*not*always best to switch doors after a gameshow host opens a door other than the one you originally pointed to, except under specific conditions which are often unstated (or even misstated) in the problem. The best strategy entirely depends on whether or not the gameshow host deliberately opened a door with a goat behind it, or whether he just happened to. Depending on the rules of the game and your knowledge of the habits of the gameshow host, it can either be best to switch, best to stay the same, or make no difference what you do. Since it's the gameshow host's behavior (and your implicitly assumed knowledge of his intentions) that clues you in to which door is the most likely to hide the car, this is the crux of the whole problem and is often overlooked. (It's intuitively clear by watching the flow of information from the host to you--perhaps if more people adopted the "conservation of information" postulate that underlies the many-worlds-interp this would be more obvious from the outset!) I've yet to see a proper explanation of this problem given on the web, and it seems that whenever I present this to someone in passing they almost never believe it on first hearing. But no-one so far who has taken the time to read my explanation below has been able to deny its validity. I feel like I should be making this an FAQ entry somewhere, but I'm just not sure where to post it. And since

**spinemasher**happened to mention the problem recently in his journal I thought it's a good time to bring it up.

#1, Classic goats and car problem:

----------------------------------------

Car is randomly placed behind one of three doors, goats are placed behind the other two. You initially pick one of the three doors.

Gameshow host opens one of the doors, but he knows where the car is and is forbidden from picking that one; he's also forbidden from picking yours. [note: these are the implicit assumptions

To simplify things, let's always label the door the car is behind "door #1", and the other two "door #2" and "door #3". (This makes it vastly more simple than labelling them based on their absolute positions; try not to confuse this labelling scheme with the "usual" one.)

4 possible situations, having unequal probabilities of occuring:

[1/6 chance] A. You pick door #1 (car), he opens door #2

[1/6 chance] B. You pick door #1 (car), he opens door #3

[1/3 chance] D. You pick door #2, he opens door #3

[1/3 chance] C. You pick door #3, he opens door #2

(notice situations A and B are just two equally likely sub-situations of you picking the car initially, which is a 1/3 chance.)

Therefore, using Strategy "Stay with same door" you will win

1/3 of the time (situations A+B).

And using Strategy "Switch doors" you will win

2/3 of the time (situations C+D).

So you should choose "Switch doors" strategy every time.

So far so good? Ok, now the tricky part...

#2, Modified goats and car problem:

--------------------------------------

Again, car is randomly placed behind one of the three doors, goats behind the other two.

Gameshow host may not pick your door, but this time he is not told where the car is ahead of time. He may use any scheme to decide which door to pick as long as it's not yours and he does not know where the car is so there is an equally likely chance of him picking that or another one (unless you've already picked the car in which case he will not pick it.)

We use the same labelling scheme as above.

Now there are 6 possible outcomes, all equally likely:

[1/6 chance] A. You pick door #1 (car), he opens door #2

[1/6 chance] B. You pick door #1 (car), he opens door #3

[1/6 chance] C. You pick door #2, he opens door #1 (car)

[1/6 chance] D. You pick door #2, he opens door #3

[1/6 chance] E. You pick door #3, he opens door #1 (car)

[1/6 chance] F. You pick door #3, he opens door #2

Now, one question is "what do we do if he opens the door with the car?" The question is merely which strategy should you use, so this shouldn't affect the answer. We can say if he picks the car, it's an automatic win and you take it home. Or we could say it's an automatic lose just like if you ended up with a goat. Or, we could say it's a do-over and you redo the whole thing with new doors, goats and car. The interesting thing is, no matter which of these 3 ways you do it, the two strategies are exactly equivalent and have no advantage over each other.

I'll do this 4 different ways. All are equivalent ways of showing that the strategies are the same. First, let's say you win if he opens the door with the car:

With Strategy "Stay the Same", you win in situations A, B, C and E.

So you win 1/6+1/6+1/6+1/6 = 2/3 of the time.

With Strategy "Change doors", you win in situations C,D,E, and F.

So you win 1/6+1/6+1/6+1/6 = 2/3 of the time. Same!

Now suppose you automatically lose if he opens the one with the car:

With Strategy "Stay the Same", you win in situations A and B.

So you win 1/6+1/6 = 1/3 of the time.

With Strategy "Change doors", you win in situations D and F.

So you win 1/6+1/6 = 1/3 of the time. Again, same.

Now suppose you do this whole experiment thousands of times, but you don't count the ones where he opened the car. In other words, you're only looking at your chances of winning with your strategy _after_ you see that he happened to not pick the car.

Of the six possible situations, 2 involve him picking the car (C and E). So we eliminate those and look at the other 4 situations which each happen an equal amount of the time. Out of these four, each happens 25% of the time (not 25% of the total time, just out of the times where he doesn't pick the car.)

With the "Stay the same" strategy, you win in situations A and B.

A + B

------- = 50% of the times when he doesn't open the car.

A+B+D+F

With the "Change doors" strategy, you win in situations D and F.

D + F

------- = 50% of the times when he doesn't open the car.

A+B+D+F

Again, the strategies are the same!

The 4th way of doing this, is if you assume you do it over every time he accidentally opens the one with the car. This may require several do-overs on the same game-show. But eventually you go home either with a car or a goat.

With "Stay the same" strategy, for the first try you win with situations A and B, and you do-over with situations C and E. So you have

1/3 chance of win

1/3 chance of do-over

1/3 chance of lose

With "Change doors" strategy, for the first try you win with situations D and F, and do-over with situations C and E. So you have

1/3 chance of win

1/3 chance of do-over

1/3 chance of lose

Either way, if you keep doing this over until you eventually win or lose, it will eventually come out that you will have

1/2 chance of win

1/2 chance of lose

(this should be obvious, but if you want I can show you the infinite series you'd add up to get it.)

So in conclusion, this is one of the most bizarre things I've run into in probability for the following reason: if you're on a game show and you know they've put a car behind one of the three doors randomly, you point to one, and the host opens another and it's a goat--what you should do depends on if he deliberately opened the one with the goat or if he randomly happened to open the one with the goat. And if you don't know what the rules are, or whether he was told where the car was, it's not at all clear whether switching doors will benefit you any. Even more bizarre, suppose he does know where the car is, but he hasn't made up his mind whether he's definitely going to open a goat or if he might open the one with the car. In fact, if you somehow know he

Unless you know what's going through his head the second he decides what to pick, there is really no way to assess whether you are better off switching.

-Jeff L. Jones, Sept 29, 2002

#1, Classic goats and car problem:

----------------------------------------

Car is randomly placed behind one of three doors, goats are placed behind the other two. You initially pick one of the three doors.

Gameshow host opens one of the doors, but he knows where the car is and is forbidden from picking that one; he's also forbidden from picking yours. [note: these are the implicit assumptions

*required*in order for the standard explanation to have any validity]

To simplify things, let's always label the door the car is behind "door #1", and the other two "door #2" and "door #3". (This makes it vastly more simple than labelling them based on their absolute positions; try not to confuse this labelling scheme with the "usual" one.)

4 possible situations, having unequal probabilities of occuring:

[1/6 chance] A. You pick door #1 (car), he opens door #2

[1/6 chance] B. You pick door #1 (car), he opens door #3

[1/3 chance] D. You pick door #2, he opens door #3

[1/3 chance] C. You pick door #3, he opens door #2

(notice situations A and B are just two equally likely sub-situations of you picking the car initially, which is a 1/3 chance.)

Therefore, using Strategy "Stay with same door" you will win

1/3 of the time (situations A+B).

And using Strategy "Switch doors" you will win

2/3 of the time (situations C+D).

So you should choose "Switch doors" strategy every time.

So far so good? Ok, now the tricky part...

#2, Modified goats and car problem:

--------------------------------------

Again, car is randomly placed behind one of the three doors, goats behind the other two.

Gameshow host may not pick your door, but this time he is not told where the car is ahead of time. He may use any scheme to decide which door to pick as long as it's not yours and he does not know where the car is so there is an equally likely chance of him picking that or another one (unless you've already picked the car in which case he will not pick it.)

We use the same labelling scheme as above.

Now there are 6 possible outcomes, all equally likely:

[1/6 chance] A. You pick door #1 (car), he opens door #2

[1/6 chance] B. You pick door #1 (car), he opens door #3

[1/6 chance] C. You pick door #2, he opens door #1 (car)

[1/6 chance] D. You pick door #2, he opens door #3

[1/6 chance] E. You pick door #3, he opens door #1 (car)

[1/6 chance] F. You pick door #3, he opens door #2

Now, one question is "what do we do if he opens the door with the car?" The question is merely which strategy should you use, so this shouldn't affect the answer. We can say if he picks the car, it's an automatic win and you take it home. Or we could say it's an automatic lose just like if you ended up with a goat. Or, we could say it's a do-over and you redo the whole thing with new doors, goats and car. The interesting thing is, no matter which of these 3 ways you do it, the two strategies are exactly equivalent and have no advantage over each other.

I'll do this 4 different ways. All are equivalent ways of showing that the strategies are the same. First, let's say you win if he opens the door with the car:

With Strategy "Stay the Same", you win in situations A, B, C and E.

So you win 1/6+1/6+1/6+1/6 = 2/3 of the time.

With Strategy "Change doors", you win in situations C,D,E, and F.

So you win 1/6+1/6+1/6+1/6 = 2/3 of the time. Same!

Now suppose you automatically lose if he opens the one with the car:

With Strategy "Stay the Same", you win in situations A and B.

So you win 1/6+1/6 = 1/3 of the time.

With Strategy "Change doors", you win in situations D and F.

So you win 1/6+1/6 = 1/3 of the time. Again, same.

Now suppose you do this whole experiment thousands of times, but you don't count the ones where he opened the car. In other words, you're only looking at your chances of winning with your strategy _after_ you see that he happened to not pick the car.

Of the six possible situations, 2 involve him picking the car (C and E). So we eliminate those and look at the other 4 situations which each happen an equal amount of the time. Out of these four, each happens 25% of the time (not 25% of the total time, just out of the times where he doesn't pick the car.)

With the "Stay the same" strategy, you win in situations A and B.

A + B

------- = 50% of the times when he doesn't open the car.

A+B+D+F

With the "Change doors" strategy, you win in situations D and F.

D + F

------- = 50% of the times when he doesn't open the car.

A+B+D+F

Again, the strategies are the same!

The 4th way of doing this, is if you assume you do it over every time he accidentally opens the one with the car. This may require several do-overs on the same game-show. But eventually you go home either with a car or a goat.

With "Stay the same" strategy, for the first try you win with situations A and B, and you do-over with situations C and E. So you have

1/3 chance of win

1/3 chance of do-over

1/3 chance of lose

With "Change doors" strategy, for the first try you win with situations D and F, and do-over with situations C and E. So you have

1/3 chance of win

1/3 chance of do-over

1/3 chance of lose

Either way, if you keep doing this over until you eventually win or lose, it will eventually come out that you will have

1/2 chance of win

1/2 chance of lose

(this should be obvious, but if you want I can show you the infinite series you'd add up to get it.)

So in conclusion, this is one of the most bizarre things I've run into in probability for the following reason: if you're on a game show and you know they've put a car behind one of the three doors randomly, you point to one, and the host opens another and it's a goat--what you should do depends on if he deliberately opened the one with the goat or if he randomly happened to open the one with the goat. And if you don't know what the rules are, or whether he was told where the car was, it's not at all clear whether switching doors will benefit you any. Even more bizarre, suppose he does know where the car is, but he hasn't made up his mind whether he's definitely going to open a goat or if he might open the one with the car. In fact, if you somehow know he

*usually*opens the one with the car if it's there... and you see him open a goat, you're best strategy is to "Stay the same" because it's more likely that you've picked the car which forced him to open a goat.

Unless you know what's going through his head the second he decides what to pick, there is really no way to assess whether you are better off switching.

-Jeff L. Jones, Sept 29, 2002

Addendum:

It's also worth mentioning that there's another scenario which I didn't describe in this 2002 version: if the host does know where the car is and isn't allowed to open it, but he

*is*allowed to open the door you first point to, then it

*also*doesn't matter whether you switch or not (unless of course he happens to open your door and reveal a goat so you

*know*you must switch). I think the reason I didn't bother mentioning this was the person I was originally trying to convince already believed me on that point. And it is in some ways more obvious, because the host doesn't even have to know then which door you're pointing to... so he can open a door first and then you can point to one and then move your finger over to the other which is obviously not going to make any difference. What all of this amounts to is that if you drop either the assumption that he can't open your door

*or*the assumption that he can't open the door with the car behind it, the standard answer that "switching is better" is invalid. Only in the case where the not-opening-your-door and not-opening-the-car are correlated do you gain information about the correlation between your door and the door with the car. Which I think is probably the most intuitive way to say it.

## Comments

firmamentI like your explanation. One further comment: if you don't know whether the host knows where the car is, and you can assign non-zero probabilities to either case, then there is an advantage to switching.

spoonlessI think I searched pretty thoroughly a few years ago for this and all the references I found seemed to say switching was always better without pointing out the subtleties involved and the important unstated conditions required for that to be true.

Although I would probably go one step further than that page and say that Marilyn vos Savant was wrong even the way it was worded in the newspaper. It may be a technicality, but part of my point is that even if the host

knowswhere the car is, he can still choose to act as if he doesn't know. (Although I suppose from a running-a-game-show perspective that doesn't make much sense for him to do that.)I like your explanation. One further comment: if you don't know whether the host knows where the car is, and you can assign non-zero probabilities to either case, then there is an advantage to switching.

Ah... you'd think so, but even this is false. Because there are some ways in which the host could act which makes you better off staying the same. If he chooses to always open the door with the car if it's one of the two you're not pointing to, and you see him open a door which is a goat (and not yours) then you are

guaranteedto be pointing to the door with the car so you should always stay the same. If you back off a bit and say he is simply more prone to open a door with a car rather than a goat, then you are still better off staying the same. And if you're not sure whether this is the case or whether he's more likely to open a door with a goat (based on whatever his demeaner is that day) then you can come to no conclusion... if you assign weighted probabilities to each possible scenario, depending on how you weight them you could convince yourself you're better of staying, better of switching, or they could cancel out and be even either way.I would agree, though, that on an actual game show, he's probably not likely to open a door with a car since that would probably imply that you automatically go home with the car (depending on the rules of the gameshow)... which would be bad business practice. But nobody said this game show host is rational! :)

spoonlessAnd if you're not sure whether this is the case or whether he's more likely to open a door with a goat (based on whatever his demeaner is that day) then you can come to no conclusion.

Incidentally, this is the flaw with so-called "Bayesian reasoning"; it requires you to assume a prior distribution, but if that distribution is something you know nothing about then there is no way to rigorously define probabilities for it at all. My suspicion is that this is the reason why Nick Bostrom's "Doomsday Argument" is flawed (if indeed it is), but I've never quite gotten around to formulating an argument against him based on it. There are valid ways to use Anthropic reasoning, but there are also very sketchy ways... and this is why the Anthropic Principle has come under such high scrutiny. It all comes down to what the most "natural" prior is. And not everybody agrees.

sid_icarus(He wouldn't discuss the Monty Hall problem with me without prefacing that we have to state our assumptions. I think he said something about how the actual behavior we could expect from a game show (host) is probably different than the problem assumes.)