Probability and the three door problem

[0112358]

I’m going to attempt to explain a somewhat legendary probability problem which @ChetAtkins asked about today:

The game show problem with three doors and one good prize.

There are plenty of explanations of the answer online and you can read them and nod but I think they often don’t highlight what specifically means the correct answer is correct. So I’m going to hopefully illustrate it with not one, but three different games where we can see what change to the game is critical.

Variant 1 (as described by Chet):

“So there are three doors to choose from. Behind two of the doors are goats, and behind the other door is a brand new car. You pick door number one. The game show host then tells you that you can change your mind a choose another door if you want. He doesn't tell you that you picked the wrong door, he just throws out the offer? What would you do?”

The player in this situation has a 1/3 chance of guessing correctly. The reason for this is not just because there are three doors - it’s because we believe the game has an even chance of placing the prize behind each. The player makes a guess and has a 1/3 chance of being correct. They are asked if they want to change their guess. It doesn’t matter probability wise what they do as each door still has a 1/3 chance of being correct.

Variant 2:

After choosing a door, let’s assume a friend of the player picks one of the two remaining doors to open. They don’t know what’s behind any door. The door they choose has a goat. Should the player now switch?

In this case the remaining doors improve to having a 1/2 chance of being correct but neither is better than the other. The friend didn’t know any extra information. We ruled out one door at random. It’s important that the friend might have picked the “car” door but didn’t, either by chance or because the original player chose the car door.

Variant 3 (the classic problem and actual game show play):

In this version after the player picks a door, the host, who knows where the car is, opens one of the remaining doors and shows a goat. Should the player now switch?

The answer is yes. I will describe the numbers and then hammer on “why”. Let’s look at how often a “switch always” strategy wins.

- 1/3 of the time the player guessed correctly to begin with. The host shows a random goat. The player switches and loses.

- 2/3 of the time the players first guess was wrong. The host -never- shows the car and -always- shows which of the other two doors is a goat. In this case switching results in winning.

So switch always wins 2/3 of the time. The key here is that 1/3 of the time the player loses, but 2/3 of the time the host effectively tells the player the answer. This is different from variant 2, because in variant 2 we can’t tell if the friend who randomly showed a goat simply got lucky.

The critical difference is that the host -often- but not always is actually providing new information that changes the odds, as opposed to purely ruling out one choice.

goat!!

[IlyaTheImpaler]

Nice post, I wrote about this when I a while ago

Do you stay with door number one? ↗

There is one important assumption in variant 3 that is often forgotten, that if the host has a choice between 2 remaining doors he would pick randomly with equal odds. You can replace this assumption by another: the host prefers to open the door closer to him, maybe he doesn't like walking? so if he has a choice between 2 doors he would favor the closer door with probability p (between 1/2 and 1). In that case the probability switching wins you the game given the door nearer to the host is opened, is 1/(1+p). The same probability given the other door was opened, is 1/(2-p). With p = 1/2 you recover the answer of variant 3.

Variant 2 could be modified similarly (if the friend favors one door over the other)

[0112358]

That would still be 50/50. The host could always open a closer door or wouldn’t change the odds so long as the actual starting setup was 1/3 for each door. The absolutely critical fact is that the host chooses with knowledge.

[IlyaTheImpaler]

No, the host could never 'always open a closer door'. He could only have a preference. He has to avoid the door with the car behind it, so he might not have a choice, but when he does, his preference matters. Make sure you understand you are talking about conditional probability, with extra knowledge like the host's preference, the probability will change. You did point out an important assumption that the host knows where the car is, everyone knows this assumption, but that's not the only important assumption. The one I mentioned actually decides whether someone truly understands this problem.

To make it easier to understand, let p = 1: meaning, suppose you know for sure the host will never go to the door farther to him if he has a choice. Now you picked a door, two doors left. If you see the host picks the door farther away, then you know with certainty which door the car is behind: the one closer to him, because otherwise he would've picked it due to his preference. And the probability switching wins is 1/(2-1) = 1. This shows the host's preference matters.

[0112358]

When I said he always opened the closet door I took it as obvious that meant he didn’t know where the car was and would potentially show it.

If the host knows where car is, won’t show it, but if given a choice chooses the closest door with p = 1, we have a new situation. In this one, if the host shows the far door we know with absolute certainty that the goat is behind the “switch” door. On the other hand I’d the host shows us the “close” door we’ve learned nothing new. There are still equal odds between first choice being correct (which always triggers close door shown) and far door correct (which also always results on close door shown).

If the host has a preference when given then option to choose the closer door with a probability p where.5 < p < 1, a switch always strategy is superior but we recognize it’s only somewhat superior when the close door is shown and highly superior when the far door is shown.

[IlyaTheImpaler]

Yup your last two paragraphs are basically what the formulas I wrote meant, intuitively!

[Secretgardenblood]

Good take.

[NearlyNapping]

I'm not buying it. After a door is opened, you are now posing an entirely new question. The fact that there were originally three doors is irrelevant. It is now a question with two doors. You have a 50-50 chance.

This is not so different than the question "You flip a coin and get heads ten times in a row. What are the odds of the next flip being heads?"

The past events are irrelevant. It's no different than the host saying "There are three doors, two of which have a goat, and one has a car. Oh wait, I read the script wrong. There are two doors, one of which has a goat, the other has a car."

[0112358]

What you’re describing is scenario 2 and in that case it is essentially a brand new question with 50/50 odds. (The actual math behind is a bit more complicated because it’s similar anyway).

Scenario 3 is fundamentally different because the host isn’t behaving randomly. If the player picked car originally the host randomly removes one of the two goat doors. But 2/3 of the time the player picked a “goat” door and in that case the host -isn’t random-

Maybe think of it this way. 1/3 of the time the host tries to trick you and make you switch. 2/3 of the time the host tells you where the car is.

[NearlyNapping]

The only differences between scenarios 2 and 3 is psychology or foreknowledge of past patterns of the host. The way the problem is normally stated, there is no foreknowledge of previous patterns. Meaning you don't know things like "Host always offers to switch if you pick a car" or "always offers a switch if you pick a goat" etc.

" If the player picked car originally the host randomly removes one of the two goat doors. But 2/3 of the time the player picked a “goat” door and in that case the host -isn’t random- "

It's irrelevant. Because from the doors you didn't pick, there will always be a goat. So the host picking a goat gives you no new information. Saying it's random or not random depending on your pick falls into the category of foreknowledge. But whether it's random or not, it gives you no information.

The only think you know, without foreknowledge, is that you have two doors, one of which contains a car, the other a goat.

[0112358]

“The only differences between scenarios 2 and 3 is psychology or foreknowledge of past patterns of the host. The way the problem is normally stated, there is no foreknowledge of previous patterns. Meaning you don't know things like "Host always offers to switch if you pick a car" or "always offers a switch if you pick a goat" etc.”

That’s incorrect. You know that the host never, ever, ever shows you the car then asks you to switch from one ghost to the other. This IS a critical point. If the host sometimes showed car, then there’s no reason to switch. The very first time the game show was run in real life people would not have known this but it rapidly became apparent.

If you feel that this distinction is often not clearly stated then I will concede that and even wish I’d done it better myself. Whoever decides which door to open first knows where the car is and avoids it if necessary.

[NearlyNapping]

I have never seen this problem stated with foreknowledge. But it doesn't matter if you know the host never picks the car door.

If you pick a goat, the host picks a goat.
If you pick a car, the host picks a goat.

The host picks a goat in either case, so it gives you no new information.

[0112358]

Let’s say one goat is black the other goat is white.

Let’s assign probabilities.
1/3 you pick white goat.
1/3 you pick black goat
1/3 you pick car.

So far so good?
Now IF you pick car, the host picks white goat 50% black goat 50%.

If you pick white goat the host picks. Black goat 100%
And vice versa.

We can build the whole grid now.
1/6 you pick car then host shows white goat
1/6 you pick car then host shows black goat
1/3 you pick white goat host picks black goat
1/3 you pick black goat host picks white goat.

That’s the full possibility set.
So let’s say the host shows you a black goat. Well we eliminate the first and last possibility. It’s either the second or third possibility. But the fact that we’ve eliminated the first and last doesn’t change the relative odds of the second and third. Possibility 3 ( you picked white goat host picked black goat) is still twice as likely as possibility 2 (you picked car, host picked black goat). So switching is better. Of course the same is also true if you see a white goat. Or if I replace the colours with names - a goat named moe and a goat named Larry.

[0112358]

Since i'm finally not doing this by phone, I drew grids with probabilities, which might help:

[ChetAtkins]

If my ex-wife was 1/10th as smart as you, we'd still be married. lol
But I'm not going to dwell on that. I'm just going for the whole "boob" thing. Those are just as nice!

Not very many people on this site impress me, but you pulled it off.

[Halucinator]

What I didn't get is how is it still different?
The probability should still be 1/2 in the variant 3. There is still no certainity if the door you've chosen first has the car. Probability should still remain 1/2

[0112358]

Here's the full grid of possibilities for variants 2 and 3, before and after seeing goat:


In every case, there's a 1/3 chance of you picking car, goat 1, or goat 2. In variant 2, there's going to be a 50% chance WITHIN each of those 1/3 chances of the friend revealing each of the other two possibilities. So all 6 possibilities, at the start of the game, will have a 1/6 or 17% chance of happening. After the friend opens a door and you see a goat, you can rule out the two possibilities where a car was revealed. That leaves the four remaining possibilities, each equal, so each rises from 17% to 25%. 2 of those choices involve you having chosen car and staying put, 2 of those involve switching. 50%/50%

In the variant 3 case, the initial grid is different. It's still 33% chance for you to pick each item. But now, we actually know that if you choose goat 1, the host -always- picks goat 2. That doesn't magically change the odds of you choosing goat 1: there's still a 33% chance you pick car, 33% chance you pick goat 1, 33% chance you pick goat 2. If you pick car, we have two equal chances for host: goat 1 or goat 2, each 16.5% (17% chance) overall. But if you pick goat 1, which happens 33% of the time, there's an absolute certainty host picks goat 2. So that path gets a full 33% chance of happening. So does the similar 'you pick goat 2 host picks goat 1'. In variant 3, the host opens a door and shows a goat - which we knew would happen, it eliminates some things that had 0% chance of happening. We still have 4 possible paths, but the 4 do NOT have equal odds. In this case, we have two paths where you lose by switching, each ADDING to 33%, and two other paths that are EACH 33% chance where switching wins.

[abdallahdaif]

Damn probabilities are the hardest thing to understand

[RedRobin]

Great take

[Nivinxus]

Monty hall?

[hakon96]

21 huh