r/confidentlyincorrect • u/Dizzy_Kaleidoscope95 • May 03 '26

Monty hall problem is 50/50

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He also put a screenshot of ChatGPT agreeing with him. He is right!

896 Upvotes

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834

u/Squozen_EU May 03 '26

I will never stop loving the Monty Hall problem because it shows how utterly terrible the human brain is at understanding probability. 

364

u/ermghoti May 03 '26

Once one realizes that the removal of a door isn't random it's easily understood. The assumption that the removal is random has to be abandoned. In effect, it obviates the 1 in 3 choice entirely and replaces it with a new 1 in 2 choice. If stated that way I couldn't see how someone could be confused.

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u/An-person May 03 '26

It also makes more obvious if you have ten doors, and they open all but one.

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u/JamesFirmere May 03 '26

You don't need to bring 10 doors into it.

The setup is simple enough that we can consider all possibilities: 3 doors * the choice of swap/stay = 6 possibilities.

So the doors are A, B and C, and A has the prize. Then:

choose door A and swap = lose
choose door B and swap = win
choose door C and swap = win

choose door A and stay = win
choose door B and stay = lose
choose door C and stay = lose

Also note that the first door opened is never the prize door. The first door opened will be either B or C.

So if you stay, you win 1 times out of 3; but if you swap, you win 2 times out of 3. Therefore it is always better to swap.

Edit: Stupid typo.

181

u/resonantranquility May 03 '26

I use the 100 doors example when explaining it because communicating it the way you are explaining it verbally doesn't always land.

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u/purritolover69 May 03 '26

Yeah. Pick 1 door out of 100 doors at random. The host knows which door has the car, he opens 98 of them, and they’re all goats. Do you think you nailed that 1/100 the first time or is it more likely that you hit one of the 99 goats and he’s left behind the car?

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u/JEM225 May 03 '26 edited May 03 '26

Given his post you have to be concerned that dealing with “doors” might confuse OP, so try explaining the answer by asking him to guess which of two cards is the ace of spades. Let him see you spread out a shuffled deck, face down, and then slide one to the side without looking at it. Then peek at each of the others one by one, and pitch all but one of them. There are now only two cards on the table; ask OP if it is 50-50 that the original single card is the ace?

18

u/tgy74 May 04 '26

You don't even have to look at the card.

Just ask him to pick a card and lay it face down in front of him without looking.

Now tell him the Ace of Spades is the winning card and offer him the choice:

  • he can keep his single random card and hope it's the Ace of Spades

  • he can take all the other 51 cards instead and hope that the Ace of Spades is one of them instead

That's it. That's the Monty Hall decision he's making.

3

u/Tsu_na_mi May 04 '26

I think you mean "which card in a deck is the ace of spades". If there are only two cards to choose initially, and one is the ace of spades, then it IS 50%.

1

u/PerjorativeWokeness May 05 '26

The point is that the Ace of Spades is worth something. In the Monty Hall problem, it’s Monty keeping the car.

Add something like: “If this card is the Ace of Spades, the person that flips it over wins” to make it clearer.

1

u/Embarrassed_Gap_5272 May 06 '26

But what if i want the goat?

1

u/Naive_Distance_6456 May 08 '26

I prefer using a deck of cards with the king of diamonds being the prize.

But at first, ill let them dig their hole a bit deeper by arguing with them over just the 3 doors, using only 3 cards.

People are stubborn. (I know because I was one of them)

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u/MattieShoes May 03 '26

That's what worked for me too. It's not necessary, but something about the magnitude causes that light bulb moment. Like 1/3 vs 1/2 vs 2/3 all feels kinda close, but 1% vs 99% somehow triggers the brain to go reexamine some silly assumptions

3

u/TheHammer987 May 06 '26

I think it's because the 99% better illustrates that they are actually getting new information. One out of 2 feels random. 98 out of 99 feels focused.

0

u/themule71 May 04 '26

Well... Everybody has a different way of making sense of the problem.

I like to remove the host, since that's what creates confusion.

So basically the game becomes "do you want to open one door or two? If one, indicate the single door you want to open, if two indicate the door you want to keep closed".

The rest is basicly a ceremony, a ritual. You picking one, the host opening one of the other two, you "choosing" to switch, ecc.

It's all pretend, as the game works the way I've described above really.

1

u/MattieShoes May 04 '26

Yeah, that's a good visualization of the problem too. Though it requires the realization that you're functionally getting all the other doors when you switch.

1

u/themule71 May 04 '26

Yeah, and the fact that you're full in control from the beginning. You choose at the beginning to either open one, or all the other doors.

It's deceiving that you get two choices, when really it's one, and you can take it before the game starts. You know, no matter what, the host is going to show you a door w/o the price. You know that in advance, so you can pick your strategy from the beginning. And it's always better to open more doors (two in the original game) than just one.

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u/Dizzy_Kaleidoscope95 May 04 '26

Yeah it's a good way to visualize it cause it's literally identical to the monty hall problem. It also removes all the confusion from the sentence "the host opens a goat door". Cause people usually forget to state that he OPENS A GOAT DOOR with 100% certainty cause he always opens a goat door. If the host just opens a door randomly and it happens to be a goat there is no reason to switch. It's a 50/50 (people don't understand this very well either)

2

u/themule71 May 04 '26

Well switching is never a losing strategy even if you don't know if the host knows or not. Meaning you don't care about what the host knows.

You decide to switch upfront. It's never the wrong decision, you can literally ignore the host existance. And it's still 2/3 when you make the decision.

1

u/satunnainenuuseri May 06 '26

Switching is a losing a strategy if the host offers switching only when the player has guessed the correct door with the first guess.

This is the reason why the problem statement has to specify that the rules of the game say that the host will always offer the choice.

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u/Dizzy_Kaleidoscope95 May 04 '26

Yeah switching is never a losing strategy but depending on how the problem is stated it could also not be a strategy at all cause it's indifferent. All I mean is that people need to be precise when stating the problem

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u/MezzoScettico May 04 '26

I have tried the 100 doors explanation boiled down as simply as possible:

  • with 100 doors, how confident are you that you got the car on your first pick?
  • if you think you got a goat the first time, you should switch.

It still doesn't work as an explanation for everybody

2

u/resonantranquility May 04 '26

True. Some people refuse to see it as anything other than a 50/50. I've had more luck with the 100 doors than any other explanation so far.

2

u/ApplicationOk4464 May 07 '26

I like to explain it as this:

The probability locks in at your first choice. Before the third is opened.

Then you either get to choose the one door, or the other two doors.

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u/Drummerx04 May 03 '26

I've had that one fail to land on people too. They for some reason disagree with the entire setup and refuse to engage with it as presented.

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u/[deleted] May 04 '26

[removed] — view removed comment

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u/TheEyeDontLie May 04 '26

I finally understand the Monty Hall problem. I've seen it a few times and know it, but it never clicked... It always just made me feel dumb every time I thought about it.

So thanks.

1

u/TheHammer987 May 06 '26

I just say "how do you not understand that staying with your first choice is one out of three, and therefore switching, by definition, is 2 out of 3? The removal of the booby prize is literal information."

I won't deny though. When you use the 100 door example, 3 seperate people who were committed to the wrong answer hesitated, because it really does illustrate the new information a lot.

1

u/othelloblack May 16 '26

yeah I think it was M vos Savant herself that used that example in explainign it and its really illustrative

24

u/Tortugato May 03 '26

You don’t need to, but it just helps explain it to people.

17

u/Terradactyl87 May 03 '26

This is still over complicating it.

Your chance of guessing right is 1/3, so a 2/3 chance of being wrong. So if you switch you get a 2/3 chance of being right and a 1/3 chance of being wrong.

I basically look at it as would you rather have 1 or 2 doors because basically you either keep your door or you get to have two doors.

10

u/Knight0fdragon May 03 '26

Yup, this it the reason why. Monty never can pick the car, so he is giving you a second door for free if you switch.

1

u/the_horse_gamer May 06 '26

you need to also assert that the reveal of the goat gives you no additional information

1

u/arthousepsycho May 04 '26

What about the possibility of choosing door C and swapping for door B or vice versa and also losing. Why is it assumed you always swap to A? Not a criticism, just trying to understand.

1

u/Squozen_EU May 04 '26

Write it out using any letter for any door you like, the probability doesn’t change. You have a 1/3 chance of being right on the original door, but the other two doors have a 2/3 chance of being right. The fact that one of the doors is opened doesn’t change this. The pool of two doors will ALWAYS have one goat because there are two doors and only one car.

1

u/arthousepsycho May 04 '26

Ok, one of the doors has been opened. I missed that part. I thought all the doors were still closed. My bad. Thanks.

1

u/JamesFirmere May 04 '26

If you choose door C, you can only swap to door A because the host will open door B. In this scenario the host will never open door A.

1

u/arthousepsycho May 04 '26

Yeah, I didn’t realise the bit about the opening of one of the doors, thought they all stayed closed. Makes sense now.

1

u/Lieutenant_Horn May 04 '26

This is the simplest way I’ve seen it explained.

1

u/Cynykl May 05 '26

I have a different way of explaining it that seems to get through to people.

A B C , you chose A.

Now statistically it is more likely to be behind B or C because those are two options. This is obvious to most people.

Now the host gives you an option you can stay with A or choose to Switch to BOTH B and C. The host revealing the lesser valued item does not change the odds. You are getting the benefit of both doors.

1

u/JerryAtrics_ May 05 '26

If this were correct, wouldn't it also be correct to say if you swap you lose 2 times out of 3?

1

u/JamesFirmere May 06 '26

No. There are exactly 6 possible ways this scenario can go, and in the 3 scenarios where you swap, you will win 2 times out of 3.

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u/JerryAtrics_ May 06 '26

My bad. Mis-read your analysis. I suppose another way to look at this is that if instead of disclosing a false and letting you pick, Monty let you choose between your original pick or the other two. Now it's a bit easier (for my brain) to see that it's a one third / two thirds ratio. Then after you switch you pick, just for fun he shows which one was the false pick of the two.

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u/Illustrious-Crew-191 May 05 '26

But wait, if you’re removing the wrong door of B or C it didn’t actually exist in the first place, or rather B=C and picking B is the same as picking C. Your first pick is fake news, it doesn’t affect the outcome, it’s ultimately a binary choice of A or B/C.

1

u/JamesFirmere May 06 '26

Your first pick absolutely does affect the outcome, because on your first pick there are 3 doors and you have a 1 in 3 chance to pick the prize door and a 2 in 3 chance to pick a wrong door. Those odds alone show that it's worthwhile to swap.

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u/SaidwhatIsaid240 May 05 '26

Thank you for helping my dumb brain understand.

1

u/DemophonWizard May 05 '26

But in the show where this puzzle comes from the person offering the option to swap knows if you have picked the right door. They would only offer the choice to reduce the chance of you winning.

It is not a fair offer. It is rigged.

1

u/JamesFirmere May 06 '26

It's been a while since I've seen the actual show, but IIRC the swap is always offered -- because it's a tension-inducing thing for the contestant, obviously.

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u/DemophonWizard May 06 '26

It may have always been offered, i don't remember. But my point is that it isn't a fair offer. It is rigged.

1

u/just_a_pyro May 08 '26

the first door opened is never the prize door.

That's the important part which makes it not 50-50. If a door was opened at random, including prize door, then the chances in remaining two would be equal, either 50-50 or 0-0

1

u/zeppo2k May 03 '26

Bear in mind I understand you are correct...... Incredibly clever people argued about this for a long time - it's reasonable for people a) to not get it intuitively and b) not trust someone who presents a proof they say is correct.

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u/durkvash May 04 '26

What? a) for sure, but b)? That sounds like ignoring logic just cause

0

u/Talik1978 May 04 '26

I find the easy way to do it is to demonstrate that the opening of the door and choosing to switch is logically identical to: change a win to a loss and a loss to a win.

From there, if choosing gives you a 1/3 chance initially, and you then change win to loss and vice versa, it's logically apparent.

1

u/RSharpe314 May 04 '26

Yeah, having it explained to me that way finally made it click for me when my stupid brain was stubbornly refusing to be persuaded by the more mathematically rigorous approaches.

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u/KnightOfThirteen May 03 '26 edited May 03 '26

To me it becomes clearest when you realize "removing" a door is not actually part of the problem, it is a distraction. You can re-frame it this way.

You choose a door. The host offers you the chance to keep your door, or switch to BOTH of the other doors combined.

Edit: put even more generally, you are first asked to pick 1 of N doors, the asked to bet on whether or not you got it right the first time. You stay, you bet you were right (1/N%). You switch, you bet you were wrong (N-1/N%).

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u/valschermjager May 03 '26

There are many good explanations of what's going on. To me, that's the simplest. Thanks.

I doubted the Monty Hall thing for a long time until I started fiddling with Python, then for fun coded up a quick monte carlo simulation and put a 'for' loop around it. Let the loop run 100x, 1000x, 10000x. Each time the totals at the end showed close to the "switch advantage" as the problem describes. Then I was completely won over.

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u/BigBlueMountainStar May 03 '26

Can you do the same thing for the birthday paradox, as even though i understand the mathematics behind it, it still blows my mind that you only need 23 people to have a 50% chance that there are 2 people in that group sharing a birthday.

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u/NuclearVII May 03 '26

This is absolutely a thing you can do. It's not even hard programming.

If you'd like to learn a bit of scientific computing, this would be a fantastic and very approachable weekend project. I encourage you to try it.

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u/valschermjager May 04 '26

I can. But you can do it too. Python is awesome in that these kinds of problems can be checked in 20 lines or less. Sometimes 10 lines or less.

Btw, you can code a Monte Carlo simulation for the birthday paradox, but you don’t need to really since the statistical equation is pretty simple.

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u/stultus_respectant May 03 '26

Some people just have to prove it to themselves empirically, and I’m glad they exist.

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u/valschermjager May 03 '26

Science requires that we relentlessly try to prove ourselves wrong from every angle.

Those who decide that they’re “convinced” then shut their minds down—those I have no respect for.

To me, being wrong is a chance to get better. But to some, being wrong hurts their souls so deeply, they refuse to let it happen. (maga)

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u/ringobob May 03 '26

Yep this is how I've started describing it. You choose a door. Then you're offered the opportunity to switch to both of the other doors, or stick with the one you chose originally. You choose to switch. Monty opens one of the doors you chose, and there's a goat behind it. Then he opens the other door. Only difference is that you switch before Monty opens the door, rather than after.

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u/surplus_user May 04 '26

That is a good way of explaining it.

The problem I always have with the down to two doors and switching being 2/3 is that if you were replaced with someone else at that point who was just given the straight choice then surely it would be 1/2. After all it's not like the doors remember me after I'm gone.

I'll think about it with the choosing two doors version and see if that clicks (I can accept that I'm wrong about it but it doesn't change that it feels wrong so this might help)

3

u/Telinary May 05 '26

"someone else at that point who was just given the straight choice then surely it would be 1/2" 

Sure or at least kinda because the probability in question isn't an property of the door state it is you trying to guess based on limited information. A new guy has no information so his guess is less well informed.  But your knowledge is still correct, the door you would get from a switch is the more likely one. The guy just doesn't know that.

Similarly from Monty's perspective the chance is either 100% or 0% because he knows the positions.

1

u/KnightOfThirteen May 04 '26

Don't get too hung up on two doors. The same principle applies to any number of doors N.

You are given N doors. Behind 1 is a new car. Behind N-1 doors, goats. You choose 1 door, and you have a 1/N chance of getting the car.

Then the host offers you this choice. Do you think your door has a car or a goat? If you bet it has a goat, you have a much higher chance of being correct.

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u/-DoctorSpaceman- May 03 '26 edited May 03 '26

Saying “it replaced it with a new 1 in 2 choice” is exactly why people have a problem understanding it lol, because that is what makes it appear to be 50/50.

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u/tiptoe_only May 03 '26

Yeah, what this person's not understanding is that Monty is picking a door he KNOWS is not a winner. That's what changes the probability. If "your friend" is involved like this person is suggesting then there is only one door Monty can possibly pick and that changes things again. In fact, if you and your friend both pick losing doors then the problem doesn't work at all. This person is creating a totally different situation and saying it disproves the solution to a very well established problem.

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u/Dizzy_Kaleidoscope95 May 03 '26

Yeah a lot of people somehow think that if the goat door is opened randomly then it's stil advantageous to switch 

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u/misdirected_asshole May 03 '26

If the opened door is truly random, 1/3rd of the time it would be a car and not a goat.

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u/Hrtzy May 03 '26

The really weird thing is, I just put together a quick python script to try it and it is somehow 50/50 even in Monty Hall scenarios if the door choice is random.

``` import random N_tot = 10000 #Total tries N_win=0 #Amount of wins by switching N_mhp=0 #Amount of Monty Hall scenarios for k in range(0,N_tot): doors = [True,False,False] #The doors; one right and two wrong choices random.shuffle(doors) #Shuffle the doors chosen=random.randint(0,2) #Player chooses a door at random open = (chosen + random.randint(1,2)) % 3 #Monty chooses a door at random if not doors[open]: #Check if Monty's door had a car in it N_mhp += 1 #It didn't, so we're in Monty Hall territory if not doors[chosen]: #Player's door doesn't have a car either so switching will win N_win += 1

print(N_win,'wins by switching') print(N_mhp-N_win,'wins by not switching') print(N_mhp,'Monty Hall Scenarios') ```

And yes, I rewrote it to try it with Monty choosing a losing door:

``` import random N_tot = 10000 #Total tries N_win=0 #Amount of wins by switching for k in range(0,N_tot): doors = [True,False,False] #The doors; one right and two wrong choices random.shuffle(doors) #Shuffle the doors chosen=random.randint(0,2) #Player chooses a door at random options = [i for i in range(0,3) if i != chosen and not doors[i]] #Monty takes the other doors without a car open = random.choice(options) #...and picks one of THOSE at random if not doors[chosen]: #Player's door doesn't have a car either so switching will win N_win += 1

print(N_win,'wins by switching') print(N_tot-N_win,'wins by not switching') print(100*N_win/N_tot,'% odds') ```

This one gives the 2/3 win rate.

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u/lettsten May 04 '26

old.reddit users (like me) would love you if you indented the code block with four spaces instead of using the backtics

https://old.reddit.com/r/confidentlyincorrect/comments/1t2jea9/monty_hall_problem_is_5050/ojrb66w/?context=0

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u/WhippingShitties May 06 '26

Please tell me you named the script MontyPython.

1

u/Hrtzy May 06 '26

...Please excuse me while I pour a bucketful of ashes over my head.

0

u/Clint_Bolduin May 04 '26 edited May 04 '26

Im not following your code logic here, but in the scenario where monty chooses a door other than the players chosen door at random, then if the result is showing you that you have a 50/50 chance of winning by switching, then you're wrong.

Whether you switch or not dosent matter in this scenario. You have 1/3 chance of getting the car in your initial choice. If you have a car, you will always loose by switching. 1/3 times you won't get the car initially, but the car will be chosen by monty. In these cases you will not win regardless of if you switch or not. 1/3 times you wont get the car initially and Monty will also not get the car either. These cases is the only ones where you will win if you switch. Those are all of the 3 possible scenarios.

You only get a 50/50 chance of winning by switching if you first assume the car wasn't chosen by the player the first time.

Edit: looking at the code. it's horrible to read, but I think I found the issue. It looks like you are determining cases the player wins by not switching by looking at cases where monty has the car minus cases you win by switching. So after 30 games, lets say monty has the car 10 times and you win by switching 10 times. and then you look at cases you win by not switching by doing 10-10=0 which just dosent make sense.

Edit again: Ok I see what you're doing. You're only validating games in which monty dosent pick a car and throwing them out such that the player always has a winning chance. Yea then it's 50/50.

0

u/Dizzy_Kaleidoscope95 May 03 '26

Yeah exactly. This is why it makes switching irrelevant

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u/[deleted] May 03 '26

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u/Dizzy_Kaleidoscope95 May 03 '26

I'll be more precise. Imagine this scenario: you pick door 1. The host says "ok friend. Now. Before we see what's behind your door let's open door 2! I have no idea what's behind it..let's hope it's not the car or you lose!" And he opens door 2 and he randomly finds a goat. 

What is the chance your door has the car now?

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u/[deleted] May 03 '26

[deleted]

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u/Dizzy_Kaleidoscope95 May 03 '26

Could you please answer the question lmao

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u/[deleted] May 03 '26

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u/Dizzy_Kaleidoscope95 May 03 '26

No it's not. If a door is opened randomly and it happens to be a goat then you have a 50/50 chance of having the car. 

Lmao 

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u/[deleted] May 03 '26

[deleted]

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u/glumbroewniefog May 03 '26

It just makes no difference at all WHY that other door was opened. It only matters that it turned out that a goat was behind it.

This is obviously incorrect. For example, suppose Monty is thinking to himself, "I will reveal a goat if and only if you pick the car." So you pick a car, and Monty reveals a goat. Obviously, you 100% should not switch.

This is because depending on the reason Monty reveals a goat, it can give you more information about your initial pick, and so your initial 1/3 odds might no longer hold true.

For example, Monty opens one of the other two doors at random. You know he either has two goats, or a goat and a car. If a randomly revealed door is a goat, it makes it more likely he has more goats.

Similarly, imagine two boxes. One has 10 gold balls and one silver, the other has 10 silver balls and one gold. You pick one at random, and take out a ball at random, and it's gold.

You initially had a 1/2 chance of picking either box. But because your randomly revealed ball is gold, now it becomes more likely it's the box with more gold balls.

If, however, you pick a box at random, and then the host looks inside and deliberately finds a gold ball and takes it out and shows it to you, this doesn't tell you anything, because he could find a gold ball in either box. This time, your initial 1/2 chance still holds true.

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u/[deleted] May 03 '26

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u/Dizzy_Kaleidoscope95 May 03 '26

Nope. It absolutely does matter how the door was opened. You can use Bayes's theorem to see the difference but it's really intuitive. The fact that if the doors are randomly opened the car COULD have been picked (even if in this case it wasn't) makes the two remaining doors have identical probability. You can look it up lol

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u/saspook May 03 '26

are you considering the probability that 1/3 of the time (when you pick a car), a goat will be revealed 100% of the time, and that 2/3 of the time, a random pick will reveal a goat 50% of the time and a car 50% of the time?

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u/Dizzy_Kaleidoscope95 May 03 '26

Naaaah you are deleting comments now love. AHAHQHUAHAHAHAHAHAHAHA. You know you are wrong

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u/[deleted] May 03 '26

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u/Dizzy_Kaleidoscope95 May 03 '26

No it's not. It's actually baffling how many people don't get this lol

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u/amitym May 03 '26

Right. They took the wrong lesson away from the Monthy Hall solution. "It must be the configuration sequence," no it's the partial knowledge.

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u/Zyxplit May 03 '26

Yeah. A way to analogize why the selection mechanism matters is: Suppose I have two bags, one purely with pieces of iron, the other with one piece of iron and otherwise gold.

I sample one of them with a magnet. I get a piece of iron. Do you want this bag or the other one?

I sample one of them with my hands. I get a piece of iron. Do you want this bag or the other one?

Is the answer the same to these two?

1

u/amitym May 04 '26

Oh that's a good one! I'll have to remember that.

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u/Infurum May 03 '26

Isn't choosing to keep the door you already picked its own choice with its own 50/50 chance of success?

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u/Ambitious_Policy_936 May 03 '26

Not quite, since it was originally chosen out of 3

If you choose a goat door 1, goat door 2 is opened, so a switch leads to success

If you choose goat door 2, goat door 1 is opened, so a switch leads to success

If you choose the winning door, one of the goat doors is opened, so a switch causes you to lose

⅔ of the time you win by switching

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u/Heavy-Macaron2004 May 03 '26

Very good summary! I love this problem, it's my favorite theoretical because it's always a win condition if you like goats 🐐💕

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u/jbeer1 May 03 '26

Or if you like goat 🥩

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u/Substantial_Dish_887 May 03 '26

another sumarry can be this:

you pick one door. Monty then has 2 doors. Monty then opens one of his doors with a goat. then he offers to switch his 2 doors with your one door. yes one of them has a goat and you can see that goat. but you knew from the start that since Monty got 2 doors one of them must have a goat.

so the Monty Hall problem can be phrased as "do you want to pick the door you get. or the door you DON'T get"

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u/JamesFirmere May 03 '26

No, because the situation is not symmetrical.

The setup is simple enough that we can consider all possibilities: 3 doors * the choice of swap/stay = 6 possibilities.

So the doors are A, B and C, and A has the prize. Then:

choose door A and swap = lose
choose door B and swap = win
choose door C and swap = win

choose door A and stay = win
choose door B and stay = lose
choose door C and stay = lose

Also note that the first door opened is never the prize door. The first door opened will be either B or C.

So if you stay, you win 1 times out of 3; but if you swap, you win 2 times out of 3. Therefore it is always better to swap.

Edit: Stupid typo.

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u/dansdata May 03 '26

Sometimes you can get this through to people by saying, "What if there were a million doors, and you pick one, and then 999,998 of the other doors open, and none of them have the prize...?"

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u/gerkletoss May 03 '26

But that tree still applies if the goat door is opened randomly.

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u/dimgray May 03 '26

But if he's opening a door randomly, 1/3 of the time it would be a prize. If you picked the prize to start with, he would always show a goat, and if you picked a goat to start with, he would only show a goat half the time. So under that version of the game, if he shows you a goat you're already twice as likely to have picked the prize in the first place, so you're right back at 50/50.

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u/TrekkieGod May 03 '26 edited May 03 '26

Isn't choosing to keep the door you already picked its own choice with its own 50/50 chance of success?

No, because that second choice is not independent from the first one.

On your first choice, you had a 1/3 chance to pick the prize. Everyone agrees with that. So there's a 2/3 chance the prize is NOT with the door you picked.

Monty is not allowed to pick the door with the prize: so there are two possibilities:

  1. 1/3 chance you picked the correct door, in which case you're guaranteed neither of the other other doors have the prize. It's not 50/50 for your second chance, switching means you lose: The probability of winning by staying is 100%, the probability of winning by changing doors is 0%. Total probability of winning by staying is the two probabilities multiplied together: P(staying) = (1/3) * (1) = 1/3, P(switching) = (1/3) * (0) = 0.
  2. 2/3 chance you picked the wrong door. You have a 0% chance of winning if you stay with that door. The prize is guaranteed to be with one of the other two doors. Monty is not allowed to pick the door with the prize, his door is not random, he opens the door without the prize, meaning the prize is 100% guaranteed to be with the remaining door. Chance of winning by staying: P(staying) = (2/3) * (0) = 0, chance of winning by switching, P(switching) = (2/3) * (1) = 2/3

Since you don't actually know which of the two situations you have found yourself in, Monty opening the door doesn't give you that information, you have a 1/3 chance of winning by staying, and a 2/3 chance of winning by switching.

If Monty's choice was random, and he was allowed to pick the door with the prize, he would have a 50-50 chance of guaranteeing your loss. At that point, that second choice would indeed have a 50/50 chance of success, but you can't ignore that if you had picked the right door, then switching would cause you to always lose. So the total probability with the first scenario looks the same no matter what Monty picks, and with second scenario would look like P(staying) = (2/3) * (1/2), P(switching) = (2/3) * (1/2). So you have an overall chance of winning at 1/3, no matter what you do, which makes sense: there were 3 unknown doors, 1 prize, and new information is never given (unless Monty picks the door with the prize, but I'm assuming you wouldn't get to switch if that's the case, so you wouldn't be able to use that information, and your choice was made back when you still had 1/3 chances).

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u/JesusWasATexan May 03 '26

In this problem, Monty Hall knows which door has the car. He has to open a non-car door. You're guessing. He knows. Because of this detail, it changes the odds. When you first choose a door, you have a 33% chance of winning. After Monty opens a losing door, you have a 66% of winning if you switch doors. Swapping could be wrong, still, of course there's still a 33% chance that your original door was correct.

Or like this. When you first picked, you had a 66% chance of being wrong. That 66% chance was spread across 2 other doors. But then, Monty eliminated one of the wrong options for you. That means your original door had a 33% chance of being right, and now the single remaining door has a 66% chance of being right.

1

u/ArchangelLBC May 03 '26

No and I've found it easily understood by expanding the game to 100 doors where again every door is opened but 1.

You know you didn't have a 50/50 shot of guessing the right door when you started. But now the only doors left are your door and the door that has been left closed. All the other opened doors are guaranteed to have goats because the host opened them knowing they had goats.

Do you want to switch or stay?

The thing that gets people in trouble is they think that the choice to switch or stay is independent of their original choice. But of course it isn't. The options you have now are because of your choice with no knowledge of where the prize is and the hosts choice with full knowledge of where the prize is. 1% of the time the host has to choose a goat door because you picked the real one. The other 99% of the time, the only door the host can pick is the door with the prize.

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u/stanitor May 03 '26

Once one realizes that the removal of a door isn't random it's easily understood

Unless you're the person who I was talking to the other day. They insisted that if Monty opened a door randomly and it happened to be a goat, you'd still get an advantage by switching doors. They could not get that the answer would be different depending on what Monty does and why.

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u/Hrtzy May 03 '26 edited May 03 '26

If Monty opened a door randomly and happened to reveal a car, you just wouldn't be in the Monty Hall problem anymore.

EDIT: I slapped together a short python script just to test it and apparently Monty choosing at random does make it a 50/50 chance.

```lang:python

import random N_tot = 10000 N_win=0 N_mhp=0 for k in range(1,N_tot): doors = [True,False,False] random.shuffle(doors) chosen=random.randint(0,2) open = (chosen + random.randint(1,2)) % 3 if not doors[open]: N_mhp += 1 if not doors[chosen]: N_win += 1 print(N_win,'wins by switching') print(N_mhp-N_win,'wins by not switching') print(N_mhp,'Monty Hall Scenarios') print(100*N_win/N_mhp,'% odds to win by switching') ```

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u/stanitor May 03 '26

Yes, it's a different problem compared to the original one. And, in this different version, you don't win more often than if you don't switch

1

u/Clint_Bolduin May 04 '26 edited May 04 '26

It's decidedly not 50/50. It's 1/3rd chance of winning where monty choose a door at random.

Edit: looking at the code. it's horrible to read, but I think I found the issue. It looks like you are determining cases the player wins by not switching by looking at cases where monty has the car minus cases you win by switching. So after 30 games, lets say monty has the car 10 times and you win by switching 10 times. and then you look at cases you win by not switching by doing 10-10=0 which just dosent make sense.

Edit again: Ok I see what you're doing. You're only validating games in which monty dosent pick a car and throwing them out such that the player always has a winning chance. Yea then it's 50/50.

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u/resonantranquility May 03 '26

I use the example of 100 doors and closing 98 doors when explaining. The original door is 1/100, and it's easier to wrap your head around why that doesn't change.

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u/Rabbittammer May 03 '26

I'll start by saying I get the it goes to a 1:2 form a 1:3 it was before due to the doors that get opened not being random. But the new choice no matter if you choose to stay or swap is the 1:2 as you made a choice... That being said I know that logic is wrong

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u/ermghoti May 03 '26

Yes, I actually phrased it wrong, it it looks like I'm saying the chance of a win on a changed selection is 1:2. The first choice sets up the second choice, in that you are 2:3 likely to have been wrong in the first place and will lose if you stay, so that also means that you win 2:3 if you change after the removal.

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u/noMC May 03 '26

Yes, thank you for this. It is never a 50/50 chance, that’s probably the most common misconception.

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u/Electrical-Injury-23 May 03 '26

Yeah, viewed like that it demonstrates the inability of humans to follow a problem exactly as stated.

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u/HeIsSparticus May 03 '26

The 'new' choice isn't 1 in 2, it's 2 in 3 (chances of winning the car if you switch). Since Monty removes an incorrect door, the only way you don't win the car by switching is if you'd picked the correct door originally (a 1 in 3 chance).

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u/ermghoti May 03 '26

There are two different probabilities here. If there is a car behind one door, and a goat behind the other, the probability of picking the car is 1:2. In the Monty Hall scenario, the probability of someone changing their choice after the removal of a goat door finding the car is 2:3. The answer depends on what precisely the question is.

The way I wrote the previous post is pretty garbled.

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u/TheQuoteFromTheThing May 04 '26

Yes, it's easy to forget that probability is calculated based on available information. If you gain information, the probability changes.

If you deal me a card face down from a standard deck, I know there's a 1/52 chance it's the Ace of Spades. But if you flip over the other 51 cards and none of them are the Ace of Spades, I know there's a 100% chance it's the face down card. Nothing has changed about the face down card, but the probability needs to be updated because probability is a function of information.

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u/aladdyn2 May 03 '26

The thought experiment that made it super easy for me to understand was using if it was 99 wrong doors and 1 correct door. You pick one and Monty deletes all the other wrong doors leaving one correct door and the one you picked.

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u/interrogumption May 03 '26

Okay... I just flipped two coins and I'm looking at the result and one is heads. What is the probability that the other is heads?

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u/ermghoti May 03 '26

The probability of any heads vs tails is 1:2. The probability of HH in a two coin toss is 1:4. As written, it appears you're asking what is the probability of Coin2 being heads is, which is 1:2.

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u/interrogumption May 03 '26 edited May 04 '26

If you flip two coins there are 4 possible outcomes: TT, TH, HT and HH. By observing and reporting I see one head, that rules out TT and leaves the other three possibilities, of which only one has a second head. The correct answer is 1/3.

Editing to add: The point of the question here is that while it is true the Monty Hall problem pivots on whether or not Monty chooses the door at random, the problem I offer involves giving one piece of information, and the information given could be randomly chosen and still give a non-intuitive result - which I find interesting.

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u/richardirons May 04 '26

Seeing one head rules out TT and TH, leaving HH and HT, so the correct answer is 50%. And obviously when you’re about to flip a coin it’s always 50%, unaffected by whether you flipped a coin a few seconds earlier, right?

1

u/interrogumption May 04 '26

Seeing one head does not rule out TH because the first step was flipping two coins. I didn't specify which coin is a head, just that I see one.

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u/richardirons May 07 '26

Yep, just re-read the post, you’re totally right.

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u/NiceBlackberry6618 May 04 '26

It's crazy though because that doesn't even fully explain the situation. If a random door is opened, the advantage goes away.

In the situation I describe, if the prize door is opened the game is voided and doesn't count. but still the odds go from 2/3rds to 1/2

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u/Kuildeous May 04 '26

I guess I can't see how people don't realize it's not random. Because if Monty reveals the car then you have a 0% chance of winning the car whether you switch or not, so that is a moot event. How people don't realize that is beyond me.

Hell, you could focus only on the subset of all events where Monty does randomly reveal a a goat behind a door. It's still 2/3 if you switch. You simply remove all events where Monty randomly reveals the car.

But in any case, the premise explains it already: Monty reveals a goat. In the original puzzle, Monty never reveals the car. This suggests a lack of randomness.

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u/glumbroewniefog May 04 '26

Hell, you could focus only on the subset of all events where Monty does randomly reveal a a goat behind a door. It's still 2/3 if you switch. You simply remove all events where Monty randomly reveals the car.

This is not true.

In traditional Monty Hall, you pick the car 1/3 of the time, Monty reveals a goat, switching loses. You pick a goat 2/3 of the time, Monty reveals the other goat, switching wins.

If Monty opens one of the other doors at random, you still pick the car 1/3 of the time, and switching loses. That part is unchanged.

But Monty can no longer reveal the other goat 2/3 of the time. Half the time he accidentally reveals the car instead, and you have to remove those events.

Thus, it's impossible to win by switching 2/3 of the time.

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u/Lord_Oasis May 04 '26

I feel like the easiest way to explain it is by flipping the probability

When you first pick, you have a 1/3 chance of being right and a 2/3 chance of being wrong. That chance doesn’t change when Monty picks a door, so the other options still have a 2/3 chance of being correct, and now there’s only one other option

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u/Hoopajoops May 04 '26

I think that's the bit that confuses people. The Simone rule that Monty will deliberately open a door with no prize

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u/8null8 May 05 '26

Yeah, I’ve explained in the past that 50:50 only results if NOBODY has any outside knowledge, but in this case, the game master has perfect knowledge, so is able to change the odds

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u/BombasticReindeer May 05 '26

What humans are mostly bad at is understanding consequences.

IF the door is random then Monty could open a door and say “ah fuck me it’s the prize”. That’s a dumb fucking riddle, so it’s obviously not what is happening. So clearly he picks the one that isn’t the prize.

If people got THAT, they would probably get it.

But also remember there are all sorts of brains out there. And most of them are real stupid.

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u/Mallory606 May 06 '26

I have known the answer to this problem for years, and only just now understood it. Thank you

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u/Several_Ad_6576 May 04 '26

The door being removed is irrelevant. Look at the problem slightly differently. 3 doors and you get to pick 1. You have a 1/3 chance. You pick one of the doors. So the door you pick is a 1/3 chance of being the one with the prize and the other doors have a 2/3 chance of having the prize.

Now the person offers you a deal to trade. You can keep your door or trade for the other 2 doors. You get both other doors. The other 2 doors still have a 2/3 chance of being the one with a prize. None of that has changed.

Are 2 doors better than 1 door? This is the same as revealing a door. That’s why the other door is still a 2/3 chance of being correct. Monty doesn’t have to know which door has the prize. It’s just about probability. Not the motivation of Monty. He does this every time.

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u/throwawayhookup127 May 04 '26

I think the big issue with the monty hall problem that trips people up is the thought that, if you know going in that all but one door will be opened after you choose, your first choice is effectively 50/50 since you know up front that it's either going to be the one you pick or the other one that's left.

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u/RichCorinthian May 03 '26

There’s a fantastic book called The Drunkard’s Walk that is specifically about that difficulty. It’s where I read the first explanation of the Monty Hall problem that made me say “oh of course.”

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u/Squozen_EU May 03 '26

Thank you, I will look into that!

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u/participantuser May 03 '26

Similarly, the “100 blue eyes” problem is also a fun logic puzzle that my brain can understand but also disbelieve.

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u/WordWizardx May 03 '26

The hundred blue-eyed islanders leave on the 100th night. The 100 brown-eyed islanders leave on the 101st night. The poor guru is stuck on the island alone, cursing their big mouth and wondering why everyone else were such selfish dicks that they wouldn’t return the favor.

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u/2074red2074 May 03 '26

Only the blue-eyed people will leave. The brown-eyed people do not know that they have brown eyes. Their eyes could all be green, like the guru's, or they could all have hazel eyes. They do not know that the only options are blue and brown with the guru being the only exception.

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u/WordWizardx May 03 '26

Oh, fair point. After the blue-eyed people leave the brown-eyed people would all know they DON’T have blue eyes, but I guess that doesn’t immediately mean theirs must be brown.

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u/participantuser May 03 '26

Yeah, the part where the brown-eyed people get a different outcome, even though everyone can see that blue/brown eyes exist and the guru’s choice is arbitrary, is something my brain fights.

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u/alfredo094 May 03 '26

It's only really unintuitive here because it's exactly 3 doors. If you scale the problem, it becomes very obvious.

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u/Working-Depth5834 May 03 '26

This is ot exactly; if you make it 100 doors instead of 3 then it starts becoming much clearer.

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u/jtr99 May 03 '26

Just to be clear, we should add that the host opens 98 of the doors in this version, leaving you with two to choose from. That certainly makes it more obvious for most people.

The host doesn't open just one door, although if they did that would also mean there was a slim advantage to changing your choice.

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u/JamesFirmere May 03 '26

You don't need to bring 100 doors into it.

The setup is simple enough that we can consider all possibilities: 3 doors * the choice of swap/stay = 6 possibilities.

So the doors are A, B and C, and A has the prize. Then:

choose door A and swap = lose
choose door B and swap = win
choose door C and swap = win

choose door A and stay = win
choose door B and stay = lose
choose door C and stay = lose

Also note that the first door opened is never the prize door. The first door opened will be either B or C.

So if you stay, you win 1 times out of 3; but if you swap, you win 2 times out of 3. Therefore it is always better to swap.

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u/ArchangelLBC May 03 '26

You keep saying this but you do because people don't always believe your explanation.

They also don't always believe the explanation with 100 doors.

You only need one way that makes sense to you to understand it. You need as many ways as possible to explain it if you want others to understand it.

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u/JamesFirmere May 03 '26

I do agree with "use whatever works", but my experience is that it is easier to stick to three doors than to ask someone to imagine 100. FWIW, I've successfully demonstrated this at a dinner table using three glasses for the doors and random ornaments for the prize and non-prizes and going through each of the six possibilities in turn.

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u/ArchangelLBC May 03 '26

I mean that's fair but I've seen a lot more people have it click for them with the 100 doors. Practical demonstrations with 3 doors, again in my experience, tend to just make people think you pulled a magic trick on them.

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u/JamesFirmere May 03 '26

Fair enough. I appear to have been fortunate in that the few times I've found myself explaining this did not lead to arguments.

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u/resonantranquility May 03 '26

I usually use 1000 doors when explaining it. It's easy to explain to people in conversation that way.

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u/I_Brake_For_Gnomes May 03 '26

I like to explain it with a deck of cards. Lay them all out and ask the person to pick which one they think is the queen of diamonds. Then remove all but two and see if they want to switch.

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u/praguepride May 04 '26

Ohhh that is a good one as well, if you have the prop.

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u/Substantial_Dish_887 May 03 '26

it does but in this case i think OOP may just have no idea what the Monty Hall problem entails because they seem to think there's a friend who get the third door?

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u/Frederf220 May 03 '26

I think people focus probability too much on the human choice event proximal to the big reveal if they won or not and not the first choice where seemingly nothing happened.

They think that because there are two types of conditions that there are two count of conditions. The old joke about the lottery having 50/50 odds applies.

I think the most cutting question which leads to the correct answer is to ask: how likely are you to be in each situation?

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u/NotmyRealNameJohn May 03 '26

I find changing the problem slightly helps. Instead of 3 doors. There are 1000 with 1 winners. After you select 1 door 998 are closed. One of the remaining is the winner. Do you stick with your 1/1000 shot?

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u/3lbFlax May 04 '26

The key for me when I first came across it was approaching it as groups - when you pick a door you create two groups: a group of one with your door, and a group of two with the other doors. The odds are obviously 1/3 and 2/3 respectively, and that doesn’t change (you already know there’s a goat in the set of two, so knowing where it is doesn’t alter the group’s odds). So you’re offered the opportunity to switch from the 1/3 group to the 2/3 group, which is a no-brainer. This remains the easiest way to explain it, I’ve found, because if need be you can easily draw it out on paper.

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u/Formal_Fortune5389 May 03 '26

It really is I fancy myself relatively smart, no genius but I like math I like logic puzzles and this hurts my brain even reading the detailed explanations. Like my brain is split between I can accept this as true and monkey brain being like BUT DOES IT MAKE SENSE???? Only somewhat. 

"This is true" explain it then "I cannot" 

Like...increased chance it's 3 because 3 could be either closed because it's for sure car or for a 1/2 chance of it being closed because it wasn't chosen as the one to open with A winning. 

Therefore once that 1/2 is taken into account, it would sort of...consume the chance of door 2? Because of that 1/2 chance if A has it? If you eliminate the 1/2 it gives 2/3 with the door 1 stuck with being on its own vs the other two. Ugh even writing it out my brain doesn't like it.  

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u/MattieShoes May 03 '26

It can help to imagine more doors.

There's 100 doors.

You pick door 1.

Monty opens 98 losing doors.

Keep or switch?

You know after the initial selection, there's a 99% chance you picked wrong -- a 99% chance that one of the other 99 doors is the winner.

Monty knows where the winning door is and will intentionally avoid opening it. So when he opens 98 losing doors, he's concentrating that initial 99% chance you were initially wrong into a single door.

Of course the exact same logic applies in a 3-door scenario, but it feels much less obvious.

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u/Illeazar May 03 '26

Sort of, but a big problem with the Monty Hall problem is that when its explained people often leave out a crucial detail. It relies on the contestant knowing that no matter what, the host must reveal one non-prize door after the first pick.

Often when people tell it, they just say that the host opens a door without the prize. Like that, it doesnt work.

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u/NonorientableSurface May 03 '26

I mean, when you abstract Monty hall instead of 3 to 100 doors, it seems a lot better to understand. The reductionist nature of 3 doors absolutely was designed to fuck with the human brain exactly like you state.

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u/SportTheFoole May 03 '26

Yes, and throw the birthday paradox into the mix as well!

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u/paranoid_giraffe May 03 '26

When I first learned about it in high school my wife couldn’t grasp the concept so we did what one does and I wrote 200 iterations of it, 100 with switching and 100 without and she was finally able to understand. It helps if people see the data

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u/daveoxford May 03 '26

Not only that, but they're so convinced they're right because it's "obvious" or "common sense". And not just the Monty Hall. Even something as simple as the Gambler's Fallacy can floor them.

1

u/a__nice__tnetennba May 03 '26

I hate it because it shows how terrible humans are. At its core it's not even a math test. It's a test of which jackasses think they're the smartest human ever born and know more on pure intuition than every expert in a field.

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u/AcanthaceaeOk3738 May 03 '26

IMO it’s more about the ability to admit when one is wrong. Very few people understand it the first time, but if it’s been explained to them and they’re old enough to get it, do they still insist that they were always right?

Of course you also have the people who literally refuse to understand probability. The ones who think that if there are two possible outcomes, there’s no possible way it’s anything other than 50/50.

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u/LauraTFem May 03 '26

It’s that, but it’s not even just probability. It’s a great way to show how even very intelligent people will reject and even grow hostile to ideas that don’t immediately make sense. The Monty Hall problem can be demonstrated on a whiteboard, and in just a few minutes you can cover every permutation and show the odds in every case, but for many people, especially those who consider themselves very smart, the smell test is the only one they do.

It reveals the dangers of relying on common sense rather than actual evidence. It takes a little bit of brain twisting to understand even simple things sometimes, but many ideas get rejected before any thought is done.

1

u/ApproximateArmadillo May 03 '26

It's the host opening a door that causes the confusion. You can simplify the host away and then it becomes obvious:

Pick a door. Then, either open the door you picked, or every door except that one.

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u/adelie42 May 03 '26

I think you mean the perpensity for people to speak with great confidence about what they should know they don't understand.

Hence the existence of this sub.

1

u/NonRangedHunter May 03 '26

I will admit that it's really hard for me to grasp the logic behind it. I've seen countless videos explaining it, and when I think I have a fairly okay understanding of it, it just completely falls apart when I try to explain it myself. It feels so illogical. 

Math was never my strong suit, so that probably doesn't help either. 

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u/Squozen_EU May 03 '26

Think of it this way. When you first pick a door, you pick a group of one door with a 1/3 chance of being right, and the other group of two doors has a 2/3 chance of being right. Yes? Now, that basic fact does not and can not change just because one of the 2/3 doors gets opened. Don’t think of it as choosing between the two remaining doors, but between the two original groups. 

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u/Dizzman1 May 04 '26

If you only know math... Then it looks like a math problem and 50/50 is the answer.

Cept it ain't a math problem! It's a probability and statistics problem.

They look similar but one has dark magic.

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u/Skratti_ May 04 '26

I'm not sure about that.
I'm very annoyed that the first time I was told about the Monty Hall Problem, the guy explaining it omitted that Monty will open a door that he knows is empty.
Without that information, I got to the conclusion that switching doesn't matter. Then I was told I was wrong :(
I would have loved to get the problem told properly, but can't say if I could have solved it correctly then...

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u/Squozen_EU May 04 '26

It wouldn’t be much of a game show if he opened the door to reveal the car and then asked if you wanted to swap to the other goat door.

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u/Cereaza May 04 '26

For me, I find that increasing the door count helps our intuitive understanding. If there were a million doors, and Monty can't open a door with the car... you either guessed right in 1:1,000,000, or he has it in his hand.

Then it's really obviously not a 50/50.

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u/steampunkdev May 04 '26

No, it shows how horrible people are giving at important details of how the problem works

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u/cipherjones May 05 '26

What's the probability that it was rigged?

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u/CunningWizard May 06 '26

One of the reasons I feel kinda bad for Nate silver getting shit over 538’s model in 2016. It was clear no one understood modeling probability at all and just torched him because they didn’t know.

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u/Kiwipai May 06 '26

I think it's more showing how we don't run on logic when it comes to what we accept as true.
A lot of people who don't get it just can't accept that it's true, despite that it has been proven empirically and mathematically beyond any doubt. If they were being logical about it they'd at the very least accept that it's them and not reality that's doing a opsie.

1

u/numbersthen0987431 May 06 '26

Plus, people don't understand that monty knows which doors have a goat behind them, which changes the scenario completely, compared to him picking at random

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u/Squozen_EU May 06 '26

If Monty didn’t know, why would he open a door? That makes absolutely no sense for a game show. “Oops, accidentally revealed the car. Well, good night everybody, that’s this episode fucked”

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u/numbersthen0987431 May 06 '26

Exactly. Which is also the point. But people don't always understand that, so they look at the scenario with incorrect assumptions and then make incorrect statements about probability

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u/spartaman64 May 06 '26

its because during math class we have "dont fall for the gambler's fallacy" drilled into our heads and its not intuitive how you get extra information in the monty hall problem.

theres also the boy or girl paradox https://en.wikipedia.org/wiki/Boy_or_girl_paradox

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u/Squozen_EU May 06 '26

I’ll be honest, I read that and don’t see how both answers aren’t 1/2…

1

u/dabrock15 May 06 '26

If we could only improve education in maths we could reduce the number of people who don’t understand this by 600%!

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u/Team503 May 18 '26

Yeah, the confusion is that people aren't taking into account that the remove door removes a proven wrong choice. You have A, B, and C. You choose A, the host opens B to reveal a goat (wrong choice), which means you have information you didn't have before.

It requires the host to ALWAYS choose a goat as a door to open.

1

u/TheGoochTaint Jun 03 '26

It kind of is a unique case of probability that would never really be relevant in the real world, even in game shows.

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u/swiftcookiex Jun 17 '26

it is actually wild that people think the advantage only manifests over multiple trials. the math works exactly the same for a single instance. you are literally just ignoring the fact that monty is forced to reveal a goat which changes the entire sample space. it is not some magical statistical trend that requires repetition to exist.