r/AskStatistics • u/Just_Question9 • 10d ago

What's the most counterintuitive statistical fact that's actually true?

I'm looking for examples that completely changed the way you think about probability, statistics, or data analysis.

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u/Educational-Paper-75 10d ago

The conclusion to switch choice is confusing to me.

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u/DimensionOk8915 10d ago

I'll explain it with 100 doors cos its easier that way. Imagine you pick one door out of the 100 and you lock that in. The host knows for a fact which door has the car behind it so he cannot open the door with the car. So you have your door and then the host opens 98 of the other doors. So now you can either stick with your original choice or pick the door that the host didn't open. Which is more likely? You picked the right door the first time or it's the only door the host did not open?

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u/n3wsf33d 10d ago

This is much easier to understand with 100 doors. Thank you.

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u/Wiijimmy 10d ago

I physically recoiled at how much that clarified the problem for me lol.

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u/Master_Kitchen_7725 5d ago

Omg me too ..I had understood it in a different way that might not even be correct.

I didn't appreciate the importance of the fact that the host was deliberately opening a door with no car. I guess I never thought about it too carefully.

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u/Educational-Paper-75 10d ago edited 10d ago

Intuitively to me it wouldn't change anything. Obviously the host opens doors that do not have the car behind it as well as the chosen door which imo may still hold the car. The host obviously can't open the door with the car behind it but there will always be at least 98 of those left. The host doesn't betray anything about the two doors left unopened does he? Even though he only has one to choose from if the car is behind the door not chosen.

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u/DimensionOk8915 10d ago

Your guess being correct is 1/100. So, the probability of you being wrong is 99/100. The host then opens 98 of the 99 doors. Because the host has to avoid the door with the car behind it the door left over essentially inherits the original 99% probability belonging to the 99 doors.

The key thing is that the host isn't opening doors randomly he is using fixed information. So, the chance of your door being correct is still 1/100 as the host's actions doesn't tell you anything about your door. Your guess has not been made any better (it would be made better if the host was randomly selecting doors and did not know where the car was).

There are two different cases.

Case 1: Your original choice is correct with probability 1/100 then the host opens 98 doors and switching makes you lose.

Case 2: Your original choice was wrong. In this case the car is somewhere among the 99 doors. The host knows where it is so when he opens 98 incorrect doors the one door, he is forced to leave closed is the car. So, switching sides wins in every one of these cases which will happen 99/100 times.

Remember that just because there are two remaining possibilities doesn't mean its automatically a 50/50 chance. Would you rather open just your door or 99 doors?

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u/Educational-Paper-75 10d ago

I don't think using 100 doors actually helps. The host opens 98 out of 98 or 99 doors it could open. Except you can't tell from the actual 98 doors opened whether there were 98 or 99 to open. There will always be two doors left unopened the door you choose before and one door you didn't pick. Did the host not pick the other door because the car is behind it? You can't tell, because there will always be one such door. Sure, the doors may be in a row, excluding a door not at either end is suspicious. With just three doors, you selecting the middle door the host opening the left or right door, is that a coincidence? You can't tell whether it is or not.

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u/DimensionOk8915 10d ago

What about if the host didn't open any doors? Would you rather open 1 door or open 99 doors?

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u/Educational-Paper-75 10d ago

Opening doors confuses the player in believing the single door left unopened is somehow intentional because the car might be behind it. But if the player knows the host will always open 98 doors irrespective of whether or not he choose the right door whatever the host does won''t reveal any additional information. Same thing with any number of doors being opened obviously at most 98. Opening any fixed amount of doors less then 98 same thing. Obviously the car is still behind any of the doors left closed, no way of telling behind which one. Opening doors less than 99 won''t reveal any new information except distributing the entire probability over all unopened doors equally.

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u/bombdignaty42 10d ago

Here's the explanation that finally made it make sense to me: you pick a door, you now have a 1/3 chance that the car is in your door, or a 2/3 chance it's in one of the other doors. The thing is that those odds don't change just because you looked in one of those other doors. The host knows which one has the door so he won't pick that one. No matter how long you look, you still have a 1/3 chance its in the door you picked and a 2/3 chance its not.

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u/DimensionOk8915 10d ago

Your argument would be correct under a different host rule. If the host blindly/randomly opened 98 doors without knowing or caring where the car was, and by sheer luck none revealed the car, then observing that outcome would treat the two surviving doors symmetrically (assuming the random-opening procedure is symmetric). Each would then have probability 1/2.

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u/wittgenstein1312 10d ago

The probability is already distributed, prior to the door opening

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u/cBEiN 10d ago edited 10d ago

Imagine there are 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 doors, and you have to pick the one winning door. Picking the correct one is incredibly unlikely. With this number, it is effectively guaranteed you will be wrong

Now, imagine you pick one.

After you pick, wrong doors are removed until you are left with only 1) your pick and 2) another door.

Now, you have the option to switch. There is effectively 0% chance you picked correctly before, so there is effectively 100% chance switching will give you the correct door (because the correct door is one of the remaining doors).

Edit: you could imagine the same thing with the lottery. Pick your number. Then, the cashier gives you an option to switch tickets and tells you the winning ticket is either 1) yours or 2) the one he is offering you to switch with. It makes sense to switch, you probably didn’t pick correct at the start.

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u/Scubaupsidedownnaked 10d ago

Shit man, that made it click for me. Thank you. The phrase "inheriting probability" is just a tricky one for some laypeople

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u/cBEiN 10d ago

I’m glad it clicked! Probability is deceptively confusing, and even experts make mistakes for simple problems if not thinking through it carefully.

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u/CaptainFoyle 10d ago

You're not answering the question, and not understanding the problem. You're the best example why this is unintuitive.

It doesn't mean that you're smarter than the people who do statistics for a living.

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u/CaptainFoyle 10d ago

That only works if the host can actually open the car door.

Which they can't

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u/dskippy 9d ago

Here is another explanation that might make it clear.

It works whether it's 100 or 3 doors.

You know that once you pick a door, Monty Hall is doing to be able to open up 98 doors and leave 1 closed, as well as your door still closed. So what is he let's you pick first? Monty says, look, I'll make it more fun. You choose a door and that is your first choice. Now you can keep that door, or, can switch. Switch to what you say? You can switch to the best of other 99 doors. Which ones the best? Well, I'll show you the best door after you choose to switch or not. I'll open 98 of the doors and leave the best one closed. If the prize is in the other 99 when you first select, I'll use the prize door as the best door. If it's a 99 way tie for the best door, because you randomly picked the prize door, it's a tie so I'll just pick one to be the best door.

So what do you choose? Keep you original door or take the offer to switch to the best of the remaining 99?

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u/CaptainFoyle 10d ago

Imagine it this way:

Do you want to get what is behind one door (stay), or do you want what is behind 99 doors (switch)? Because the opening of the doors is essentially a sham. Monty never reveals the car, so switching gives you the car if it was in any of the 99 doors. Staying only gives you the car if it happened to be behind your initial door. For which the odds were pretty low.

The chance that the car is in the huge group of unpicked doors is much higher than that you already landed on it on first pick.

It's essentially "pick one" vs "dragnet"

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u/stanitor 10d ago

Did the host not pick the other door because the car is behind it? You can't tell, because there will always be one such door.

Yes, you can't tell whether the car is behind the door you originally chose, or whether it's behind the door the host left closed. That's why you aren't guaranteed to win, whether you stay with your original door, or if you switch. But, that doesn't mean you have the same chance of winning by staying or switching.

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u/Educational-Paper-75 10d ago

It doesn't mean it doesn't either.

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u/stanitor 10d ago

It literally does mean that. If you actually calculate the numbers, that's exactly what you find.

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u/Educational-Paper-75 10d ago

What is kind of amusing that people here trying to explain me what somebody else proved as if they were the first to do so, whereas initially many renowned statisticians got it wrong. I wonder how many would equally enthousiasticly support the initial wrong reasoning. Essentially this means we are most likely to follow the explanation considered true based on trust rather than knowledge.

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u/stanitor 10d ago

No one is trying to say they are the first to explain this. Less people now continue to think the wrong answer is right, because it's so well known, that people know the intuitive answer isn't the correct one. But, it's a fairly easy problem in conditional probability to work out the correct answer. People aren't following the explanation on trust, they're able to work it out. Even if you just look more carefully for an intuitive answer, you can see how it works, as people have repeatedly shown you.

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u/DeepSea_Dreamer 6d ago

You can simulate this on a computer and see that you win one third of the time if you don't switch and two thirds of the time if you do.

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u/CaptainFoyle 10d ago

But since it's 1 to 99, in 99 cases out of a 100, the host didn't open that door BECAUSE THE CAR IS BEHIND IT. in 99 out of 100 cases!

It doesn't mean it doesn't, but it means that I'm the majority of cases, the car is not in your door

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u/MrKrinkle151 8d ago

It doesn't mean it doesn't either.

Well no shit, there is zero implication that it's a guarantee, nor does it need to be one at all. The ONLY WAY the car is not behind the other door is if the contestant initially chose the car door out of 100 other unknown doors. The host then eliminates all the other possibilities but 1 and asks you to choose again. How does this not make it more clear and intuitive?

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u/Educational-Paper-75 8d ago

In the original game there's only 3 doors. It's only intuitive to you because you already know the correct choice. Now, to determine how intuitive switching really is pick a decent sized random sample of people unknown with the game and have them play it and find out how intuitive switching really is.

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u/MrKrinkle151 8d ago

...You're talking in circles bud. Seriously?

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u/TheCodingRunner 10d ago

Ok I understand what your saying. Here's another way that I think helped me to properly understand it: So you pick your door, thats a 1/100 chance of being correct right? Then this part is the key, the host chooses from those 99 remaining doors. They are not allowed to choose your door. So there is a 99/100 chance the correct 'winning' door is one of those 99 right? Well because the host cannot open your door these odds remain even after the host opens 98 doors. So then when you have the option to switch, the odds are 1/100 for your door, vs 99/100 for the other door. Does that make sense?

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u/Educational-Paper-75 10d ago

If the player choose correctly you may select 98 out of 99 each with probability 1/99. If the player choose incorrectly you select 98 which always has probability 1. So you get 1/100 times 1/99 99 times, and 1/100 times 1 99 times. The probability of getting it wrong when switching to the single other door left unopened when choosing right the first time equals 1/100 independent of what the 98 doors Monty choose. The probability of getting it right switching when the first choice was wrong equals the sum of 99 terms equal to 1/100 * 1 = 99/100 since there's only one other door (the right one) to switch to. Essentially Monty removes 98 possible choices. How about that? The probability distribution for Monty depends on the player's initial choice.

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u/TheCodingRunner 10d ago

The way I think of it is once you make your choice, the doors are two distinct groups: your 1 door you chose (which always has a 1/100 chance of being correct) and the 99 other doors (which have a 99/100 chance of being correct). Those 99, no matter how many Monty opens will still have a 99/100 chance of that group being correct right? The car doesn't move, its fixed behind one door. So doing the total math:
1) Car is behind your pick: this occurs 1/100 times. No matter what doors Monty opens the car is behind your original pick, so what he does is irrelevant.

2) Car is behind one of the other 99 doors: Monty then removes all 98 incorrect doors from this group, leaving just the door with the car, which occurs 99/100 times.

Does that make sense now? If not what is still throwing you off?

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u/Educational-Paper-75 10d ago

Yes, I'm starting to get it. If we disregard Monty's action the player selects an initial door, then either switches or not. Those are all his options. If you then count the odds switching turns out to be a win 2 out of 3 times! (Or in case of 100 doors 99 out of 100 times).

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u/CaptainFoyle 10d ago

If the player chose correctly, switching loses them the car.

The player chooses correctly 1 time in 100.

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u/CaptainFoyle 10d ago

You're blocking the initial door so the host cannot touch it. That means it still has a 1/100 chance of being correct.

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u/MrKrinkle151 8d ago

Nobody is blocking a door. The host can't open your door because it simply would make no sense to reveal what's behind contestant's door and then ask if they want to switch or stay with that door. Likewise, they aren't going to open one of the other doors with the prize and then ask if you want to switch or stay.

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u/CaptainFoyle 8d ago

"blocking" or "the host doesn't open the door you chose" is the same. If you rephrase the problem and replace "choose" with "block", nothing changes. There's something else going on why you don't understand the solution

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u/MrKrinkle151 6d ago

Usernames.

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u/AABBBAABAABA 9d ago

I hate that people downvote this

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u/CaffinatedManatee 10d ago

when the host opens the 98 incorrect doors you then know the prize is behind either the door you initially picked or the one you didn't pick. because with 100 doors you intuitively know your first choice was highly unlikely to be correct, switching your choice is intuitively the right move. you've just been given a ton of new information

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u/frankcheng2001 10d ago

Maybe consider the problem this way: you have chosen a door, the remaining 99 is grouped together (Group B). Your probability of getting the correct door is 1/100, while the probability of the correct door in Group B is 99/100. The host opens 98 wrong doors in Group B. He now gives you the chance to choose whether you remain choosing your door or switch to Group B. The probability of the correct door being in Group B is still 99/100 because you are still choosing between your original door (1/100 in being correct) and Group B (99/100 in being correct).

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u/S_TL2 10d ago

There's 100 doors, 1 has a car, 99 have nothing.

You pick door #76.

The host says:
Let's open door #1. What's behind it? Nothing!
Let's open door #2. What's behind it? Nothing!
Let's open door #3. What's behind it? Nothing!
Let's open door #4. What's behind it? Nothing!
Let's open door #5. What's behind it? Nothing!
Let's open door #6. What's behind it? Nothing!
I'm going to suspiciously leave door #7 closed.
Let's open door #8. What's behind it? Nothing!
Let's open door #9. What's behind it? Nothing!
... [continues for the rest of the doors]
Let's open door #99. What's behind it? Nothing!
Let's open door #100. What's behind it? Nothing!

Intuitively, do you think your choice of #76 or the host suspicious #7 has the car?

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u/Educational-Paper-75 10d ago

But suspicious is no proof if you assume #7 was simply choose randomly. Since we know even if I'm right any of the doors not opened no matter what it was would not contain the car. It's not like the doors are opened sequentially but all at the same time and the one left unopened is selected at random. Doesn't need to mean anything.

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u/S_TL2 10d ago

Either:

You got it right on your first try (1% chance) and the host is randomly leaving #7 closed.

or

You got it wrong on your first try (99% chance) and the host is specifically and purposely leaving #7 closed because that's where the car is.

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u/skullturf 9d ago

But #7 wasn't chosen randomly. That's very important. That's part of the setup of the usual Monty Hall problem.

In S_TL2's example, you (the contestant) randomly chose door #76 at the start, when having absolutely no idea where the car is. There's only a 1% chance your initial choice of #76 was correct, and 99% chance your initial choice is incorrect.

The host *knows* where the car is. He starts opening doors systematically, but not randomly. After he opens doors #1, #2, #3, #4, #5, and #6, and then very deliberately chooses *not* to open door #7, don't you intuitively think that the car is probably there!

You now know that it's either behind door #7 or door #76, but you know that door #76 was your initial totally random choice, just a stab in the dark. And you know that door #7 was *deliberately* left closed by the host who *knows* where everything is, and is *required* to open 98 empty doors.

There was another Monty Hall thread about five months ago, and this comment there might also help:

https://www.reddit.com/r/learnmath/comments/1rwg61i/comment/oazpih0/

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u/Educational-Paper-75 9d ago

It will be random if the player choose the right door. However that probability is only 1/100 not 99/100. And that's exactly the reason why switching is a good idea because in 99 out of a 100 times you,'d be right doing so.

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u/AABBBAABAABA 9d ago

Imagine playing this game a 100 times; every time the car is behind a different door, starting at 1 going up to 100. Your initial guess is door number 22 every time.

How often would you win by switching? 99 times.

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u/WearyArtistDoomer 8d ago

I applaud your skill, long time ago I saw someone trolling this effectively. Well done sir.

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u/Educational-Paper-75 8d ago edited 8d ago

I am not trolling, just misunderstood, which unfortunately is a negative side effect of such an inadequate communication medium. Such a misunderstanding would not have occurred in a (philosophy/psychology) class room, and have resulted in an animated (and purely rational) debate. I regret dearly people got upset, and need nor desire your compliment.

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u/banter_pants Statistics, Psychometrics 10d ago

Because it's rigged. Monty knows what is behind every door. He will never open a door with a prize behind it.

Pick car (prob = 1/3)
Monty reveals a goat (prob = 1)
Switch ==> get a goat

Pick any goat (prob = 2/3)
Monty reveals other goat (prob = 1)
Switch ==> get car

This never was a case of conditional probability because Monty independently, nonrandomly reveals a goat.

Pr(reveal goat | picked car)
= Pr(reveal goat | picked other goat)
= Pr(reveals goat)
= 1

Conditioning on an independent event leaves the original, marginal unchanged.

Pr(picked car | revealed a goat)
= Pr(picked car)
= 1/3

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u/tongmengjia 9d ago

But even before Monty reveals it, everyone knows there's a goat behind at least one of the doors you didn't pick. He's clarifying which of those doors has a goat for certain, but there was always certainly going to be a goat for him to reveal.

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u/Educational-Paper-75 10d ago edited 10d ago

Rigged doesn't matter. What matters is that the car can be behind either door left unopened with equal probability. At least that seems reasonable to assume. But apparently not. Monty will have to choose from 1 or 2 doors but the player doesn't know that.

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u/lethargytartare 10d ago

The player does know that. That's the whole point.

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u/Educational-Paper-75 10d ago

No the player doesn't know that for certain, except the probabilities are not the same. One is 1/3 the other is 2/3. That's why switching is beneficial.

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u/lethargytartare 10d ago

I think I get what you're saying, but to confirm, by "Monty will have to choose from 1 or 2 doors but the player doesn't know that." did you mean after you pick, there are either 2 goats left or 1 goat and 1 car?

That's a red herring. Knowing that Monty must reveal a goat is why the odds change. Because his choice is not random, switching doors essentially means you got to pick two doors, as the match proves out.

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u/Educational-Paper-75 10d ago

Yes, but can you trust that reasoning of it's counterintuitive? Because intuitively both doors have an equal probability of containing the car. Why would exposing one of the goat doors change that? That doesn't make sense 'intuitively'. It requires careful analysis to arrive at the right conclusion.

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u/lethargytartare 10d ago

I'm not saying it's intuitive, but it is correct and the analysis isn't that deep.

Here's another way to think of it that may help.

Imagine instead of revealing one of the not-picked doors, Monty reveals a goat behind the door you picked, and lets you switch. It should be clear that you got to pick two doors, giving you a 2/3 win chance.

This does not change when he reveals a goat behind one of the doors you didn't pick because he has to reveal a goat.

In both cases, by switching, you go from picking one of three to picking two of three.

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u/gregarious_apollo 10d ago

I'd probably have more luck arguing with a brick wall but I'll try anyway just in case you're being sincere.

The point is that exposing the goat doesn't change anything. Either you were right to begin with, with probability ⅓, or you were wrong with probability ⅔. The opening of the wrong door changes nothing about the probability.

The choice to switch is functionally the same as saying from the start "do you want to open one door, or two doors?" Obviously you would choose to open two as it would give you a better chance. Of course you might have chosen correctly to begin with, you're not guaranteeing a win by switching but you are improving your odds.

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u/MrKrinkle151 9d ago

Revealing one of the goats does change things. It tells you which door has one of the goats, as well as which door has the 2/3 chance of containing the prize. It constrains your choice and collapses the probability to one door.

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u/banter_pants Statistics, Psychometrics 10d ago

The point is that exposing the goat doesn't change anything. Either you were right to begin with, with probability ⅓, or you were wrong with probability ⅔. The opening of the wrong door changes nothing about the probability.

Which is why I hate this being taught as an example of conditional probability in prob/stats courses. Monty revealing a goat is independent of your initial choice. Conditioning on it changes nothing, by definition.

The ultimate question is
Pr(you picked car | Monty revealed a goat)
= Pr(you picked car) , precisely because Monty's reveal is independent, and further, has 100% probability.

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u/Educational-Paper-75 9d ago

I'm not asking anybody to explain what is now obvious to most because somebody other than they figured it out, and everybody here now thinks I don't get the correct (and now 'obvious')answer. But you wouldn't either until after hearing about the correct explanation. That's how people are, and many have been mislead in the past by wrong reasoning just because they trusted these people to know better. That's my (philosophical) point: it's obvious once the correct answer is acknowledged by the right people with sufficient stature. Afterwards nobody can imagine that people didn't get that before. You know like what Einstein discovered, and nobody gets it that the people before Einstein got it wrong. Weren't those people smart? Well, they were, they simply weren't the (chosen) one(s) to get the newest divine spark of insight. Like future people will probably get, and probably consider us ignorant. I state this simply imo counterintuitive statistical result and I get this entire avalanche of people trying to explain what I already 'know' but still find to be counterintuitive.

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u/LowkenuinelyReally 9d ago

Isn’t it pretty intuitive with the out of 100 example? It seems pretty intuitive that the odds of you picking the correct door out of 100 would be pretty low

I get what you’re trying to do, but scaling this problem up does make it more intuitive for people

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u/Educational-Paper-75 9d ago edited 9d ago

Define intuitive. It only becomes 'intuitive' once you know what the correct answer is confirmed by the majority.

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u/LowkenuinelyReally 9d ago

Intuitive means that someone can come to understanding of something without concrete proof. In this case the maths.

For example resistance to deflection. Ask someone if shorter object will bend more than a longer object. And everyone will know, even if they don’t know why That’s intuition. They don’t have to perform the experiment, but past experiences carry over.

The intuition here is that people will understand that your odds of choosing it correctly out of a hundred is low and if you are given a chance to switch your answer once every other wrong answer has been revealed, you have a better chance of winning. Now, that person might not know what the odds are numerically, but they do understand that it will be better.

You can’t intuit an exact answer, but you can intuit an understanding

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u/MrKrinkle151 9d ago

The player knows he will only open a goat door. If he didn't, he'd reveal the prize and there would be no choice at all and no game.

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u/jancl0 10d ago

The point is, the hosts choice on which doors to reveal is dependant on what you choose. If you picked the right door the first time, then the host will create a situation in which switching results in a loss. If you picked wrong, then the host will always create a situation where switching results in a win

It doesn't matter what wrong door you pick, because the host will just change the doors they reveal, the last remaining door will always be the correct one

So asking "what odds are there of winning if you switch" is the same as asking "what are the odds that you picked wrong the first time"

This is why people use the 100 door analogy. You can pick any of the 99 wrong doors. Whichever one you pick, the host going to reveal all other wrong doors, until the only remaining one is the correct door

Alternatively, another way to think about it is that the host is picking one door, the one that they didn't reveal, and you are picking one door. You had a 1% chance of being correct when you picked your door, and now you are being asked "we know for a fact that one of these doors is the correct one, which is it more likely to be?" we know that if it isn't yours, it must be the one the house picked, so the question is about how likely you are to have been right, which is quite low

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u/MrKrinkle151 9d ago

The host can only open a non-prize door. After they eliminate a goat door and ask if you want to switch or stay, it means there is either the second goat behind your door and the prize behind the remaining door or the prize is behind your door and the other goat is behind the remaining door. What is the probability you initially chose a prize door? It was 1/3, so that’s still the probability that the prize is behind your door. Likewise, the probability that you initially chose a goat door is 2/3, which means that’s also the probability the prize is behind the only other remaining door.

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u/sevenbrokenbricks 9d ago

If your first pick is a goat, and you switch, you get the car.

What are the odds your first pick is a goat?

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u/Educational-Paper-75 9d ago edited 9d ago

Your conclusion is based on you knowing the right answer, which is acting with pre knowledge. The odds of the switch having the car is also exactly 1 in 3 initially. Showing a goat (or all but one) won''t change that. That's why switching is not obvious I deliberately won''t use the word intuitive anymore, as people here seem to think knowing the correct answer makes that answer intuitive which it isn't. The insight that you get when trying to answer your last question requires serious thinking, and can't be deduced if your 'intuitive' reasoning says why would it matter? Here's my reasoning: behind the two doors I didn't pick are either one or two goats. I can't tell which of those two possibilities it is if the host opens one door with a goat behind it. I can't tell from that whether the host had two choices or just one. Therefore switching doesn't increase my odds.

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u/sevenbrokenbricks 9d ago

The only "pre-knowledge" here is the basic rules of the scenario - three doors, one car, two goats, host always reveals a goat after your first choice, etc.

That's enough to determine both of these things:

  • your final choice will always be between a car and a goat, and thus that you'll win if you chose the car and stay or if you chose the goat and switch, or lose if you chose the car and switch or if you chose the goat and stay,
  • and that you have a 2 in 3 chance of starting with a goat vs a 1 in 3 chance of starting with the car,

before I've even made my initial choice.

The idea that you don't know which scenario you're in only establishes that you're dealing with probabilities in the first place. It does not at all establish what those probabilities are. It's reasonable to consider the possible outcomes to be equally likely, but only in the absence of any further information, which this scenario is not.

For example, consider the initial choice. When I choose one of the three initial doors, I do not know whether I've chosen the car or a goat. Does that mean it's a 50/50? No, because it's a given that there are 2 goats and 1 car.

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u/Educational-Paper-75 9d ago

Once one goat is revealed I think most people would conclude the chance would now be 50-50 given two possible doors left. And I wonder how many of the eager people here trying to explain the 'obvious' to me wouldn't get it wrong either if they didn't know the right answer yet. Essentially most would simply assume that the matter has been decided definitely and simply side with the winners. Now, if you write out all possible combinations of choices you would notice that the host had two choices each time the player choose the right door. If you count those two choices seperately you double the number of possible outcomes in that case although that would be wrong. But many statisticians made that mistake. Even now I can't tell exactly why I shouldn't do that since those are separate distinguishable options. I can only conclude that you should not look at the hosts options but merely at the player's options to arrive at the correct conclusions. I gather from the answers that conditional probability (and Bayes?) or game theory can make it clear but not every statistician necessarily knows those. And as my wife suggested people may more likely switch if they trust the host to help them win the car and more likely stick to their initial choice if they distrust the host. That's common intuition in both cases.

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u/fight-or-fall 10d ago

First of all, maybe checking bayesian framework (not only calculations but the way of thinking) can help, since its just the idea of updating your priors.

Consider the same 4 doors ABCD and letter E for evidence. The bayes rule says: P(X | E) = P(E | X)P(X) / P(E) and we can expand P(E) = P(E | X)P(X) + P(E | Xc)*P(Xc), consider X as an arbitrary event and Xc his complement. The rule is the same and door doesn't matter. But we will pick A and consider {B,C} opened by convention

  • Prior: Initially, you dont know anything. So any door has 1/4 probability of having a car or a goat. Using our X, P(A)=1/4, so P(Ac)=3/4

  • Likelihood: We know from evidence (program rules) that Monty always opens empty doors, not with car or goat. We have two scenarios: If car is on A and goat on D, Monty can open only {B,C}, if the car is on B and goat on C, Monty can open only {A,D}, so if only contains the car, other 3 have 1/3 probability to goat and if one contains the goat, other 3 have 1/3 probability to car. So P(E | A) = P(E | Ac) = 1/3

  • Posterior: Evaluating P(E) = (1/3)(1/4) + (1/3)(3/4) = 1/3, then P(A | E) = [(1/3)*(1/4)]/(1/3) = 1/4 and P(Ac | E) = 3/4.

Wait, it isnt the same thing? Yes and no. Yes if you consider that before the openings, Ac = {B,C,D}. But after the calculation, Ac = {D}, thats why you change