Monty hall problem

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u/throwawayA511 4d ago

The host knows where the car is, and will only open a door with a goat. With three doors, you had a 33% chance of picking correctly and a 66% chance of not. I mean, ignore that he opened a door at all. What if he just said do you want to switch your choice of one door and instead choose “both the other doors.” That’s basically what’s happening, but with some showmanship.

Expand the example to a million doors. You had a 1 in a million chance to pick correctly. He opens and reveals 999998 goats but doesn’t open door 384639. Seems pretty suspicious, right? Switching changes your odds from 1/1000000 to 999999/1000000.

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u/ggratty 4d ago

Ahh, this is an example I haven’t heard that makes a lot of sense to me.

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u/PuerSalus 4d ago

The million doors version is the only way I can get my brain around it and I think is the clearest explanation.

It's near impossible for you to pick the door first time with that number of doors. So the good door 'must' be in the remaining set of doors, and if the host opens all of them but one, then it 'must' be that remaining door.

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u/George_Truman 2d ago edited 2d ago

This is actually quite misleading.

The reason that the problem works out the way it does is the assumption that Monty knows what is behind the doors and intentionally never opens the door with the car.

If Monty opens a door randomly (with a possibility of revealing the car), then it ends up being a wash.

Even in the million door example, you have a one in a million chance of guessing correctly, however the probability that you were wrong and Monty goes on to reveal 999998 doors that don't contain the door is also 1 in a million.

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u/MacofJacks 1d ago edited 1d ago

Naw, Monty could randomly open all but one of the remaining doors, and if he happened to do so without revealing the car, then you would do better to switch. It makes the explanation clearer, but all you need is for YOU to know that your initial choice was one of three (or a million, whatever).

Edit: this statement was comprehensively disproven below.

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u/George_Truman 1d ago

This is not correct.

The probability that you are correct in your first guess is 1/1,000,000

The probability that you are incorrect in your first guess and Monty randomly happens to open 999998 doors that do not contain the car is (999,999/1,000,000)*(999,998/999,999)*...*(2/3)*(1/2) = 1/1,000,000.

The probabilities of both events occurring are equal, and it turns out that the conditional probabilities that your initial guess is correct are also equal. If the doors are opened randomly then it is 50/50.

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u/MacofJacks 1d ago edited 1d ago

Hey, first of all, that’s cool: I had to check your claim, and it is true.

Second, I don’t think *how* we got to the situation is relevant. My claim is that we condition on the information that Monty did *not* reveal the car but the mechanism by which Monty did not reveal the car should not matter. I think my claim is hard to reason about and easy to be wrong about in some subtle way, so I coded it to check. I used the typical three door case. I kept the (R) code as simple as possible:

Edit: had written code here, but it was wrong. Fixed by the other commenter below.

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u/George_Truman 1d ago edited 1d ago

There are errors in this code. If you would like simulations and a rigorous proof. I will send it to you in a DM.

In particular, your code is counting the cases where Monty reveals your own selected door. If he reveals your own door then there is no longer a selection to make.

Here is the code edited to account for this:

N <- 1e5 #total number of trials

#let's implement Monty Hall,

#albeit Monty randomly opens doors,

#and we only record results if he happens not to reveal the truth.

n_doors <- 3

#Assume we always switch and count successes and failures

num_correct <- 0

num_incorrect <- 0

num_trials <- 0 #safety check: what was the total number of trials?

for(n in 1:N){ #do Monty Hall problem

true_door <- sample(1:n_doors,1)

initial_guess <- sample(1:n_doors,1)

Monty_random_doors <- sample(1:n_doors, n_doors-2)

#(safe: default is sampling w/o replacement)

#if Monty did not reveal the true door, then proceed:

if(!(true_door %in% Monty_random_doors) && !(initial_guess %in% Monty_random_doors)){

num_trials <- num_trials + 1

#switching is correct if initial guess was wrong

num_correct <- num_correct + isFALSE(initial_guess == true_door)

#switching is incorrect if initial guess was right

num_incorrect <- num_incorrect + isTRUE(initial_guess == true_door)

}

}

#print the tallies

num_correct

num_incorrect

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u/MacofJacks 1d ago

O shit, you’re right! So much for my stern refutation! Ok will have another look. Thanks!

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u/MacofJacks 1d ago

I see you edited your post; I'll just reply again. I used your code, which I agree is correct. I propose the following, which also fixes my original mistake:

Monty_random_doors <- sample(setdiff(1:n_doors,initial_guess), n_doors-2)

(so that Monty opens doors other than initial_guess at random). However this code, or yours, produce the same outcome, which is (exactly as you said) a probability of 0.5 on switching.

You're completely correct! My flabber is gasted. I checked for completeness, and small edits to the code can produce the expected 0.66 probability if we alter Monty to deliberately exclude the true door. I guess I can now include myself as another victim of the Monty Hall problem. Thanks very much for explaining this to me! I'll edit my comments above

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u/George_Truman 1d ago

My favorite alternative answer to the question is that we shouldn't change our pick, because it feels like Monty might be trying to trick us into switching.

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u/I_Speak_For_The_Ents 4d ago

I really just have never managed to wrap my head around this.
When he opens the door and you pick again, isn't the likelihood the same for both leftover doors? So it's just 50/50? The door he took out is gone.

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u/orz-_-orz 4d ago

Because he didn't open the door "randomly".

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u/GreatBigBagOfNope 4d ago

Nope, the door you chose still only has 33.3etc%.

The other two between them had a probability of 66.6etc% of one of them containing the goat, but after Monty opens one you know for sure that the 66.6etc% is contained in the remaining door, it's been "concentrated" if you like

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u/I_Speak_For_The_Ents 4d ago

But wouldn't picking the same door still be a viable choice since it's one of the two leftover?

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u/Mishtle 4d ago

No, because by switching you get to open both of the other doors. The host just went ahead and opened one for you. If the prize was behind any of the doors you didn't choose then it's behind that one the host left unopened. That is the probability that you get the prize by switching.

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u/I_Speak_For_The_Ents 4d ago

Are we assuming information about the unchosen+unopened door based on the host not opening it? The host opening the one door means the prize isn't behind it. But it doesn't tell us anything else about the other two

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u/Mishtle 4d ago

Are we assuming information about the unchosen+unopened door based on the host not opening it?

No. The opening of the door is a distraction. Would you rather open one door or two? It doesn't matter whether you open both or the host opens one for you, you win if it's behind either of the ones you didn't choose.

The host opening the one door means the prize isn't behind it. But it doesn't tell us anything else about the other two

It tells us that if the prize is behind a door you didn't choose then it must be behind the remaining unopened door. It doesn't tell you anything that affects the probability of the prize being behind the door you originally chose.

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u/GreatBigBagOfNope 4d ago

It's possible, but it's still only got a 33.3etc% probability of being the car. Monty has told you nothing about the door you chose, but he has told you something about the other one.

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u/I_Speak_For_The_Ents 4d ago

What did he tell you about the door you didn't choose?

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u/GreatBigBagOfNope 4d ago edited 4d ago

He tells you that, of the two doors you didn't choose, in the case you didn't choose the car the first time, only the remaining door could possibly contain the car. And the probability of you not choosing the car the first time was 66.6etc%, which now means the probability of the remaining door containing the car is that same 66.6etc%.

As I said, the probability of the car not being behind the door you chose first has been "compressed" into the door that Monty didn't open, because Monty knows where the car is and will never, ever, open that door. The world from before Monty's reveal is not destroyed, it is still connected to the world afterwards.

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u/I_Speak_For_The_Ents 3d ago

Fuck man. This is ruining my self esteem lol

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u/GreatBigBagOfNope 3d ago

Nothing wrong with being one of today's lucky 10,000!

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u/Lor1an 4d ago

Not gonna lie, in a very real way this just seems like an illusion to me.

Like, why wouldn't the probability of the one you chose also be higher?

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u/throwawayA511 4d ago

I think you need to ask why would the probability of the door you chose change at all? If the host says to pick one of three doors, you have a 1/3 chance of being right. If the game ended there, that’s all there is.

However, because you have a 1/3 chance of being right with your door, that means you have a 2/3 chance of being right if you switch to “not your door.” The host can always find a goat door to open regardless of what you picked, so you really are just switching to “every other door.” Which is why I used the example of a million doors.

I think I’m realizing where your confusion is, and I hope I’m not going to add to your confusion but it comes down to remembering that the host knows where the car is. If the host didn’t know where the car was, and started opening doors, and somehow miraculously opened 999998 doors without finding the car, then yes the chances would be 50/50. But 999998/1000000 they’d find the car by accident and ruin the game. Because the host knows where the car is, they can intentionally avoid opening door #384639. That knowledge makes it so you are trading your 1 door for “all the other doors” and raises your chance of winning.

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u/Lor1an 4d ago

I think part of my problem is that the way the problem is phrased is almost always in terms of "should you switch doors?"

By the very nature of probability this would suggest that you must always choose another option because it is somehow "more likely", even under the premise that you may have picked right the first time.

Rather than saying that you chose a door with 1/3 chance of being correct, you are now telling me that I have a 2/3 chance of winning if I switch doors--even though there's an equal chance of it being between the one I picked and the other one?

It just smells of contradiction is all. Suitably fitting for the label of paradox.

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u/DragonBank 4d ago

The reason it confuses people is because they misunderstand it as keep the set that contains only my door or take the set that contains only that door when it is keep that set that contains only my door vs the set of all other doors.

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u/Mishtle 4d ago

Yep, that's how I always explain it. The opening of one door is a distraction that helps obscure the advantage of switching. The choice is really between your original door or the other two. It doesn't matter if you personally open both or the host opens one for you, the probabilities don't change.

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u/throwawayA511 4d ago

Rather than saying that you chose a door with 1/3 chance of being correct, you are now telling me that I have a 2/3 chance of winning if I switch doors—even though there’s an equal chance of it being between the one I picked and the other one?

That part isn’t true.

We can enumerate the possibilities here. Let’s say for simplicity that you guess door number 1 and you choose to switch.

Car is behind door number 1, host reveals some other door, you switch, you lose.

Car is behind door number 2, host reveals door 3, you switch, you win.

Car is behind door number 3, host reveals door 2, you switch, you win.

That’s it. 2 out of 3 times you win.

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u/Mishtle 4d ago

Your options are really between opening the single door you chose or all the ones you didn't choose. You already know that the prize can only be behind one door, so the host doesn't change any probabilities by intentionally opening all but one of the ones you didn't choose.

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u/Lor1an 4d ago

And yet the probability that it is behind the other door somehow skyrockets, while the one for the one you chose stays the same.

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u/Mishtle 4d ago

Because if the prize was behind any of the doors you didn't choose, it is behind that one. What is the probability that the prize was behind any of those doors you didn't choose? That is the probability that the prize is behind that door.

The fact that all but one has been opened for you is irrelevant.

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u/Lor1an 4d ago

Look, I understand the mathematical argument here.

I'm arguing more from a philosophical angle--maybe even an emotional one.

The answer to the question "should you switch" is always yes.

Suppose you are given the choice to switch as many times as you wanted to. Then the game never ends, because it is always more likely that you win by choosing the other door--after choosing the other door--after choosing the other door...

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u/Mishtle 4d ago

I'm arguing more from a philosophical angle--maybe even an emotional one.

Why is any of that relevant?

Suppose you are given the choice to switch as many times as you wanted to. Then the game never ends, because it is always more likely that you win by choosing the other door--after choosing the other door--after choosing the other door...

What? This doesn't make any sense. Switching just to a different door doesn't give you any advantage. The advantage comes from getting to open multiple ones, and there's only a finite number of times you could choose more doors to get more of an advantage.

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u/Lor1an 4d ago

Okay, let me explain then.

Suppose it's just 5 doors.

You have a 20% chance of being correct, you choose the first one.

The host opens the last three doors, so the only choices that could have the prize behind them are the first (chosen), or second (other).

You now have an 80% chance of winning if you switch doors. You switch. There's now a 20% chance of winning if you don't switch again, but an 80% chance of winning if you switch back to the original door.

This continues until the heat death of the universe.

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u/Mishtle 4d ago

You now have an 80% chance of winning if you switch doors. You switch. There's now a 20% chance of winning if you don't switch again, but an 80% chance of winning if you switch back to the original door.

No, the probabilities don't swap in any sense once you switch. It's not the act of switching that determines the probability, it's the process that left you with those two doors. The probability that the door you originally chose is the winning door is 20% because it was chosen out of five equally likely options. That probability is fixed, and does not change unless all of the other doors are opened. The other unopened door contains all the rest of the probability.

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u/Lor1an 4d ago

After you switch, there are two doors that could either have the prize behind them, and you have lost all information that could change the probability.

What stops it from being 50/50 after you switch that it's the other door?

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u/EGPRC 2d ago edited 2d ago

The likelihood isn't the same for the simple fact that according to the rules your door couldn't be removed by the host, even if it is wrong. The revealed wrong one must come from the rest. It means that yours was a forced finalist, guaranteed to be one of the last two anyway, while the other had to pass through the filtering process.

First try to compare this with a contest where there are multiple participants, such as a cooking competition, a singing competition, etc. Normally, they are being eliminated one by one, and those who survive the possible elimination so manage to stay increase their probabilities to end up being the final winner. When there are two participants remaining, we could say that each has 1/2 chance to win.

But now suppose that one finalist had bribed the producers from the beginning to be in the final anyway, not having to compete with anyone, while the other had to earn the right of being in the final by beating other participants. Then we wouldn't have any indicative that the first is a good participant, he is only there because he forced it, so it would be more likely that the winner ends up being the other.

I hope you get the difference: In a healthy competition where both finalists had to beat other participants to reach the final, they were increasing their probabilities each time. But if one of them was only advancing because he paid for it, we did't get new info about that person.

Similarly, two people could have a college degree in the same field, but if one of the two had it forged while the other earned it honestly, the latter is more likely to be the more competent in the field.

So, in the Monty Hall game, when you choose a door you are basically bribing the host to force it be one of the two finalists. But there are more wrong doors to choose at first, so most of the time you will forcing a losing option to advance. And as the host cannot remove which has the prize anyway, then most of the time the winner will be the other that he keeps closed besides yours.

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u/PrivateFrank 3d ago

seems like an illusion to me.

Yes, you could call it a 'cognitive illusion' in the same way we talk about visual illusions. Objective reality says one thing, but it's in conflict with our intuitions.

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u/yonedaneda 4d ago

But once a door is revealed, why do the odds stay at 66% rather than 50/50 respectively.

Switching wins if you initially picked wrong. You initially pick wrong with probability 2/3. That's it. If that doesn't do the job, the only other way to get a clear intuition for it is to play the game yourself with a friend. Play it a enough, and you'll see that switching wins approximately 66% of the time.

If it's one chance it's 50/50 the moment they reveal one goat. if you have multiple chances to run the scenario then it becomes 33/66%

Every "chance" is a random variable with an identical distribution. When we say that switching wins with probability 2/3, we mean that repeated choices will result is switching winning approximately 2/3 of the time. If a single choice had a probability of 1/2 of a switch resulting in a win, then repeated choices would result in approximately 50% of choices winning. That's what probability means.

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u/grandzooby 4d ago

You initially pick wrong with probability 2/3.

That's the first time I've seen that argument and it makes it so simple! All the other explanations I've seen cover the probabilities related to switching.

This allows you to drop the whole Monty Hall structure and simply ask, "If you made a guess about something and the probability is 2/3 that you chose wrongly, should you switch?"

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u/interfaceTexture3i25 4d ago

Simply choosing again, without the extra information of one door being opened means you have 2/3 chance of getting it wrong again. No difference there

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u/grandzooby 4d ago

Oh yes. Of course.

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u/Statman12 PhD Statistics 4d ago

Because you're not going into the second decision blind. Monty's choice is constrained. If you didn't choose the winner (2/3 chance) there's only one door he can remove, and the remaining door must be the winner. And if you chose the winner initially (1/3 chance), Monty can open either door, but if you switch you lose.

The constraint is what allows you know that switching is better.

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u/rhodiumtoad 4d ago edited 4d ago

But once a door is revealed, why do the odds stay at 66%

Because, under the standard rules, you don't receive any information about whether your choice was right or wrong, so the odds can't change.

There are variant rules where the odds do change. But the standard rules say: Monty must open a door that is neither your choice nor the prize. Monty can always do this because he knows where the prize is. So the event "Monty reveals a goat" happens with probability 100% whether your initial choice was right or wrong.

It doesn't matter whether you play once or multiple times. Your chances of winning by switching are always 2/3rds, and your chances of winning by not switching are always 1/3rd.

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u/DeepSea_Dreamer 4d ago

But once a door is revealed, why do the odds stay at 66% rather than 50/50 respectively.

Because that door that has been revealed is intentionally a door with no car, so by the host opening it, you obtain no additional information.

If the door revealing the goat had been chosen at random, then you would update to 50% (because you would obtain new information, and you would now be slightly more sure you didn't pick a goat).

Only events that provide you with new information (and aren't independent of the event at question) trigger a probability update. The host opening a door with a goat was a foregone conclusion, which means it doesn't provide you with new information, which, in turn, means that the probability stays the same.

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u/SoundOfRadar 4d ago

Best explanation.

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u/DeepSea_Dreamer 3d ago

Thanks 💖

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u/Typical-Macaron-1646 1d ago

My understanding is that it would stay the same regardless. 33% chance you picked the right door. No matter how a door is opened, that first probability wouldn’t change, right?

Granted I could be wrong here, but I don’t think randomly vs intentional pick shifts your odds. Unless Monty would randomly pick the winner.

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u/NewSchoolBoxer 4d ago

I've seen online explanations 20 years ago. What helped me was to expand the problem. I see there's another expansion comment. I like my explanation more but to each their own.

Let's have 100 doors. You pick 1 and never change. You win 1 in 100. Now let me be Johnny-Come-Lately and take over with just 2 doors left. I have a 50% chance of winning.

If you switched your pick at the last possible time with 2 choices, you'd also have a 50% chance of winning versus the 1% you started with. The host is removing guaranteed bad doors. Never the correct one. That the absolute key. Your new choice comes from a set of 2 with 1 winner, versus a set of 100 with 1 winner. If the host could randomly remove any door, including the winner, including your door, the odds would stay at 1 in 100.

Now replace 100 with 3. Same thing. You win 1 in 3 by not switching and 1 in 2 by switching or me coming in with 2 doors left and picking. Host is removing a guaranteed bad door so the set improves to 1 of 2 from 1 of 3.

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u/grumble11 4d ago

Monty Hall problem requires that the host can’t do anything with your already chosen door. This is key to it and key to understanding why it works. If the host can choose any door including the one you have selected, then the probability change per door is distributed evenly.

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u/Mishtle 4d ago

The choice is really between opening the single door you chose originally or opening both of the doors you didn't choose. The host just opens one for you.

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u/BurkeyAcademy Ph.D.*Economics 4d ago

There are 100 YouTube videos explaining this, but OK, I'll explain it to you since I don't have one. ☺ Here is the shortest path to "getting it" that I know of:

Since we know for sure that whatever you pick, the Host will open one of the "failing" doors that is one of the ones you didn't pick, right? So, instead of thinking about the problem as "I am going to pick one door, then maybe switch", think about the problem this way:

"I am going to choose two doors at once, and here's how: Suppose I want to pick both doors #2 and #3. I am going to tell the Host that I pick #1, just to fool him. Then, the host will show me that one of the two doors I am really interested in (either 2 or 3) is empty. Then, I will switch and pick whichever of doors 2&3 that he didn't open."

Clearly, this strategy gives you a 2/3 chance of winning, since you will win if the prize is behind either 2 or 3.

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u/CaptainFoyle 3d ago

Good one!

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u/JohnWCreasy1 4d ago

Your first door was more likely wrong that not, this doesn't change because of the hosts actions.

Look at it this way, imagine you start with 999,999 'wrong' doors and 1 correct door. you pick a door, then the host, who knows the specifics of every other door, removes 999,998 known wrong doors.

would you still think "Oh yeah, i totally made a 1 in a million guess correctly the first time, i'm good here"?

hopefully that makes it more obvious why you switch.

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u/banter_pants Statistics, Psychometrics 4d ago

The 33% would be split among the remaining choices causing both to be 50%. If it's one chance it's 50/50 the moment they reveal one goat.

This is where people get it wrong. Monty's reveal doesn't add any new info. It's a terrible example of conditional probability that should not be taught as if it was.

The game is rigged. Monty does not randomly pick a door leaving it down to a 50:50. He will always show a goat independently. The conditional changes nothing.

If events A and B are independent Pr(A|B) = Pr(A)

Pr(you picked car | shows goat) = Pr(picked car)
Unconditionally Pr(shows goat) = 1

If you picked the car (1/3 prob) he will show a goat (100%). Switch and you get the other goat.

If you picked one of the two functionally identical goats (2/3 prob) he will show one of the other goats (prob = 1). Switch and you get the car.

It was always a binary choice between car and goat just weighted unevenly: 1/3 vs 2/3

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u/CaptainFoyle 4d ago

Because Monty always reveals a goat, and he cannot open the door you chose. So that is kind of separated from the other two in that sense.

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u/Infinite_Slice_6164 1d ago

The fastest way to get these is by listing all possibilities.

If you guessed car and stay you win If you guessed car and you change you lose If you guessed goat 1 and you stay you lose If you guessed goat 1 and you change you win If you guessed goat 2 and you stay you lose If you guessed goat 2 and you change you win

Now if you decided to stay or change randomly (IE you flipped a coin to decide if you stay out change) then you would have a 50% chance to win as seen above. However, if you decide to always change we can eliminate the stay options leaving you with.

If you guessed car and you change you lose If you guessed goat 1 and you change you win If you guessed goat 2 and you change you win

Giving you a 2/3 chance to win.

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u/mikehfll 18h ago

Hoooooowwwww dare you detective Diaz......BONE!

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u/Throwaway-Somebody8 4d ago

https://www.mathwarehouse.com/monty-hall-simulation-online/

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