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".