Please Explain (actually explain) The Monty Hall Problem

[FactChecker]

None of Monty's motions changes the 1/3 chance you have of having picked the prize door.

If you lose when you switch, it's because you picked the prize door and switched; but whenever (2/3 chance) you don't pick the prize door, switching always wins.

That's a good way to look at it. Your original door still has 1/3 chance of winning. That leaves 1-1/3 = 2/3 chance and only one door for the 2/3 chance. So it is best to switch to that door.

ADDED: What makes your door different from the other remaining unopened door is that the rules didn't allow Monte to open your door. So we learn nothing from the fact that he didn't open your door. On the other hand, he was allowed to open the other remaining unopened door if it doesn't have the prize. So we should adjust our judgement about that door.

 

[Hornbein]

In such a case one may resort to experiment.

 

[PeroK]

Monty doesn't open the door with the car, and he doesn't open the contestant's selected door. So he always opens a non-car non-selected door.

This is the critical assumption/characteristic of the game. First, Monty's pick is not random (he never reveals the car); and second, he always offers a switch (not only when he knows you have the car!).

A more general problem would involve Monty opening a door and offering a switch with probability $p$ if the contestant has picked the car door, and a probablity $q$ if the contestant has not picked the car door.

In that case, given that a switch has been offered, sticking wins with probablity $\frac{p}{p + 2q}$; and switching wins with probablity $\frac{2q}{p + 2q}$.

And it's better to switch if $2q > p$.

The problem you are analysing has $p = q = 1$.

If $p = 1, q = 0$, then it's better to stick.

And, if $p = 1, q = 1/2$, then sticking and switching have equal likelihood of winning. I like this twist!

 

[pinball1970]

Nice thread, I do not understand all of the explanations, especially the entropy one! (wow!)

I have an explanation of why you should always switch, it would be good if the maths guys gave a view, I would say it is more of a qualitative/ intuitive explanation, although there are some numbers in there.

At the beginning of the game you are more likely to pick a goat simply because there are more goats than cars, only 33% chance of a car. So the car is always more likely to be in the other 67% that you did not pick i.e. one of the other 2 doors.

So when half of that 67% is revealed as goat, it is then logical to switch from your door to that other door.

 

[PeroK]

At the beginning of the game you are more likely to pick a goat simply because there are more goats than cars, only 33% chance of a car. So the car is always more likely to be in the other 67% that you did not pick i.e. one of the other 2 doors.

So when half of that 67% is revealed as goat, it is then logical to switch from your door to that other door.

That argument fits the Monty Hall game, as generally understood. But, there are alternative ways to play where that argument fails. The obvious one is where Monty only offers a switch when he knows you have picked the car door.

And, the subtler one is the so-called Monty Fall problem, where Monty opens a door at random that just happens to have a goat behind it. An equivalent scenario would be where there are two players. You pick a door first. Then, your opponent picks a door, which is opened to reveal a goat. In this game there is no advantage in switching.

That's why you have to establish what is happening and why. Then calculate on that basis.

 

[FactChecker]

And, the subtler one is the so-called Monty Fall problem, where Monty opens a door at random that just happens to have a goat behind it. An equivalent scenario would be where there are two players. You pick a door first. Then, your opponent picks a door, which is opened to reveal a goat. In this game there is no advantage in switching.

That's why you have to establish what is happening and why. Then calculate on that basis.

This is the point that is often missed. Suppose Monte Hall just blindly opened one of the remaining two doors and it happened to have a goat. Then your door and the remaining closed door have the same probability of having the prize. It is the intentional avoidance of opening a door with the prize that increases the probability of the remaining closed door being more probably a winner than your door.

I like to consider more extreme cases, like if we started with 100 doors, you picked one. Your door only has 1/100 chance of having the prize. Suppose Monte Hall opened another 98 losing doors, leaving one closed door other than yours. If that was just blind luck, it was an extremely unlikely event. Given that unlikely event, both your door and the remaining door have equal 1/2 odds. On the other hand, if Monte Hall is not allowed to open a prize door, the fact that none of the 98 doors had the prize is no longer an unlikely event; it was guaranteed. That gives an obvious clue that the remaining door has the prize. In fact, it has a 99/100 chance of having the prize.

 

[javisot]

This is the point that is often missed. Suppose Monte Hall just blindly opened one of the remaining two doors and it happened to have a goat. Then your door and the remaining closed door have the same probability of having the prize. It is the intentional avoidance of opening a door with the prize that increases the probability of the remaining closed door being more probably a winner than your door.

It is not essential that Monty opens a door with a goat consciously or unconsciously, it is irrelevant, the important thing is that he opens a door with a goat.

 

[Dale]

It is not essential that Monty opens a door with a goat consciously or unconsciously, it is irrelevant, the important thing is that he opens a door with a goat.

It is relevant because it changes the conditional probabilities and the amount of information received. There is still some information obtained anyway, but it is less if Monty selects randomly.

 

[pinball1970]

It is relevant because it changes the conditional probabilities and the amount of information received. There is still some information obtained anyway, but it is less if Monty selects randomly.

How would that change the decision Dale? If Monty opened randomly and produced a goat or could see beforehand and chose a goat.

I would still change my door if a goat was revealed?

EDIT: Just to add, if Monty chose blind and revealed the car then the game is over, so the above scenario, goat revealed should mean I change.

 

[Dale]

How would that change the decision Dale?

I haven’t run the numbers for all of the variations that @PeroK describes. But my educated guess is that the optimal decision will not change, but the expected winning will change. It will still be better to switch, but most variations will have a win percentage less than 2/3.

 

[PeroK]

How would that change the decision Dale? If Monty opened randomly and produced a goat or could see beforehand and chose a goat.

It changes the probabilities because it changes the relative frequencies, which one can calculate.

In other words, you play the game many times, you count what happens and the numbers tell you the probabilities. If the measured probabilities differ in two scenarios, then you can't argue against it. Your intuition has to be brought into line with what the numbers tell you.

I would still change my door if a goat was revealed?

In this case, it makes no difference. There is a 50% chance that you have the car and a 50% chance that the car is behind the third door. This can be calculated by playing the game many times.

Alternatively, the direct calculation, without recourse to experiment is:

1/3 of the time you pick the right door
1/3 of the time Monty reveals the car and the game is over
1/3 of the time the car is behind the third door

If we exclude the cases where Monty reveals the car, then the probability of winning with your original door is equal to the probability of winning by switching.

Technically, the conditional probability of both increase to 1/2, given that a random selection of the other door revealed a goat.

The moral of the story is that random data is not the same as non-random data. In other words, it makes a difference if Monty knows where the car is and always avoids it.

 

[FactChecker]

It is not essential that Monty opens a door with a goat consciously or unconsciously, it is irrelevant, the important thing is that he opens a door with a goat.

Really? So knowing that he may have intentionally avoided that one door doesn't clue you in and change the odds? What if it started with 100 doors, he opened 98 doors, all goats, and left one unopened? Would that give you a clue?
Of course it would. Apply Bayes' Rule. There you can see the clear difference:
P(open door 1 | prize in door 1) = 0.

 

[Ibix]

Really? So knowing that he may have intentionally avoided that one door doesn't clue you in and change the odds? What if it started with 100 doors, he opened 98 doors, all goats, and left one unopened? Would that give you a clue?
Of course it would. Apply Bayes' Rule. There you can see the clear difference:
P(open door 1 | prize in door 1) = 0.

I think it depends what you do with games where he accidentally reveals the car. If you strike them out and restart the game you get a different result from if he says "you have to switch to the closed door or stick, thems the rules, so you lose".

 

[javisot]

It is relevant because it changes the conditional probabilities and the amount of information received. There is still some information obtained anyway, but it is less if Monty selects randomly.

Maybe I didn't express myself well Dale, obviously it's important that Monty chooses a door with the goat, but it's not important that he does it consciously. You can say that Monty will always open a door with a goat in it by chance, that doesn't change anything.

The Monty Hall problem shows that if you have a set A with 1 box (1/3) and a set B with 2 boxes (2/3), under the conditions described in the problem, it is incorrect to state that by removing one box from set B the probability evolves to: set A with 1 box (1/2) and set B with 1 box (1/2). The correct statement is that set A still has a probability of (1/3) and that set B, now with 1 box, still has a probability of (2/3).

The conditions of the problem include, as Perok mentions, assuming that Monty will open a door with the goat (it doesn't matter if random or not random)

 

[DaveC426913]

In such a case one may resort to experiment.

Or a simulation. :)

 

[FactChecker]

I think it depends what you do with games where he accidentally reveals the car. If you strike them out and restart the game you get a different result from if he says "you have to switch to the closed door or stick, thems the rules, so you lose".

Yes. Restarting the game is the same as making a rule that he can not open the door with the prize. Any time he violates that rule, the game is invalid and start over.

 

[FactChecker]

Maybe I didn't express myself well Dale, obviously it's important that Monty chooses a door with the goat, but it's not important that he does it consciously. You can say that Monty will always open a door with a goat in it by chance, that doesn't change anything.

Yes, it does make a difference. There is a difference between an accidental result of rare luck versus the intentional act of avoiding opening one door. Consider the example (again) of the game with 100 doors. In the first case of luck, the only inference that can be made from opening 98 doors without revealing the prize is that a rare, lucky event was just witnessed. But it might be just as lucky that the one door out of 100 that you picked has the prize. So it implies the same strange luck equally for both doors. The revised probabilities for both doors is 1/2.

In the second case, where it is a rule that Monte can not open the door with the prize, we can infer that the remaining unopened door might have been skipped because it has the prize. That does not change the odds that your door has the prize. Your door still has the initial probability of 1/100 and the remaining unopened door has a probability of 99/100.

In any case, rather than continuing to restate the same position, the calculation using conditional probabilities should be studied. There, it is clear that the rule P(Monte opens door X | door X has the prize) = 0 makes all the difference.

 

[javisot]

Yes, it does make a difference. There is a difference between an accidental result of rare luck versus the intentional act of avoiding opening one door. Consider the example (again) of the game with 100 doors. In the first case of luck, the only inference that can be made from opening 98 doors without revealing the prize is that a rare, lucky event was just witnessed. But it might be just as lucky that the one door out of 100 that you picked has the prize. So it implies the same strange luck equally for both doors. The revised probabilities for both doors is 1/2.

In the second case, where it is a rule that Monte can not open the door with the prize, we can infer that the remaining unopened door might have been skipped because it has the prize. That does not change the odds that your door has the prize. Your door still has the initial probability of 1/100 and the remaining unopened door has a probability of 99/100.

In any case, rather than continuing to restate the same position, the calculation using conditional probabilities should be studied. There, it is clear that the rule P(Monte opens door X | door X has the prize) = 0 makes all the difference.

You've explained the Monty Hall problem again, but you still haven't explained the quantitative difference you claim exists between "randomly, Monty always opens a door containing a goat" and "non-randomly, Monty always opens a door containing a goat." In both cases, Monty always opens a door containing a goat.

I fail to see the difference you're referring to.

The conditions are that Monty always open a door containing a goat and then offer a switch, regardless of whether Monty does it randomly or non-randomly.

 

[javisot]

You can do the following: we're no longer talking about doors, we're talking about trapdoors in the roof. These trapdoors are designed to withstand heavy loads, like a car, but not light loads, like a goat. After a certain time, the two goats will fall while the car remains on top. The initial distribution of the goats and the car is random.

1. Monty and the contestant stand beneath the three trapdoors.
2. Monty knows nothing about the contents of the trapdoors.
3. Monty asks the contestant to choose a trapdoor.
4. The contestant chooses a trapdoor.
5. The chosen trapdoor is then blocked to prevent its contents from falling out.
6. Monty tells the contestant to wait a while, as one of the goats will naturally fall out.
7. Now we have an open trapdoor that always contains a goat, and Monty didn't need to know the contents of the trapdoors.
8. After the goat falls out, Monty offers the contestant a switch.

9. What should the contestant do?

 

[Dale]

it's not important that he does it consciously

It is important because it changes the conditional probabilities. As a matter of notation, let’s call the door that the contestant chooses 1 and the door Monty chooses 2.

Let’s call the evidence of Monty opening door 2 $E_2$ and look at three hypotheses: $H_1$ the prize is behind door 1, $H_2$ the prize is behind door 2, and $H_3$ the prize is behind door 3.

If Monty knows which door contains the prize then the conditional probabilities are $$P(E_2|H_1)=1/2$$$$P(E_2|H_2)=0$$$$P(E_2|H_3)=1$$

If Monty does not know then the conditional probabilities are $$P(E_2|H_1)=1/2$$$$P(E_2|H_2)=1/2$$$$P(E_2|H_3)=1/2$$

 

[FactChecker]

I fail to see the difference you're referring to.

The conditions are that Monty always open a door containing a goat and then offer a switch, regardless of whether Monty does it randomly or non-randomly.

Have you done the math? Math is your friend in this. The conditional probabilities are completely different in the two situations. See Dale's post above, but you should do the complete calculation or use a probability tree.

 

[DaveC426913]

These trapdoors are designed to withstand heavy loads, like a car, but not light loads, like a goat.

So the door equivalent of a non-Newtonian fluid...

 

[javisot]

So the door equivalent of a non-Newtonian fluid...

No, that assessment doesn't make sense. You could say that above the trapdoors, the goats are hanging from ropes, supporting their own weight, so they'll eventually fall. Meanwhile, the car is securely attached to a rope and won't fall.

With that, you already have a system of three trapdoors that can drop the goats but not the car. The initial distribution of goats and cars is random for Monty; he doesn't know anything.

According to you, Dale, and Factchecker, in this version of the trapdoors, it doesn't matter whether the contestant accepts the switch or not, which doesn't make sense; they must accept the switch in this version as well.

 

[javisot]

It is important because it changes the conditional probabilities. As a matter of notation, let’s call the door that the contestant chooses 1 and the door Monty chooses 2.

Let’s call the evidence of Monty opening door 2 $E_2$ and look at three hypotheses: $H_1$ the prize is behind door 1, $H_2$ the prize is behind door 2, and $H_3$ the prize is behind door 3.

If Monty knows which door contains the prize then the conditional probabilities are $$P(E_2|H_1)=1/2$$$$P(E_2|H_2)=0$$$$P(E_2|H_3)=1$$

If Monty does not know then the conditional probabilities are $$P(E_2|H_1)=1/2$$$$P(E_2|H_2)=1/2$$$$P(E_2|H_3)=1/2$$

So Dale, are you saying that in this version with trapdoors and Monty's ignorance, the contestant is indifferent as to whether he accepts the switch or not?

 

[Ibix]

With that, you already have a system of three trapdoors that can drop the goats but not the car. The initial distribution of goats and cars is random for Monty; he doesn't know anything.

According to you, Dale, and Factchecker, in this version of the trapdoors, it doesn't matter whether the contestant accepts the switch or not, which doesn't make sense; they must accept the switch in this version as well.

I don't think this is the system they're thinking of. This version will randomly open a door containing a goat. The usual Monty Fall problem has Monty randomly opening a currently closed door - and this time he gets lucky and happens to open a goat. He could have opened a door containing a car, but by chance he didn't. That should alter your beliefs about what to do, because you no longer know that he wouldn't have opened the door with the car. Your trapdoor system shouldn't cause you to change beliefs because it's just a technological way of implementing Monty Hall's usual rules.

The long-run behaviour of Monty Fall isn't well defined, however, because no rules are specified for the case where the host trips and opens the door with the car - do you restart, do you win, or are you forced to make a losing move?

 

[javisot]

I don't think this is the system they're thinking of. This version will randomly open a door containing a goat. The usual Monty Fall problem has Monty randomly opening a currently closed door - and this time he gets lucky and happens to open a goat. He could have opened a door containing a car, but by chance he didn't. That should alter your beliefs about what to do, because you no longer know that he wouldn't have opened the door with the car. Your trapdoor system shouldn't cause you to change beliefs because it's just a technological way of implementing Monty Hall's usual rules.

You understand what I'm saying. So, Ibix, the conclusion is that the rule is what matters, not whether Monty does it randomly or not. Monty is perfectly replaceable by an automated system that doesn't depend on knowing the contents of the trapdoors.

 

[Ibix]

You understand what I'm saying. So, Ibix, the conclusion is that the rule is what matters, not whether Monty does it randomly or not. Monty is perfectly replaceable by an automated system that doesn't depend on knowing the contents of the trapdoors.

Well, the automated system defines the same rules. The strategy in any of the Monty Hall variants depends on knowing the rules and can be different if there are different rules. I agree that how the rules are implemented doesn't actually matter.

You can even play standard Monty Hall on your own with three playing cards and a coin (at least enough to identify whether you should have stuck or switched in that particular instance).

 

[PeroK]

There was a British Game show called "Deal or No Deal", which is interesting here. The essential point is that there are 22 boxes, each with a prize ranging from £0.01 to £250,000.

The main contestant picks a box at random. Then, gradually the other boxes are opened to reveal what the prize they hold - which is not available to the contestant. At various points the contestant is offered a certain amount for their box, which they may accept and the game ends. Or, they may continue opening more boxes. In general, the amount they are offered increases if none of the high-value prizes has been revealed.

On the one hand, the box they chose has a certain prize within it, which never changes. But, depending on what prizes have been revealed, the expected amount in that box does change.

This is related to the Monty Hall "paradox", although is just basic probability theory.

We can focus on the probability of having the £250,000 prize. At the start, the probability is 1/22. But, if 20 boxes have been opened without revealing this prize, then the probability that the box contains the biggest prize is now 1/2. This only occasionally happens. It would be wrong to think that the original box still only has a probability of 1/22 of having the big prize.

By contrast, if the game show host opened boxes and deliberately avoided the big prize (like Monty), then the probability that the original box contains the big prize remains 1/22 throughout.

The whole game hinges on the ever-changing probability and expected value of what is in the contestant's box. Even though what is in that box is fixed throughout. If the game is honest, then no one knows what is in that box. Neither the host, nor the "banker" who offers to buy the box and end the game.

 

[FactChecker]

You understand what I'm saying. So, Ibix, the conclusion is that the rule is what matters, not whether Monty does it randomly or not. Monty is perfectly replaceable by an automated system that doesn't depend on knowing the contents of the trapdoors.

Exactly. Good point. It is the "filtering" action of the rule that makes the difference. With the rule, you know that if the remaining door(s) might have the prize, that door will not be opened.

I like to use an even more extreme example to illustrate the process and why I like to call it a "filtering" process:
Suppose you have a bucket of sand with a diamond hidden in it. You are allowed to take a pinch out of the bucket. Your pinch has practically no chance of containing the diamond, say 1 in 1000. And the odds that the diamond is still in the bucket is 999 out of 1000. Now, a filter is placed over the bucket and all the sand except a pinch is filtered out. If the diamond was still in the bucket, the filter keeps the diamond in the pinch remaining in the bucket. The filter acts like the rule that Monte can not open a door with the prize. Now there is a 999 out of 1000 chance that the diamond remains in the pinch of sand remaining in the bucket. Clearly, if given the chance, you would exchange your pathetic 1 in 1000 pinch for the 999 in 1000 pinch remaining in the bucket.
ADDED: Now, compare that with the situation where no filter is used and all but a pinch is dumped out of the bucket. Suppose we determine that the diamond is not in the dumped sand. WHAT STRANGE LUCK! But was the strange luck that the diamond remained in the bucket or was it that your pinch has the diamond. They are equally likely.

 

[Dale]

So Dale, are you saying that in this version with trapdoors and Monty's ignorance, the contestant is indifferent as to whether he accepts the switch or not?

No. I am saying that the important thing is the conditional probabilities. And I am saying that in the regular approach Monty’s knowledge does change the conditional probabilities, so it is important.

I never made any claims that knowledge was the only way to set the conditional probabilities.