding scale of how often he's making the deliberate offer.

Rolfe.

That depends on whether he's allowed to base his deliberate choices on whether you chose right or not. Are we excluding that?

Edit to add: If he offers you a choice when you were wrong, you switch and win. If he offers you a choice when you were right, you switch and lose. So if he tailors his offers to the same number of winning and losing selections, the result can be 100% deliberate yet will be indistinguishable from his choosing randomly. In that case, of course, you can treat the game as if he always acted randomly so it makes no difference whether you switch or not.

In the end, it probably doesn't matter how many choices were random. If you have to watch a very large number of games to try to work it out, you can just count the number of times it paid to swap and get the right answer that way.
 
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Yes.

If Monty made the decision in advance, then there are three possibilities.

(1) He chooses to end the game after your choice, in which case there's no opportunity to switch.
(2) He chooses in advance to open a door at random.
(2a) If he opens the door with the car, you win or lose immediately, so there's no opportunity to switch.
(2b) He opens a door with a goat. We can argue all day over this one, but there's no rational argument that gives a worse than 50% chance if you switch.
(3) This is the classic Monty Hall formulation, in which he decides in advance to open a door with a goat behind it. In this case, your chance in switching is 67%.

Overall, your chance on switching is at least 0.5*A/(A+B) + 0.67*B/(A+B), where A and B are the probabilities of cases 2b and 3. This cannot be less than 0.5, so the winning strategy is always to switch if you get the chance.


That's what I thought. I just struck out the possibilities that were already off the table as were were only considering what you should do if you got the chance to switch.

The thing is, although you see 3 as "the classic Monty Hall formulation", the usual wording doesn't exclude the possibility that he simply opened that door at random, or indeed was deliberately trying to trick you into switching away from the prize.

With the amended wording, Monty loses the capacity for Evil. So you ought to switch if offered the choice as your chances will then be either 1/2 or 2/3 compared to your original chance of 1/3.


Of course, he also loses his capacity for benevolance, so he can't kindly offer you the switch when you chose wrong, just because he likes your bonny blue e'en.

That depends on whether he's allowed to base his deliberate choices on whether you chose right or not. Are we excluding that?


Yes, it's in the wording. He has to decide which road he's going to take (random door, end the game, or offer choice) before you choose your door.

Rolfe.
 
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Of course, he also loses his capacity for benevolance, so he can't kindly offer you the switch when you chose wrong, just because he likes your bonny blue e'en.
Rolfe.

And you say Snoopy is....
Intuitionists can never withstand a challenge to that very thing.
 
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Conclusion: this form of communication is redundant, since it cannot make people agree on basic problem as this one.

The problem itself, dines out on that. I read that just last year, an error was found in a peer-reviewed paper of 20 years ago.
 
I will prove to all that this problem is not of probability but a simple game. Not unlike the schoolboy game of "Battleships"

This reasoning is false:

"...the competitor's initial chance is 1/3, Monty reveals a goat, so the probability of the car being behind the remaining door is 2/3."

There are simulations will support the above, and but have you ever seen one that says anything else?

The Monty Hall game as posed by Vos Savant, is a two player, two stage, zero sum game.

Simple Rules
1) Monty wants to keep the car
2) The contestant (Dave) wants to win it.
3) Monty does not need to offer, but there is a second stage.


Only two cases need be examined.

Dave's first choice is correct.
Monty reasons that his only choice is to lure Dave away from that winning door.
Opening two doors is no good because that tells Dave he has won.
Opening Dave's tells Dave he has won.
That leaves only one goat door as an opportunity for Monty.

Dave's first choice is incorrect.
Monty reasons that anything he does, can only improve upon Dave's position.
Opening two doors is no good. That tells Dave he has won.
Opening the car door is no good. That tells Dave he has won.
That leaves only one goat door as an opportunity for Monty.

What should Dave do? Stick or swap?

1/3 of the time stick. 2/3 time, swap.
Isn't that the same advice?

Both competitors are employing Von Neuman's minimax game strategy.
A characteristic of such games is that if each competitor plays his optimum strategy, both can share the same information, and neither can gain from it improve upon the optimal strategy.

New rule: Monty tells all.

Only two cases need to be examined.

Dave's first choice is correct.
Monty tells Dave that.
Dave sticks, meaning that he wins.

Dave's first choice is incorrect.
Monty tells Dave that.
Dave swaps, meaning that he wins.

The first occurs 1/3 of the time, the second 2/3 of the time.

The "intuition" that one must "get", and be foolish to object to, is trivial, isn't it?

The falsehood of the Vos Savant argument can be seen if you think about what must actually happen. If it is true that 2/3 of the time, the car is behind the "other" door, it must actually be there, 2/3 of the time.
That is also why the 100 door argument is false.

If somebody writes a simulation, where the car is randomly seeded, and then employs a state machine to emulate Monty, it will "prove" the "probability" has shifted.

ETA: Vos Savant's version refers to "the other" doors meaning that Monty will not open the contestant's door, so the above optimal strategy applies to that game.
If Monty is allowed to open the contestant's door he can force a 1/2 risk onto Dave, when Dave has chosen a goat.
Again, Monty can tell Dave that he has not won, and Dave will reason he should try the other doors. Exactly the same as if Monty opens the door to reveal the goat.
 
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It's only 7 pages. They're just getting warmed up.
 
It's only 7 pages. They're just getting warmed up.

No, what have "they" being doing when trying to define the game? Falling over themselves with corrections, and then again.
Quarky says that this is like ddwfftw. That similarity may be because both use arguments around intuition - Monty Hall has been used to support the ddwfftw's intuition argument.

But this one is a bit different, and it's Vos Savant, who is the cult leader, and peddler of intuition.
Martin Gardner also posed the same puzzle (in a different form), and wrote a large number of puzzle books, yet there are no cases of academics calling him an idiot, nor any "Martin is Wrong" sites.

There have been many papers written about this problem, and they all refer at one time or another, to Vos Savant. Why?
 

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