Making the Monty Hall problem weirder but obvious | Hacker News Reader

Making the Monty Hall problem weirder but obvious

dyno-might.github.io

196 pointsdyno-might269 comments

> It’s important that Monty looked behind the doors before choosing which to open.

This is often not explicitly stated when the problem is given. It is even not a 100% clear from the statement above. Monty always chooses a door with a goat. So:

1. You choose a door.

2. Prob that there is a car behind it: 1/3

3. Prob that the car is behind the two other doors: 2/3

4. If the car was behind the two other doors (which, remember, has p=2/3), Monty will choose the door without a car for you, and the door with a car will remain closed. In this case you are guaranteed to have the car if you switched.

So with switching, the overall probability is 2/3. Without, its the original 1/3.

If you did not understand that Monty always chooses a goat door, but the person giving you the problem does, or vice versa, then what usually happens is that both of you try to explain why your intuition is correct. Because most people don't talk formal probabilities, your explanations will be so vague that the other person will not realize your different understanding. You will discuss forever, you will both be right, and you will part ways with the strange feeling that maybe the other person was right, when all along you were talking about different problems. This is why this problem is so notorious.

>> It’s important that Monty looked behind the doors before choosing which to open.

> This is often not explicitly stated when the problem is given, which imho is the whole reason this problem has the reputation of being hard to understand.

If that were the only difficulty, why have so many people continued to have trouble accepting it even after this misunderstanding has been cleared up, and even after the correct answer has been explained to them? According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation.

I recall mention of an analysis of the responses to Vos Savant's Parade article, concluding that a majority disputing the result were aware of this constraint, and I will post a link if I can find it again (though if a majority did not explain their reasoning, it may not be possible to figure out what assumptions they made. Nevertheless, the question in my first paragraph still stands.)

I agree that "Monty knowing the answer" isn't the "whole reason" MH is confusing.

IMO, people are confusing MH with a case where the door is opened randomly before your choice.

Let's say Monty shows you three doors, then knowingly opens a wrong door (B) before you choose. Do you want A or C? It's a coin toss, of course.

But suppose you pick door A first and THEN he knowingly opens a wrong door you didn't choose (B). Now the odds of winning are 2/3rds if you switch. But it feels the same as in the previous case. ("How could choosing a door first affect the odds?! It couldn't possibly change anything!!")

The answer is that your choice of a door constrains Monty.

Suppose you choose a door secretly before Monty opens a door. Suppose you choose A, and Monty opens door A, revealing a goat. Well, uh, you can't pick that any more, now, can you? You're forced to choose between B and C, a toss up.

But if you publicly pick door A then Monty can't/won't open A; he has to open one of the other doors. Since you probably chose the wrong door to begin with, and he always opens a wrong door, the other remaining closed door probably has the car.

Extending these types of examples to 100 doors with 99 goats of which Monty opens 98 makes it even easier.

Especially if you walk through what that would look like. You choose door #14. Monty then goes from door to door, opening every single one except door #67. What's special about door #67? Either you picked the door with the car behind it at 1/100 probability, and Monty picked it at random, leaving it closed. Or, with 99/100 probability, Monty is skipping opening that door because it's got the car behind it...

Never thought of it this way, but extending to 100 doors will definitely be my go-to explanation/walkthrough in the future! Thanks for the insight.

I've tried that with people with limited success, though it could be that I've explained it badly.

When you put the games in terms of "you and a friend playing the game simultaneously", most people get it.

You and a friend are playing the game simultaneously; you both decide ahead of time to: a. always pick the same initial door b. no matter what Monty does, you ALWAYS switch, your friend NEVER does.

The possible outcomes before the reveal are: * Monty wins, because he shows you the car. We agree ahead of time that this never happens.

Now, since you know one of you is going to win, since Monty didn't, who has the better chance? Your friend started with a 1/3 chance, and he hasn't switched; does his chance suddenly change because Monty opened a door? Most people (not all!) will agree that it has not.

Since one of you MUST win, and friend has 1/3 chance, what's left?

Woah. That's such a concise way of immediately making the concept clear.

What makes the three doors Monty Hall so counterintuitive is that people tend to correctly reason about the case where the second door is randomly opened and don't understand why it doesn't apply.

I believe that this example makes understanding why people don't get it easier: you are looking for someone with one of your friend. You know they are in one of three rooms. Right before you can open the first one, your friend opens the second one and say: "not there". People assume the Monty Hall problem means that it's more likely your friend is in the third room and not the one you were going to open and think it's silly. And they are right to think that. What they don't get is that the case where your friend opened the correct door is part of the switching choice in the Monty Hall situation.

Your first paragraph looks very plausible, though I am not aware of any data saying it is a prevalent one. Are you suggesting that people still look at this problem as if the choice was random, even when they know it is not? (e.g., if they think the 'random' and 'chosen' cases are equivalent.) That is also plausible, but if it is common, that would imply that the phrasing of the problem is not the root cause of their misunderstanding of the optimal strategy.

Personally, IIRC, my first reaction was to assume the second box opening was not random (perhaps only because the question is not phrased as being conditional on this act revealing a goat) but did not see how this gave any useful information.

Here's another possible way of getting it wrong, regardless of the phrasing of the problem: assuming that, after the reveal (and whether one thinks of it as random or not), one is, as it were, starting over, except with a choice between two boxes rather than three, and no other information.

I definitely don't think it is the only point of confusion, but I definitely think it is a big one. We discussed the problem in my discrete math class and the professor had stated the problem without the constraint. Many in the class, being far too young to even know who Monty Hall is, were confused, and several had their 'aha!' moment once that constraint was mentioned.

> According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation.

Interestingly, coding up my own simulation took me from understanding the problem to grokking it. I'd definitely suggest anyone who has programming ability but doesn't grok the problem to write up a simulation.

If anyone reading this uses this example in a class, or is thinking of doing so, perhaps it could be set up as an experiment to see how much the phrasing influences the result?

> So with switching, the overall probability is 2/3. Without, its the original 1/3.

Thank you for spelling this out so clearly. If explained in terms of strategies (always switch v. always stick), the probabilities should be crystal clear.

What does the switcher have to do to win? Pick the wrong door first (2/3). What does the sticker have to do to win? Pick the right door first (1/3).

Thank you this is the first explanation of Monty Hall that finally makes sense. Now it intuitively makes sense.

I was always told Monty reveals new info and that changes probabilities. Never made sense. Your 2/3 vs 1/3 explanation makes it obvious.

It's also not always emphasized that the host never picks the same door you did. The host's behavior is not done in isolation. It is a response to your choice.

What kind of game shows are people watching where the host opens a door and says, hey, there's the prize but too bad, now you can't win it? Of course he always picks a door with a goat.

The original game in Let's Make a Deal went exactly like you described. Monty Hall opened the remaining doors; no option to switch.

Then there is the ambiguity as whether the host will always offer you to switch or just when he hasn't revealed the car.

Deal or No Deal does this - the contestant selects a briefcase, then opens up briefcases randomly. It's very similar to the Monty Hall problem, with some twists iirc.

Yeah. That is literally the whole reason to use the game show analogy.

Why not? Reveal all the choices the contestant didn't make, one at a time, slowly increasing the odds of a prize. This adds to the suspense. It's used all the time.

The problem is confusing because it appears that the choices are independent when they really are dependent. That's what confuses people and what needs to be adequately explained (and thus explicitly acknowledged).

This explanation is IMO more intuitive than the linked one.

It doesn't matter at all if Monty knows which doors are winners and losers.

If Monty doesn't know, then sometimes the game will be ruined because he will expose the grand prize and then the game is moot. But if is simply lucky by showing the goat door vs he picked it with foreknowledge doesn't change the odds in any way.

It does matter. If you are looking at a single instance where Monty got lucky and chose a goat door, then the probability that the door you have chosen has the car is 1/2, and switching doesn't change anything. This can easily be tested: just write a simulation that runs the experiment with Monty choosing a random door, and discard the instances where the game was "ruined" because he picked the car. In the remaining instances, both strategies will perform the same.

Sure, here's the simulation. There are only six distinct cases, so we can "simulate" it exhaustively, and a Monte Carlo will approximate this.

Let "A", "B", and "C" represent the prizes; A is valuable, B and C are goats. They're shuffled behind doors, so the player and Monty choose "1" or "2" or "3", and after the fact we'll map ABC to 123. As such, we can assume without loss of generality that the player always chooses "1" and Monty always chooses "2".

What happens?

  123
  ABC - player was right at first, so they lose if they switch
  ACB - player was right at first, so they lose if they switch
  BAC - player was wrong, Monty chooses 2 which is the car, so the game is ruined
  BCA - player was wrong, Monty finds a goat, player wins if they switch.
  CAB - player was wrong, Monty chooses 2 which is the car, so the game is ruined
  CBA - player was wrong, Monty finds a goat, player wins if they switch.

Huh. Of the 4 non-ruined games, half of them are improved by switching, half are worsened.

I believed you were wrong, but now I believe you.

Exactly and if Monty knows of the car location (as in the original formulation), he will basically save BAC and CAB from being ruined and both of these will become "player wins if they switch" outcomes. So instead of 2 out of 4, we now have 4 out of 6 cases where switching is better.

It matters because in the case where Monty does know to avoid the door, those games end up as wins for the person who selected to switch, as opposed to broken games.

You are taking an empty result and replacing it with a strict win.

I think lqet's point is that if it's just recounted as being thrown into a situation -- you've picked a door, and Monty opens a door, and it's a goat -- you don't know if you're in a world where Monty always picks a goat or one where he just happened to be lucky. In the second scenario, it doesn't matter if you switch.

Well in that situation it's still beneficial to switch, unless the probability of being in the first scenario is exactly zero.

Indeed. It's like the argument that you may as well believe in God, because if he doesn't exist it doesn't hurt you.

That said, though, if you're trying to convince someone of the correctness of the Monty Hall Problem, and they're stuck on that misconception, saying "well you may as well switch in case I'm right" isn't a great argument.

You’re forgetting about the possibility of being in a universe where Monty only opens a door if you chose the correct door originally (otherwise he does something else like ending the game). If you know nothing about the rule set of the game show, Monty opening the door tells you nothing and you have no reason to either switch or not switch.

Can’t that be expressed even more simply as:

1/3: you picked car, switch fails

1/3: you picked goat, monty ruins. Not included

1/3: you picked goat, monty open goat. Switch wins.

Of the two non ruined scenarios, half win

We can also demonstrate by using the 100 door example. Monty opens 98 of the 99 doors you didn’t choose. 98 times out of 99, monty ruins the game.

So random choice dramatically changes things.

I think where I differ (also expressed on the parent comment) is that when Monty ruins the game, you are tossing out many cases where the "I'm sticking with my original choice" strategy failed, while I'm counting them as a failure in the strategy.

You are free to count them as a failure of the "I'm sticking with my original choice" strategy, but only if you also count them as a failure of the "I'm switching" strategy, because when Monty ruins the game there is no way to win the car by switching to the remaining closed door.

If Monty chooses randomly both strategies fail 2/3 of the time.

Precisely. Either way switch or no switch have the same success rate if Monty ruins the game. You can count it as 1/2 or 2/3 but either way both choices are the same. So it’s a very different scenario from when Monty knows the door is a goat.

Why are we discarding the games that are ruined?

Whether Monty knows for sure or is just picking at random only affects whether games are ruined, not the probability that the door the contestant picked is a winner.

If there are N doors and the player picks door 1, there is a 1/N chance he will be correct. If he sticks to that door, there is nothing that will ever change that 1/N odds that it was the winner.

Downthread someone enumerated the cases, but again, they discard the 1/3 of the cases where Monty ruined it. What that does is throw out half of the cases where contestant picked the wrong door and had the wrong strategy. Whether Monty ruined the surprise by showing the winning door or Monty got lucky and exposed a goat, that original door always has a 1/3 probability of being the right one.

Because that is a fundamental part of the "Monty Hall" problem. You are correct in the case where Monty will randomly pick doors, and sometimes he will pick the winning door and ruin the game, but that is a fundamentally different problem than what is described as the "Monty Hall problem", and therefore irrelevant.

Its like saying soccer players could score more if they picked up the ball and ran with it. While true, you are no longer describing the game of soccer.

>Whether Monty ruined the surprise by showing the winning door or Monty got lucky and exposed a goat, that original door always has a 1/3 probability of being the right one.

By that logic, so does the remaining door. 1/3 vs 1/3.

Because the problem asks what to do after a goat is shown.

In terms of the logic problem, I can't see any distinction between not counting moot games where he reveals a grand prize vs assuming he never reveals the grand prize. It's just two different ways to say that the only games under consideration are the ones where Monty reveals a goat.

So I'm not getting the point you're making, unless you're saying that would be a more grokable way to state the problem?

It makes a difference how you got there, assuming in the second one that he reliably/intentionally never reveals the grand prize.

Imagine you have two identical looking bags of marbles, one contains two red marbles, and one contains one red and one black. You reach into a random one and draw a marble. It is red. If you picked the marble randomly, you now know you're more likely to be holding the bag that contained two red marbles than the mixed bag. The red marble you randomly drew provides bayesian evidence for which bag you're holding.

      Marble   1   2
  Bag
  2R           R   R
  BR           B   R

Out of the three possible equally likely worlds (given that you know you didn't roll into BR-1), two of them have you holding bag 2R, aka a 2/3rds chance.

If, on the other hand, you use a red-marble-picking robot, you get:

      Marble   1   2
  Bag
  2R           R   R
  BR           R   R

Now you have four possible worlds and are equally likely to be holding either bag, ie 1/2.

So, if our friend Monty opens a random door and you only look at simulations where it reveals a goat... There is a 1/3rd chance you initially get the car, but if you do he's guaranteed to reveal a goat. So 1/3rd of games, you have a car. There's a 2/3rd chance you initially get a goat, but then he has only a 50/50 chance of revealing the goat. So half of the 2/3rds of games, the game is discard as invalid, and the other half of 2/3rds (aka 1/3) you have a goat. As such, the potential outcomes are split 1:1:1 between car:goat:invalid. Removing the invalid cases, it's 1:1 car:goat, despite the only 1/3rd chance of you choosing the car initially.

tl;dr: If he's randomly opening doors and you discard any where it's the car, those cases are entirely ones where you didn't initially choose the car, which introduces some bias. If he's opening doors deliberately, no cases are thrown out since it's always possible to reveal a goat, maintaining the original 1:2 probability ratio.

EDIT: formatting.

OK, so I pick a door, Monty randomly reveals a goat. Your claim is that this information should make me increase the odds that my initial pick was a car. That sounds right. In the normal Monty problem, Monty doesn't give me any information about my original pick, just about the remaining door.

Alright just working through your logic here. Let's say I pick the first door always. I'll write out all the possible configurationss with both possibilities for Monty.

CGG > Monty reveals mid (Goat)

CGG > Monty reveals right (Goat)

----

GCG > Monty reveals mid (Car)

GCG > Monty reveals right (Goat)

---

GGC > Monty reveals mid (Goat)

GGC > Monty reveals right (Car)

So there's 4 worlds where Monty reveals a goat. In two of them, I picked a car initially. In the other two, I get a car by switching.

My prior post is mistaken -- thank you!

There are only three situations. Initial pick in parentheses.

(C)GG - switch and lose

(G)CG - switch and win

(G)GC - switch and win

By the time you have the decision, yeah, that's true.

There are other situations where the game is over before you get a decision.

Both GCG and GGC are eliminated half the time before you get to make a decision. But CGG is never eliminated before you get to make a decision.

Once you're making a decision and the game isn't already over, you can use this information -- that the game didn't end -- to assign a 50% chance of being in CGG, and 25% of each of the other two.

Once upon a time I was convinced of this, for reasons.

I wrote a simulation to demonstrate how right I was.

I was wrong and you are too.

It's weirdly great that you had enough misplaced conviction to motivate you to write a simulation which ended up proving you wrong.

It's actually an interesting, possibly general story, that kind of makes me think. Perhaps for certain personality types, misplaced conviction will put you on a higher velocity trajectory toward truth than honest confusion.

>Perhaps for certain personality types, misplaced conviction will put you on a higher velocity trajectory toward truth than honest confusion.

The key part of this sentence that makes it true, is the "for certain personality types".

I would wager the majority of personality types however, it would put you on a higher velocity trajectory to willful ignorance.

For sure. It's a small fraction of the population whose first instinct to prove themselves right is to do which also has the possibility of also proving themselves wrong.

It's the story of science! Unlike lawyers or debaters, scientists are delighted when their hypotheses are clearly disproven; this is at the crux of the scientific method, and is essential for discovery of the truth. Assert something falsifiable, then experiment. The truth wins.

You only remove games as moot when Monty reveals a car, which can only happen when you haven't initially selected the car.

Therefore, no games are removed as moot when you have initially selected a car, and 50% of games are removed as moot when you have initially selected a goat.

So it ends in a coin toss whether you switch or not.

The 2/3 to 1/3 split only happens when Monty removes a goat using privileged information.

Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.

> If the host had just picked at random

The overall point is that the host is not picking at random, and is thus affecting the outcome in a statistically reliable way.

This is gonna have anthropic principle vibes but here we go. Either the host looks behind the door, or they don't and we just don't talk about games where they accidentally show you the car. The probabilities are the same. We're already conditioned on being in the "they show you a goat" universe.

They're not the same if the host chooses randomly. Check it out:

WLOG assume the car is in door A (The rest will be the same by symmetry.)

You pick A, B, or C.

The host picks one of the other two doors to show you.

The universes can be described by your choice, followed by the host's random choice. There are 6 possibilities: AB, AC, BA, BC, CA, CB.

Since we're not in a universe where the host chose a car, we eliminate BA and CA.

We are left with four possibilities: AB, AC, BC, CB.

AB = You chose the correct door. (probability 1/4)

AC = You chose the correct door. (probability 1/4)

BC = You chose the wrong door. (probability 1/4)

CB = You chose the wrong door. (probability 1/4)

Since the host's choice was random, it's 50/50 as to whether the car is behind the door you chose or the remaining door.

This same reasoning breaks down when the host knows what's behind the doors because you're no longer eliminating a couple of the universes and it's a 2/6 vs 4/6 situation.

No.

If you're playing blackjack, and you hit on a 19, but you have x-ray vision and you know a 2 is at the top of the deck, you're not playing the same game as you would be if you didn't have that knowledge.

Same with the Monty Hall problem. It is a critical distinction whether opening the door has a 0% chance, or a 33% chance, of showing you a car.

Simulate it.

I thought as you think. I wrote a simulation to show I was right. I was wrong.

I have had this interchange several times. Invariably it goes one of two ways. They have endless reasons why they have to be right and they don't need to write a goddamn simulation, or they tell me they wrote the simulation and they have learned they were wrong.

I think you’ve simply misread what they are saying. They are saying that the situation in which Monty opens a door and it reveals the grand prize, are the set of cases that we do not care about because we have already lost, and can thus discard them.

This is basically just a different way of saying that Monty looks behind the door to be sure to only reveal goats.

Yes, the probability of wins will be different in these two scenarios, but it doesn’t affect the conclusion: when Monty reveals a goat, you should always switch.

Edit: if you believe I am the confused one after this comment, I will go make the simulation as you suggest.

We have two possible games, with most of the setup the same.

In one, Monty reliably reveals a goat. In the other, Monty picks randomly between the other two doors.

We are, in parallel universes, playing both of those games. We know which. The host has just revealed a goat.

I read fouronnes3 as saying that these two situations are the same. They are not.

There is a simulation in the sibling to your comment, ready to be pasted in your browser console.

That simulates a different scenario than what’s being described here; it simulates a situation where we count the times when Monty picks the car. But those situations are irrelevant because they have no bearing on the fundamental question of whether or not to switch doors when Monty shows you a goat. Effectively, we discard all outcomes when Monty shows you the car, making it an identical “game” as when Monty simply does not ever choose the car.

As I said, the outcome per game will be different (since you suddenly have an additional opportunity to lose), but the math around whether or not to switch when shown a goat remains unchanged.

It will be different. The question comes down to P(winning by switching | monty reveals a goat). The key difference is whether P(mony reveals a goat | you chose a goat door the first time) is 50% or 100% (if you choose the car, there's a 100% chance he reveals a goat in either case). Since 'winning by switching' is the same as 'choosing a goat door first', you can apply bayes theorem to see how the results change.

To put it another way, if Monty is choosing randomly, half the time where you would win by switching, instead the game just ends/isn't counted, but the same is not true of the case where you win by not switching. From a bayesian point of view, Monty randomly revealing a goat should increase your belief that you picked the car the first time.

That said, switching is not worse than not switching unless Monty is biased towards revealing the car instead.

So let's say you and I sit down, and do the following:

    1) I roll a 3-sided die (or a 6 sided, wrapping) and keep it covered.
    2) You pick a number, 1-3
    3) I flip a coin. If it's heads, I pick the lower available number; if it's tails I pick the higher.
    4) I peak at the die.  If it's my number, we reveal and start over.
    5) I (always, at this point) offer you a wager: if the die shows your number (so you would have lost if you switch), you pay me $7; if the die doesn't show your number, I pay you $5.

Assuming you believe the die and coin are fair, etc, would you agree to play that game 1000 times? In those games, is there a reason you would turn down the wager?

Is it the same game if I flip the coin secretly, peek at the die, announce my number (picked algorithmically in the obvious way), and then pay out according to whether switching would win (as above)?

Because (assuming I've explained these games as I intend... it's getting late) I would play the former with you not the latter (but I would play the latter if we switched the payments around - I chose 5 and 7 because 7/12 is halfway between 1/2 and 2/3).

>it simulates a situation where we count the times when Monty picks the car.

No, it does not.

>> if (montys_choice === car_door) { i -= 1; continue; }

This discards the game where Monty chose a car.

>> do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door);

This makes Monty never choose a door with a car.

Neither one counts the game where Monty opened a door with a car as a win for either staying or switching.

Thank you (and everyone else in this thread) for sticking with it with me. It was a struggle to read that single line of code on a cell-phone screen, and that was compounded by the fact that it turns out that I don't understand the Monty Hall problem when I thought that I did. I'm going to sit with this one for awhile.

Specifically, it's the difference between

    var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice); if (montys_choice === car_door) { i -= 1; continue; } if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins);

and

    var num_stays_wins = 0; var num_switches_wins = 0; for (var i = 0; i < 100000; i++) { const car_door = Math.floor(Math.random() * 3); const my_choice = Math.floor(Math.random() * 3); var montys_choice; do { montys_choice = Math.floor(Math.random() * 3); } while (montys_choice === my_choice || montys_choice === car_door); if (my_choice === car_door) { num_stays_wins += 1; } else { num_switches_wins += 1; } } console.log(num_stays_wins, num_switches_wins);

(The difference is in the calculation of `montys_choice`.)

Imagine there are two contestants and each one selects a door. The host opens the remaining one and it happens to show goat. Do you think they will both increase their odds of winning by switching?

Can both select the same door? If the answer is no, you've changed the parameters of the game, because the person who chooses second can only pick from two doors.

The subset of games where they close different doors is smaller than all possible games.

And if they could have selected the same?

If both contestants choose the same door, and the host shows a goat, then yes, they both would increase their odds by switching.

And if they could have chosen the same but didn’t?

The probability that the host opens a door with a goat is always the same. (Again, under jtsiskin’s assumption that he picks randomly!)

If participants are allowed to pick the same door, even if they don't in practice, then yes: after Monty shows a goat behind the third goat then both participants should switch. This is because they are essentially playing two simultaneous but independent games, with no interference between each other.

As before, switching is essentially saying "I think I got it wrong on my first try", which is a good bet.

After they are shown a goat behind the unchosen door they are less sure that they got it wrong in the first try. It becomes a 50/50 bet.

No, your intuition is misleading you: it's not a choice between two doors. If the probabilities don't convince you, you can easily simulate this by using three cards and considering one is the car and the other two goats. Pick one at random, then pick the remaining "goat" as if you were Monty; lastly, switch your initial choice as a player. Repeat the experiment 10 times and count how many times you win; remember you must always switch. You will win more often than lose! I tried this, that's how I know this will convince you.

Monty opening a door with a goat doesn't change the probability of your initial door being right: it's always 1/3. You're not choosing between two doors; you're choosing between "I think I got it right the first time" (1/3) vs "I got it wrong the first time" (2/3) -- the second is more probable.

The game where there are two simultaneous participants, both free to open any door (including both picking the same door) only muddles things and offers no insight. It's effectively two simultaneous but independent games. How they are going to split the price if they both win, anyway? ;)

So when they don’t switch each one has probability 1/3 of getting the car? Where does the car go if no one gets it?

And if they switch both have probability 2/3 of getting the car? Is this some form of car sharing?

Maybe you should try to play with three cards and one ace. Pick one card (you are the first contestant). Pick a second card (you are also the second contestant). Flip the remaining card. If was the ace start again.

Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?

> So when they don’t switch each one has probability 1/3 of getting the car?

Yes.

> Where does the car go if no one gets it?

I don't understand your question. What does this have to do with probabilities?

> And if they switch both have probability 2/3 of getting the car?

Yes. Remember, because both can pick the same door, this is effectively two independent games running simultaneously.

They each have a 2/3 probability of getting the car because, when they first chose, they had each 1/3 of getting it right. Each contestant's choice and corresponding probability of winning is independent of the other's.

> Is this some form of car sharing?

That's what I asked :) Car sharing makes little sense, prize-wise, which is why this makes no sense as a contest. But in regards to probabilities, it makes no difference: yes, both contestants should switch.

> Maybe you should try to play with three cards and one ace.

There are only 3 doors, so you should only use 3 cards.

> Now, do you prefer card #2 to card #1 and simultaneously prefer card #1 to card #2?

I don't understand your question. Probabilities are computed independently for each contestant. There's no "simultaneous preference".

If following has happened already:

Player #1 has picked door A

Player #2 has picked door B

The host has opened door C and there was a goat

At this point:

What are the chances of winning of player #1 if he keeps his pick of door A? And if he switches to door B?

What are the chances of winning of player #2?

Player #1 had a 1/3 probability of picking the right door. Therefore, he has a 2/3 probability of winning if he switches.

Player #2 also had a 1/3 prob of having picked the right door. Therefore, she has a 2/3 prob of winning if she switches.

In summary, both should switch. This is independent of whether they chose the same door between themselves. In other words, in a game where you're not forced by the other player's choices, your probability of winning is independent of the other player's. Your strategy should always be to switch (you can verify this empirically!).

So you think that at that point (after #1 selected door A, #2 selected door B, and a goat walked out of door C):

The probability of player #1 getting the car if he keeps his door, which is equal to the probability that the car is behind door A, is 1/3

The probability of player #1 getting the car if he switches, which is equal to the probability that the car is behind door B, is 2/3

The probability of player #2 if he keeps his door, which is equal to the probability that the car is behind door B, is 1/3

The probability of player #2 getting the car if he switches, which is equal to the probability that the car is behind door A, is 2/3

In summary, the car is:

Behind door C with probability 0

Behind door A with probability 1/3

Behind door B with probability 2/3

Behind door B with probability 1/3

Behind door A with probability 2/3

No. Because the experiments are completely independent (i.e. player #1 & player #2 don't interact at all; their choices don't interfere), the events are effectively "choosing door A and switching to door B" and "choosing door A' and switching to door B'". You have doors A, B, A' & B'. This of course assumes all doors are revealed simultaenously; i.e. no player gets to see the result of the other's choice beforehand.

You don't have probabilities "behind doors", you have probabilities of "chose right the first time". The probabilities of players #1 and #2 choosing right are clearly independent.

But don't trust me: try it!

Hang on, I think you've missed something. They are not completely independent, because if they pick different doors, that forces the host to pick the last door. Removing the host's freedom to choose a non-winning door (and effectively help the player) changes the math. Having two players with the ability to choose different doors (but still 3 doors and the host must choose the third) makes it a different game with different dynamics. It doesn't work the same because it's not the same game.

Or, because they are two independent games, the host can flip different goats for different people. You would have to add an element of "no screen peaking" where the contestants cant see which door opened for the other person.

So you have Player A who chose door 1. The host flips the goat in door 3.

You have Player B who chooses door 2. For them the host flips open the goat in door 1.

Player A should switch, and wins the car. Player B should switch, but had the car and loses it. The right move in both their cases was to switch.

Really, its no different than playing two separate games. Unless, like you said, you somehow change the rules and limit someones choices, in which case Im not sure why its a part of the conversation, other than being introduced through confusion.

Who says they are two independent games?

https://news.ycombinator.com/item?id=24713352

If the players pick different doors, the host will open the third door. That prevents him from being sure to get a goat, in the same way that opening a door at random in the single-player case prevents him from being sure to get a goat.

Well, you're right: in fact, in some cases Monty cannot open any door! That is, if players A & B both chose goats, then the remaining door is the car, and Monty cannot open it without ruining the game.

For this reason, both players should be playing "blind", without seeing the other's choices or the door Monty opens for the other player. Anything else doesn't make sense for the game.

> Removing the host's freedom to choose a non-winning door (and effectively help the player) changes the math.

As it does removing the host's ability to choose a non-winning door in the variant where "the host had just picked at random from the two doors, and it happened to show a goat". That doesn't work the same either because it's not the same game.

Why don’t you try to give an answer to the following questions?

After player #1 has picked door A, player #2 has picked door B and door C has been opened revealing a goat:

a) if player #1 stays with door A his probability of winning is ___

b) if player #1 switches to door B his probability of winning is ___

c) if player #2 stays with door B his probability of winning is ___

d) if player #2 switches to door A his probability of winning is ___

e) the probabilty that the car is behind door A is ___

f) the probabilty that the car is behind door B is ___

g) the probabilty that the car is behind door C is ___

These are two independent games, so for every player:

a) The probability of winning if staying: 1/3

b) The probability of winning if switching: 2/3

c) The probability of the car being behind door the player chose is the same as a): 1/3

d) The probability of the car being behind the door Monty didn't open is 2/3

e) The probability of the car being behind the door Monty did open is zero: we know he chose a goat!

Note that the game doesn't work if it's not run independently for both players, because if they both choose goats, Monty would be forced to open a door hiding a car, ruining the game (it makes no sense at that point to either stay or switch). In order for the game to work the way I imagine you want, we must assume one player always picks a goat and the other a car... which cannot be guaranteed.

There is one single game. It's my version and I make the rules!

  Rules2: "One player picks a door, the second picks a door (maybe the same), the host opens a door that has not been picked (if there are two he picks at random). Then the players can either keep their choice or switch to the other closed door."

  Scenario2: "Player #1 picked one door, player #2 picked another door, the third door was opened and there was a goat behind it."

In that scenario, what are the probabilities?

I agree it's different from the original problem, that's the point!

  Rules0: "The player picks a door, the host opens a door that has not been picked with full knowledge that there will be a goat behind it. Then the player can either keep his choice or switch to the other closed door."

  Scenario0: "The player picked one door, the host opened another door and there was a goat behind it."

But the game I proposed is equivalent to the version of the problem that some people, including you apparently [0], insist in this thread that is equivalent to the original problem.

  Rules1: "The player picks a door, the host opens another door selected at random. Then the player can either keep his choice or switch to the other closed door."

  Scenario1: "The player picked one door, the host opened another door and there was a goat behind it."

The latter is not equivalent to the original problem. The situation is similar but the rules are different. The solution is different.

However the solution of Problems 1 and 2 is the same. In both cases the rules allow for the car being unveiled when the host opens the door (it happens with probability 1/3). In both cases under the scenarios proposed there is no point in switching doors.

They are both different to the original problem where everybody knows beforehand that the door opened will hide a goat.

[0] The problem I proposed at https://news.ycombinator.com/item?id=24707305 is also equivalent:

    Rules1': "The player picks a door, and opens another door selected at random. Then he can either keep his choice or switch to the other closed door."

    Scenario1': "The player picked one door, then opened another door and there was a goat behind it."

This breaks most people's brains because there WILL be games where the host can't show a goat in the third door. That has to be accounted for.

If he picks at random (the hypothesis in this subthread) there will be (one third of) games where he doesn’t show a goat.

In your example, the host didn't get to choose their door, which kept them from choosing goat. As I follow this discussion, the key is the knowledge that given a choice, the host will pick a non-winning door. If the host can't choose, you lose the benefit of that fact.

Of course. I was replying to someone who says that the host could pick randomly and it wouldn’t change anything.

The host has no choice in this case so his action has no effect. The chance to win is 1/3 for all. And after the host gets a goat one of the contestant has to wins so its 1/2 for both whether they switch or not.

That’s indeed what happens when the host picks at random (the hypothesis in this subthread). He has a 1/3 chance of showing the car. If he doesn’t, the car is behind any of the other doors with probability 1/2.

> Is it important the host looked behind the doors? If the host had just picked at random from the two doors, and it happened to show a goat, the odds would be the same.

If the host picked a door at random, and it happened to show the prize, then the game would be ruined. That would make some terrible television. Therefore, he can't be picking a door at random.

To make it more obvious, assume there are 100 doors with 99 goats. And after you select a door, the host opens 98 goats doors

I get what you are saying, but if I look at it differently I still intuitively feel like it should be p=1/2 if they switch instead of 2/3. For example:

- the bottom line is Monty will always eliminate one wrong choice

- therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2)

(edit: thanks i see, no need for any more answers)

I think if you really read over the article, you'll see this is wrong. If not, tell us on which step you disagree with the author.

A more common re-telling of the story:

Monty has 1 million doors, with only one car behind one. When you pick a door, Monty will always open 999,998 other doors, all of which show goats, leaving one other door conspicuously closed.

Do you still think you have 50% chance of being right by sticking with your choice?

This got me closer to understanding. But, after 999,998 doors are opened, although the one other door is conspicuously closed, my door is also closed. That seems conspicuous, too. In that moment, there are 2 doors to pick from and so 50/50 makes sense.

As some other comments have mentioned, I think this isn't a fully satisfying explanation because it's hard to reason about the difference between a general strategy vs. an in-the-moment choice; why does it matter that I've watched Monty narrow the choices down to 2 doors, vs. seeing 2 doors from the start?

I think the 1 million doors explanation does a very poor job of explaining why switching is the correct choice and only gives a more intuitive answer, not more intuitive understanding.

This is how I understand the Monty Hall problem (Using 3 doors):

1) You are asked to choose a door from A, B or C. You choose A.

    Probabilities:
    p(A = Car) = 1/3
    p(not A = Car) = p(B or C = Car) = 2/3

2) Monty opens a door and shows you a goat. Let's say he opens door B and gives you the choice to switch from door A.

    Probabilities with B eliminated:
    p(A = Car) = 1/3
    p(not A = Car) = p(C = Car) = 2/3

3) The probability of the car being behind door A remains unchanged at 1/3, therefore the probability of the car being behind not A is also still 2/3. Since door B has been eliminated, this means the probability that the car is behind door C _must_ be 2/3! Therefore, we choose to switch from door A to door C.

Bravo! This explanation wins in every way.

Thank you!

Monty never opens your door until the end of the game. So Monty opening 999,998 other doors doesn’t give you any information about your door, just the other one that remains closed.

If you made a one in a million guess the first time, your door has a car. If you didn’t, the other remaining door must have the car. Which do you choose?

It’s easier to think that you aren’t opening 1 door. You are opening 999,999 doors, someone else just did it for you.

Great way of phrasing it

Let's say Monty doesn't open any door, but after you've chosen one, he offers the chance to switch to both of the other doors. Pretty clear then that switching gives you a 2/3 chance of getting the car, and that is essentially the choice Monty is offering.

It’s not regardless of what came before. That’s like saying, if there were two doors and Monty simply tells you which one the car is behind, there’s still a 50/50 chance of each door being right. No, there’s a 100% chance the one he told you is right. You learn real new information when Monty opens a door just like when he tells you something.

> therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2)

I get why you intuit this - I had a lot of trouble with it at first as well. Common sense is lying to you, however, because you are still dealing with the original odds (1/3 vs 2/3). Monty's choice is always to choose a Goat (probability 1.0), therefore his effect on the probabilities will always be with a weighting of 1.0. You end up with a probably spread of (1/3 * 1.0 vs 2/3 * 1.0).

Therefore it is always in your interests to choose Monty's remaining door.

> Monty always chooses a door with a goat.

Yes, this is the single most critical piece of information, and when people struggle to understand the Monty Hall problem, its almost always because this wasn't made clear. When it is, the problem becomes much more intuitive.

This carries over to a general problem with wordy riddles and puzzles. They rest on a slew of assumptions and tacit knowledge. If the riddler fails to pluck out the essential parts, then there's a good chance the interaction will fail.

Your explanation was great! You should become a teacher. It really nails the problem at hand

Here's what worked for me:

Imagine there are 1,000 doors and you pick 1. All other doors except 1 are opened and you're given the offer: keep the door you picked, or pick this other door. What are the chances you picked the right door (vs. this other door)?

People seem to intuitively understand that having only one door unopened is a massive "hint" to where the prize is.

(I learned this idea from Better Explained: https://betterexplained.com/articles/understanding-the-monty...)

I was lucky enough to get this explanation from my high school physics teacher, who first presented the classic Monty Hall problem and then illustrated the changing of the odds by substituting all of the lockers in our school for the three doors. Switching gives a clear advantage.

The rest of this post is an anecdote from the same class that this brought to mind, and is unrelated to the topic. Maybe we can say it shows how good teachers engage their students or something, but really it’s just a good yarn.

We were learning about inelastic vs elastic collisions, and how an elastic collision has 2x the energy of an inelastic one. The teacher asked for a volunteer, and a bright-eyed student rose to the occasion. The teacher gave him some safety glasses and told him to lie down on the floor.

The teacher took the inelastic ball and said, “Okay, I’m gonna drop this on your forehead now, ready?” PLONK. “Ow.”

“Remember that feeling! This is the elastic one, and it has the same mass, so it should hurt twice as much.” PLONK. “Ow.”

The teacher asked, “So, did the second one hurt more than the first?” The rest of us anticipated the experimental confirmation of what we’d just learned about.

“...I couldn’t really tell the difference,” said the student.

“Yeah,” said the teacher, “I knew you wouldn’t. I just wanted to see if you’d let me do it.”

I don’t think this helps me understand it.

In the 1,000 doors problem, my odds of being right initially were something like 1/1000 and then it changes to something like 998/1000 or 999/1000 for switching, I can’t intuitively grasp exactly what the odds become of winning if I switch, I just know it’s high. Bringing it down to 3 doors doesn’t help me much — it’s still something like 1/2 or 1/3.

Try thinking of it this way. It may or may not be any more useful. I've seen different people come to understand the problem from different examples.

    1. Observe that 3/3 = 1. Pedantic, yes, but good for frame of mind here.
    2. Pick one of three doors. (1/3 odds)
    3. Gain information that one of the three doors is a loser.
    4. Note your odds on choosing the original door correctly are still 1/3.
    5. Note that if you change doors, there are still 2/3 doors there to choose.
    5. Note you're not going to switch to the known loser door, so if you change doors you know 100% which of the other 2/3 of doors to choose.

The intuition usually is that you're down to two doors after the loser door is opened, but that's not the case. There are still three doors. The host has just told you that if you trade doors, you know which door to trade for. So trade for it.

Note there's a newer version of "Let's Make a Deal", hosted by Wayne Brady, but there is no option to switch after a losing door has been shown in that version.

Step 4, kind of a mystery. Why are my odds still the same even though the situation is different?

You choosing a door in the first place gave you 1-in-3 chance. But your 1-in-3 choice is deducted from the 3-of-3 choices that Monty could have had, so Monty only had a 2-in-3 choice: there's a 2/3 chance that the car is in the pool of doors from which Monty could select. Monty has a 100% chance of choosing a door without a car. Therefore, you inherit the 2-in-3 chances if you change your selection.

The first door will have the car 1/3 of the time. The second door's chances had been expanded to the remaining 2/3 percent thanks to Monty always choosing the last 1/3 door which is guaranteed to not have a car.

Because it isn't different really. It is always a goat door that is opened, so you don't gain any information about your door by the opening of the goat door.

I'm thinking of a number between 1 and 10, guess it. If I now tell you a number I promise is not the one I was thinking and not your number, you have no more information about if you were correct.

Because it really centers on the initial premise: Monty will always open a goat door after your choice, no matter what.

So, you make a totally random choice. That choice must be 1/3 right, right? Now the thing that you already knew would definitely happen happens: Monty opens a goat door. How can your odds suddenly jump to 1/2?

Are you saying every single time you play the game, you always have a 1/2 chance of getting it right first time?

What you knew about the door you initially picked hasn't changed at all. What you now know about the other doors has. By giving you information about 1/2 of the other 2/3 doors, Monty has given you an extra 1/3 chance if you pick among those.

Assume there are seven billion people on the planet. One of them knows the location of a specific hidden treasure. I know who it is and I ask you to guess who it is, but you have no possible way of knowing or even getting a hint about it.

You pick some random person. I then bring in another stranger and tell you that the person who knows where the hidden treasure is is either the random person you chose or the one I brought in.

At this point, there are only two possibilities:

1. You happened to randomly choose the right person on Earth and in my surprise, I had to pick some other random stranger to pretend they knew the secret.

2. You chose a total rando who has no idea what's going on and the person I brought in is in fact the one who knows where the treasure is

You don't need to know the exact odds to understand that it's higher. I think that's the main takeaway of making it a 1,000 door problem. It makes it intuitive that the correct solution is to switch. The exact probability doesn't matter.

IMO what helps is to imagine slightly changing the order.

First step is still that you pick a door. There's a 1/3 chance it has the car. Now you can either keep that single door (with a 1/3 chance of a car), or switch and get both of the other two doors (each with a 1/3 chance of the car, for a total of 2/3 chance). After you pick, I'll reveal all the goats.

probability of winning if you switch is 1 - (1 / n) aka (n-1)/n.

probability of winning if you don't is 1 / n.

By switching, you are simply betting that your original guess of 1/n was wrong.

Try watching this video below, which explains the 1/100 analogy with a real world example:

https://youtu.be/GPoPSNxV1D4?t=365

I've timestamped the relevant bit - but you should watch the full thing from the start, it's very entertaining :)

Wow. I "understood" the reasoning behind the original and knew that was the right answer, but your anecdote just made it totally click. Of course if I choose one random locker there is a 1/1000 chance of getting a prize, and if Monty opens 998 other empty lockers, of course I should switch. That would make switching be the correct choice in 999/1000 times.

For someone who doesn't get it, the problem is then whether the correct extension is all the doors being opened. With 3 doors it's the same.

For me the most sensible explanation requires you to know that a dud door is always opened, thus the probability from the 2/3 is the one you are switching to.

This is the most intuitive explanation by far. I'd even say 1 million doors to really drive the point home.

You choose a door with only 1 in a million odds of it being the door with a prize. Monty Hall know where the prize is and will only open the remaining doors he KNOWS doesn't have the prize. If he then opens up 999,998 doors without a prize behind them and asks if you want to keep your original door or switch, you'd obviously know that Monty's last remaining door must be the one with the prize.

When I first heard about the problem, I struggled with the seemingly 50/50 chances. either you picked the right door and now switch and lose, or you picked a wrong door and switch and win. Switching seemed a zero sum game.

The explanation that works best for me is that you were more likely to have picked a wrong door in the first place, so while the impacts are opposite equals, the likelihoods are not equivalent.

That's the explanation that helped me, and led me to realize something about this problem. The problem asks what is the best strategy. It doesn't ask what should you do at that moment. At that moment implies you should process the information available to you right then and there. The information available to you at that time - two doors, ignores prior information, which I think is the counterintuitive aspect that trips people up.

That explanation never worked for me, because you can turn it around to the situation where Monty does not know where the car is. Say there are 1000 doors, and you pick door 429. On his way to open door 429, Monty stumbles, falls, and accidentally knocks open every door except door 128. If by some coincidence all opened doors happened to contain goats, you will have nothing to gain from switching. Very counter-intuitive, but just as true as the original problem.

A possible intuition here is that Universes where your first pick was the door with the car, which initially were just 1 in a 1000 compared to Universes in which you picked a goat, will suddenly become massively overrepresented. After all, in these types of Universe Monty's Fall couldn't possibly have shown a car, whereas most of the other Universes will not survive to the next "round".

Of course, if this happened in real life, Bayesian thinking would increase the likelihood of hypotheses such as, for example, "The door containing the car has a better lock" to such an extent that I would switch.

You can't really turn it around, because Monty knowing and using his knowledge of where the car is to reveal only goats is what makes switching advantageous.

In the case of the clumsy Monty of your example, it goes like this:

1. There is a 1/1000 chance door 429 has the car.

2a. If it has the car, then when Monty accidentally opens 998 doors no car will be revealed. This does not change the chances that 429 has that car, which remain 1/1000.

2b. If 429 does NOT have the car, then 998/999 times that Monty accidentally opens 998 doors, he will reveal a car, which presumably ends that game. There is only a 1/999 chance that he will not reveal the car and the game proceeds.

3. Thus, there are two cases where the game reaches the point of two remaining doors, with 998 revealed, the car is behind one of the two, and you have a chance to switch.

3a. Your door has the car, which happens 1/1000 games.

3b. Your door does not have the car, which happens 999/1000 x 1/999 games, or 1/1000 games.

In other words, if the clumsy Monty version is played repeatedly, 998 out of 1000 games end without even getting too the point you get a chance to switch, and 2 get to where you get the chance. In those two, one has the car in your door, one not. There is no advantage to switching.

In the case of the systematic Monty who knows where everything is and ALWAYS opens 998 goats, it goes like this:

1. There is a 1/1000 chance your door, 429, has the car.

2a. If it had the car, Monty opens 998 doors that do not have the car, leaving one door besides your yours.

2b. If your door did not have the car, it is one of the 999, and Monty systematically opens the 998 of those 999 that do not have the car.

3. You always reach the choice stage. You can either get there via 2a, which always results in the car being behind your door, or via 2b, which always results in the car being behind the other door.

3a. You get there via 2a in 1/1000 games.

3b. You get there via 2b in 999/1000 games.

If you do not switch, you only if and only if you got there via 2a, so you only win 1/1000 games. If you always switch, you win if and only if you get there via 2b, so you win 999/1000 games.

Your proposal is changing the scenario, though.

In the original problem, Monty always opens a door to reveal a goat. This strongly implies that he knows which doors have goats.

Your alternative is different. If the doors opened are truly selected randomly, then the odds are high that he would reveal a car, especially in the 1000-door version. What are the odds that Monty could choose 998 doors randomly out of 1000 and NOT pick the one with the car?

If Monty doesn't know where the car is, then you're arguing with a completely different version of the game that could result in him opening the door with the car, so the strategy is going to be different.

Changing the scenario is the exactly the point of why I am unsure increasing the number of doors gives true intuition. In the original problem with the original scenario, people intuitively think switching doesn't matter, when it provably does. With 1000 doors and the Monty Fall scenario, again the intuition is wrong. So are we gaining true intuition for the Monty Hall scenario by expanding the setup, or is it just some subconscious Bayesian thinking accidentally steering us toward the right answer.

Most of the subtlety of the problem lies in this small sentence in the Side Notes: "He deliberately chooses to show you goats."

This is not made as explicit in the standard formulation "the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat."

Which could be read as "which happens to have a goat".

It's the ambiguity in Monte's door opening strategy that leads to different answers.

Whether or not he knew or whether it was luck, it doesnt exactly matter, although traditional stats knowledge would make you think it does.

The important thing to understand is that the premise of the question says he will show you a goat. If you rerun the experiment 10,000 times, he will show you a goat 100% of the time, either through peaking, premonition, or consistent luck.

The problem gets trickier because people start applying domain knowledge of stats, and treating it as a simulation with random events. The goat being chosen is not random, it is an event that occurs 100% of the time in the premise of the thought experiment. Thinking about the random chance of him choosing the car is outside the bounds of the axiom/postulate we start with.

tldr: it doesnt matter how he opened a goat door, all that matters is that he did.

But... it does matter how he opened a goat door. Specifically, whether there was a chance that he might not have.

I originally thought it couldn't matter. I wrote a simulation to demonstrate how right I was. I was wrong. I encourage duplicating the experience.

> Specifically, whether there was a chance that he might not have.

That is false. Him opening the car is outside the constraints of the premise of the thought experiment. He can't not open a goat. The chance of him not opening a goat is irrelevant, and the probability is 0%.

So in a new, different, and entirely unrelated thought experiment:

You pick a door. Monty picks another door at random - he has clearly committed to doing so, flips a coin in front of you or whatever. God only knows what would have happened if he'd revealed a car, but this time he didn't. Having found yourself in that situation, what are your odds if you switch doors?

If you simulate this procedure, you'll find that the number of outcomes where "switch" wins is about 1/3 of the total games; "stick" wins in about 1/3 of the total games; and 1/3 of the total games Monty revealed the car. When Monty does pick the goat reliably, as in the Monty Hall Problem as it was intended to be understood, Monty is doing the work to convert those "revealed" games into wins for "switch".

If you disagree, I'll happily take your money.

I'm sorry, I understand the difference you're implying, but can't figure out how it matters in practice. How can I even differentiate if Monty opens the door to reveal a goat by knowledge, or by chance?

I mean that literally - how can I write this into a simulation to test that there's a difference?

It depends on the context. If it's a riddle, you could see if it's made clear by the wording, interrogate the person asking the riddle, or state the assumption you're making in your answer.

If you find yourself actually in the situation, you need to make your best guess from what information is available.

As for simulation, write a function that handles a single play through of the game and returns the outcome if you switch. Make Monty's strategy a parameter. Remember that "Monty revealed the car" is a separate outcome from "win" or "lose". Run that function a whole bunch of times, counting the outcomes. Compare the ratio between wins and losses, at various choices of strategy.

To really drive home the point that strategy can matter, consider another possibility: maybe this season the studio executives have decided that cutting costs is more important than having an interesting game show, and Monty's strategy is now "reveal the car if you haven't already chosen it". Monty has revealed a goat. Is there still a 2/3 chance that you didn't pick the car?

How does it matter how he opened a goat door? If you allow the possibility of Monty opening a car door, you change the problem entirely, and you before stating whether it changes or doesn't change the odds, you need to tell us your new interpretation of the problem.

Problem-1: "You pick a door. Then Monty opens a door he knows having the goat. Calculate probability of winning if staying."

Problem-2: "You pick a door. Monty opens a door at random, which just happens to be a goat. Calculate probability of winning if staying."

Both are conditional probability P(A|B) (that is, probability of A happening, under the assumption B has happened).

   - A is probability of picking the correct door at the first try (or switching if you prefer)
   - B is the probability that Monty picked a door with a goat

P(A|B) is defined as: P(A|B) = P(A and B) / P(B)

In both problems P(A) is 1/3.

In Problem-1, P(B) is 1 because Monty knows and it's not a random event and P(A and B) = P(A), so P(A|B) = P(A)*P(B)/P(B) = P(A) = 1/3.

In Problem-2, P(A) is 1/3, P(B) is 2/3, P(A and B) is 1/3 (if you pick a car, Monty is guaranteed to pick a goat), so P(A|B) = 1/3 / 2/3 = 1/2.

If you want to visualize them in "enumeration of cases".

The complete cases for the choices is:

Problem-1:

    1) You pick goat#1, Monty opens goat#2
    2) You pick goat#2, Monty opens goat#1
    3) You pick car, Monty chooses a goat of his liking

Problem-2:

    1) You pick goat#1, Monty opens goat#2
    2) You pick goat#2, Monty opens goat#1
    3) You pick car, Monty opens goat#1
    4) You pick car, Monty opens goat#2
    5) You pick goat#1, Monty opens car
    6) You pick goat#2, Monty opens car

We know Monty didn't pick a car, that reduces it to:

Problem-1:

    1) You pick goat#1, Monty opens goat#2
    2) You pick goat#2, Monty opens goat#1
    3) You pick car, Monty chooses a goat of his liking

Problem-2:

    1) You pick goat#1, Monty opens goat#2
    2) You pick goat#2, Monty opens goat#1
    3) You pick car, Monty opens goat#1
    4) You pick car, Monty opens goat#2

Or, 1/3 for Problem-1, 2/4 (=1/2) for Problem-2

Imagine he picks one of the following ways:

    1) picks randomly and opens the door, or
    2) picks randomly and then, if that door has a car, switches.

His application of knowledge is converting those cases that would have been thrown away in case 1 into victories for "switch".

The counts bear this out. If you simulate it, 1/3 of all games go to 'stick' regardless. Either 1/3 or 2/3 go to 'switch', and either 1/3 or 0/3 (respectively) end up with Monty revealing the car.

You said it well: "he will show you a goat."

But the standard formulation just says Monte opened a door and it had a goat behind it. No explicit mention of intention.

It matters not that he happened to do. It matters that he will.

It doesn't, as long as he always shows a goat.

Nope. https://news.ycombinator.com/item?id=24714612

See the response to the comment. You don't understand it. I'm sorry.

https://news.ycombinator.com/item?id=24716753

It's a Conditional Probability of non independent events.

What response? The only one, which agrees with it?

Simulations where he opens a goat door are a subset of all simulations. The thought experiment is only concerned with the subset.

What if he didn't know and just happened to open the goat door? Should you still switch?

It doesn't matter if you switch in that case. See https://news.ycombinator.com/item?id=24714612

One of the realities we face here is that different people have different brains which work differently. Thus different phraseology is going to help. Adding variant phrases such as "He will never show you a car", will probably help for some.

It's also possible that Monty only gives you the option to switch if you chose the car. Or only if you chose door No 1. All of which affect the answer. Therefore, it must be made fully explicit what Monty can and cannot do.

What door Monty opens is irrelevant. He shows you a door with a goat, you’re better off switching. He shows you a door with a car, you’re still better off switching. What matters is that when you first choose a door, you have most likely chosen a goat.

First caveat: If he shows you a door with a car, there is literally no point switching. There's a goat behind both doors you're allowed to choose from, which are the remaining two.

(Below I use non-chosen to mean non-chosen by the contestants initial choice.)

Second point: If we are in the subset of all possible histories where Monty picked randomly revealed a goat, then we will have 50% of histories where both non-chosen doors contain 1 goat selected by our history subset, and 100% of histories where both non-chosen doors contain 2 goats selected by our history subset.

Since there are twice as many possible histories where the non-chosen doors contains 1 goat vs 2 goats, after selection, we have an equal number of histories in our sample where we have 1 goat or 2 goats behind the non-chosen doors. Or equivalently, we have a 50% chance that the non-chosen doors contain a car.

Therefore it is irrelevant whether you switch.

Monty needs to make an intelligent selection to change the game.

I interpreted the original game as follows: you can freely choose any of the three doors. Monty then opens one of the doors that you didn't pick, and then you may again freely choose any of the three doors. (In the standard problem, there is no reason why you would choose Monty's door). The question is: does the strategy "choose a new door" have better odds of winning?

We can work out odds for this "always switch away from original door" strategy: suppose you initially chose a door with the car (1/3 probability). Then choosing a new door makes you surely lose regardless. On the other hand, suppose you initially chose a door with a goat (2/3 probability). Then regardless of which of the two doors Monty opens, you can choose the car (if he revealed the car, choose that. If he revealed the goat, choose the other door). So our odds of winning with this strategy are still 2/3.

So it's up to the interpretation of the modified game I guess.

> So our odds of winning with this strategy are still 2/3.

That’s true before he opens a door. Then either

A) he shows a car and the odds of winning with this strategy are 100%

B) he shows a goat and the odds of winning with this strategy are between 1/2 and 2/3 depending on how the choice of door was made

If the choice is random, he shows a car with probability 1/3. The probability of winning is 1/3 x 1 + 2/3 x 1/2 = 2/3

But that analysis is for the unconditional problem and what we’re asked is what to do in case B, after a goat has been unveiled.

If he shows always a goat, the probability of winning is 0 x 1 + 1 x 2/3 = 2/3. Here there is no difference between the unconditional and the conditional problems because A never happens.

Okay, let's go with your interpretation.

Monty has now randomly chosen a door, and not revealed a car. If he had revealed a car you would have changed doors to it and won. But he didn't.[0] Now you have a choice to make.

His probability of not revealing a car in the case that both doors you didn't pick had goats behind them, is 100%.

His probability of not revealing a car in the case that one door you didn't pick had a car behind it, is 50%.

We can reason then, that of the three possible scenarios for the two non picked doors {g,c}, {c,g}, {g,g}, that it is equally likely now that since he didn't randomly reveal a car, {g,g} must be weighted twice as much as the other two.

Therefore, we must be as likely to have a goat behind the remaining door as we didn't.

In conclusion, Monty only changes the problem if he selects with information.

[0] Notice that at this point, we have to throw away from the decision tree 1/3 of the histories, and in all of them switching was beneficial. In what remains then, switching is less beneficial than it was before this trimming.

The opening of the door seems irrelevant and only serves to confuse. There will always be a goat door to open, who cares if it gets opened prior to the choice? The choice you are being given is "keep one door" or "choose both of the other doors." That is functionally the choice because Monty always opens a goat door. The chance of the other two doors containing a car is not affected.

That's what I dont understand about all the other explanations trying to "simplify" the situation. Just ask people, would you prefer 1 door or 2 doors. I've never understood how expanding it to 100 or 1000 doors is simplier than asking "which is better odds 1/3rds or 2/3rds."

I also dont believe the original intent of the question was ever meant to be ambiguous with regard to whether he had knowledge of the goat door or whether he chose at random. The intent was for him to have prior knowledge or impeccable luck, and the wordsmithing of the question came later as, in my opinion, a failed rebuttal to the simplicity of the question. The question might have been worded to not be immediately obvious, but it was not intended to have different correct outcomes depending on interpretation.

Understanding the probability tree is fairly interesting though.

If he opens a door and tells you he knows it's a goat, you double your likelihood of winning by switching to the remaining door.

If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.

> If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching. Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.

What do you mean "opens a door randomly"? Is he picking from all three doors? Yours, and the two others?

In that cases you get interesting but trivially-obvious-what-move-to-make scenarios like "he chose your door and showed you you were right" "he chose your door and showed you you were wrong" "he chose a different door which had the car"...

Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because you no longer are also capturing the cases that would otherwise be the "he chose randomly and opened the one with the car that you did not choose" that are included in the original "switch or not" decision because he always goes to a goat?

> Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because you no longer are also capturing the cases that would otherwise be the "he chose randomly and opened the one with the car that you did not choose" that are included in the original "switch or not" decision because he always goes to a goat?

He chooses randomly from the doors you did not choose, and draws a goat. When offered the choice to switch to the third door, it offers you no benefit. Both are equally likely to contain the car.

>Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because

Yes switching is. Switching is beneficial IF he shows you a goat out of the doors you didnt choose.

Nope.

Assume you choose the car (1/3). He will always choose a goat. If you switch, you have a 100% chance of losing. (1/3) * (1/1) = 1/3 chance of losing if you switch.

Assume you choose a goat (2/3). There is a 50% chance that he opens a goat if he chooses randomly. If this happens, there's a 100% chance that you win if you switch. (2/3) * (1/2) * 100% = 1/3 chance that you win if you switch.

The remaining 1/3 chance is voided because we've conditioned the game on him randomly choosing a goat, and not a car. So the probability of winning if you switch is (1/3) / ((1/3) + (1/3)) = (1/2)

exactly 50/50.

In the traditional monty hall, switching gives a 2/3 chance of winning. This is still true in the random variant but only if you're allowed to choose the car door in monty opens that one by chance.

If the host opens a door truly at random...

- you choose door 1, goat

-- host chooses door 1, goat

--- switch: 1/2 chance of being right

--- stay: you're wrong

-- host chooses door 2, goat

--- switch: you are right

--- stay: you are wrong

-- host chooses door 3, car

--- switch: you are right

--- stay: you are wrong

- you choose door 2, goat

-- host chooses door 1, goat

--- switch: you are right

--- stay: you're wrong

-- host chooses door 2, goat

--- switch: 1/2 chance of being right

--- stay: you are wrong

-- host chooses door 3, car

--- switch: you are right

--- stay: you are wrong

- you choose door 3, car

-- host chooses door 1, goat

--- switch: you are wrong

--- stay: you are right

-- host chooses door 2, goat

--- switch: you are wrong

--- stay: you are right

-- host chooses door 3, car

--- switch: you are wrong

--- stay: you are right

so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times.

there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too.

this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.

> so there are 6 times out of 9 where the host shows you a goat total.

No. He picks goat 100% of the time. Choosing not goat is outside the bounds/parameters of the initial constraints.

https://news.ycombinator.com/item?id=24713352

>If he opens a door randomly, and gets a goat, you don't modify your likelihood at all by switching.

At best, that was intended as a red herring to make the correct solution less obvious. It was not intended as an alternate correct answer. It's not how the game ever worked.

The initial version of the question said he opens a goat door. Whether he knew it would be a goat door or not is slightly irrelevant, because your odds of being right the first time were 1/3. As far as the premise of the thought experiment, a goat door always gets opened, either through peaking, premonition, or consistent luck.

>Saying 1 door or 2 doors doesn't mean you actually grasp the entirety of the problem.

I disagree, and if you don't see it as that simple, you are falling for the trap that makes the question fun.

I don't have a particularly strong opinion on this, but if we accept that its interesting as a thought experiment and not a useful strat should one find themselves time traveled into the monte hall game show, I think it's worth exploring the very closely related variants of the thought experiment.

If I time traveled into the show, and he opened a goat door, I would switch.

There is no better strategy than switching. Only under certain rules is there an equally good strategy in staying.

Not if switching costs a dollar :)

Not important, and just barely relevant, but "Let's Make a Deal" was Monty Hall's gameshow and it's still being made.

It probably went like this:

Hey lets put a car behind three doors, have people choose, open one of the other two doors, then ask them to switch.

Sounds good Bob, but wait, won't the show end prematurely 1/3 of the time because you will randomly open a car?

Huh, good point. Um, let's sneak a peek before opening so we never open a car?

Sounds good Bob.

Too bad Let’s Make a Deal never actually let you switch, IIRC

> I also dont believe the original intent of the question was ever meant to be ambiguous with regard to whether he had knowledge of the goat door or whether he chose at random.

That’s extremely obvious! The entire reason to phrase the problem as a game show with a game show host is to make it abundantly clear that the host “knows the answers.”

It might be semantics, but it doesn't matter if he knows or just always chooses goat.

Reordering the sentence makes it clearer. "In all games where Monty chooses goat...". The premise creates a subset.

No, if Monty sometimes opens the door with the prize, and you ignore those games, then the result is different, and you don't benefit from switching doors.

I like this phraseology. How to make the answer intuitive is the objective, and underlining that you're choosing two doors, one of which is going to be wrong, really helps.

Yes I finally understood it. At the beginning you are choosing one door vs "the others". If you could you would already chose "the others" because they have a 67% win probability but you are only allowed to chose one single door. Then he opens all the wrong doors of "the other doors". The "other doors" still have 67% probability and now only one door is remaining in the "other doors". Obviously that last remaining door now has 67%.

This is the comment that has a car behind it.

Monty’s knowledge is irrelevant. The original problem can be modified with no change in probabilities to:

Pick one of three doors. You can have whatever is behind that door, or you can change your pick to both of the other doors and win whatever is behind both of them. Should you switch? Obviously.

Step 1: Take three cards face down, one of which is an ace.

Step 2: Divide it into a pair of cards, and a single card.

Step 3: Reveal one of the pairs of cards. If you reveal and ace, return to step 1 since this history is eliminated from the story we've been told: we know this wasn't a possible path to our endgame. Otherwise continue to step 4.

Step 4: Which of the remaining two cards is more likely to be an ace?

Step 5: Realize that they're just as likely as each other to be the ace. Step 2 didn't magically imbue the card remaining in the pair with extra probability juice.

Long story short, you definitely need Monty to make an intelligent selection, so Monty's knowledge is far from irrelevant. It matters whether he revealed a goat by luck or by knowledge, because he's 2x as likely to get "lucky" in the case where there are 2 goats behind the doors you didn't choose in your original guess.

I mean, step 2 did "imbue the card remaining in the pair with extra probability juice." It went from 1/3 to 1/2. But so did the card you picked initially.

That you will never be worse by switching doesn’t mean that you will always be better.

https://news.ycombinator.com/item?id=24713352

When he shows (randomly from the doors you didn’t pick) the car, switching to open door (where the car is) and the other door (which hides a goat) you increase your chances from 0% to 100%.

When he shows a goat, switching to the open door (where there is a goat) and the other door (which may or may not hide the car) your chances stay at 50%.

And what happens if he opens the door you picked and the car is there? Is still obvious that you should switch?

Switching is the best ex-ante strategy, not necessarily so after a door has been opened.

Monty hall is a conditional probability problem. Conditional probability problems are typically not intuitive and require very precise definition of the condition. Rule of thumb: don't meddle with the condition.

An algebraic, Bayes-theorem solution to the problem:

The relevant events:

  A_x = "contestant first picks door x"
  B_y = "Monty opens door y, revealing a goat"
  C_z = "price is behind door z"

We want to calculate P(C_z|A_x & B_y) for certain combinations of x,y and z. I assume x=1, y=2 for the following calculations (A = A_1, B = B_2).

Assumptions:

  P(C_z) = 1/3, the price can be behind any door with equal probabilities
  A and C are independent, the contestant has no prior knowledge of the placement of the price
  P(B|A&C_3)=1, that is Monty opens door 2 with probability 1 if the contestant first opened door 1 and the price is behind door 3, Monty deliberately picks the door with the goat, very important!
  P(B|A&C_2)=0, Monty never opens the door with the price.
  P(B|A&C_1)=1/2, Monty equally randomly picks between two doors when he can.

Now substitute into all the probabilities:

  P(C_3|A & B) //probability for winning when switching
  = P(A & B|C_3)*P(C_3) / P(A & B)
  = P(B|A & C_3)*P(A|C_3)*P(C_3)
    /(P(B|A)*P(A)) // P(A|C_3) = P(A) due to independence
  = P(B|A & C_3)*P(C_3)
    /P(B|A) //expand denominator
  = P(B|A & C_3)*P(C_3)
    /( P(B|A & C_1)*P(C_1)
       + P(B|A & C_2)*P(C_2)
       + P(B|A & C_3)*P(C_3) ) // use P(C_1) = P(C_2) = P(C_3) = 1/3
  = P(B|A & C_3)
    /( P(B|A & C_1)
       + P(B|A & C_2)
       + P(B|A & C_3) ) // substitute all our assumptions above
  = 1 / (1 + 0 + 1/2)
  = 2/3

We can see that the condition B is not trivial and requires precise knowledge of Monty's strategy. Meddling with this condition results in different outcomes.

I got out three playing cards and did the experiment myself over and over. That was many years ago and now whenever I see something on the Monty Hall problem it seems so obvious as if I can no longer even see why people think its unintuitive. The reason its unintuitive is that people don't really have experience with anything that works this way. I'm not sure that the linked explanation will be that helpful to people because of that, although it does make it plainly obvious that Monty Hall knows which door has the prize and he's telegraphing to you which door it might be by only opening doors which don't have it in there. I suppose if that's the missing part of the puzzle for you, then that will help.

> I suppose if that's the missing part of the puzzle for you, then that will help.

Of course it will help, because it changes the entire scenario from two people playing a game of chance to one person playing a game of deduction against secret knowledge.

The odds don't change because of some quirk of the Universe, they change because one player is changing the parameters due to his knowledge.

> I got out three playing cards and did the experiment myself over and over.

How did you do that when you didn't know which cards were 'goats'?

I've always thought of it like the 'denominator' of probability for the second stage became 2/3 (vs. the usual/first stage 1).

So when he reveals a goat door you know that door is certainly not a winner, 0 probability (of 2/3), and the switch door is certainly (if it were one of those) the winner, 1 probability (of 2/3).

But I still used to manage to confuse myself thinking it's intuitively 'more likely' to be the original door 'now' that it isn't one of the others.

Until I studied information theory at university and it really clicked - Monty's door choice has lower entropy than your initial pick!

(But I acknowledge you can't say to the masses 'look look let's simplify this, if we just step back and take an information theoretic approach -')

There is another important assumption: what if Monty knows which door has a car in advance and:

1) if you pick the car, Monty opens up other doors as in the regular problem

2) if you do not pick a car, Monty just opens your door and says "unfortunate pick, contestant".

If Monty is an adversarial agent (which is not an unfair assumption in television land), then your strategy changes. So an important assumption to make for the Monty Hall problem to work, is that Monty announced what he will do after you pick your door _before_ you pick your door. Only in that case do you have your counterfactuals correct and in that case it is indeed better to swap doors. It is a hidden assumption in most probability theory type of answers.

This is at the heart of why so many people find this unintuitive. People who know the right answer are assuming that this is a repeated game, where every time Monty Hall does the same thing.

Every instance given in the linked article includes the assumption that you will always have a choice. If you encounter this situation, how can you possibly know whether you would be offered the choice had you picked the wrong one?

I've had issues understanding the Monty Hall problem for years until now. What made it click was that, at the end of the article, it's explained that the doors are not opened by Monty at random. My previous reads about the problem did not disclose this, so I had made the assumption that the door opening was random. I've also never seen the show, for what it's worth.

The problem explicitly says that he opens the door to reveal a goat, though. If he was opening the door at random, then he would have a 1/3 chance to reveal the car. Since he always reveals the goat, to me that implies that it's not random.

No, it states only what is happening this time. It does not say "always" or talk about what happens/happened for other contestants.

An accurate and precise description of the problem is essential to understanding. Without explicitly stating that the door choice is non-random, I did not make the assumption.

Key observation: you lose when switching the door only if your first pick was winning.

Your first pick had 1/3 chance of winning. So switching has 1/3 chance of losing. Thus it has 1 - 1/3 == 2/3 chance of winning.

Yeah, this is pretty close to how I see it.

The key is making it clear that switching doors will always switch you from Lose to Win and vice versa. Switching doors is equivalent to switching outcomes. That's the critical fact, IMHO.

Once that's accepted, the rest falls into place: there's a 2/3 chance that you picked a Losing door, so there's a 2/3 chance that you'll benefit from switching outcomes, and since you're guaranteed to switch outcomes if you switch doors, there's a 2/3 chance that you'll benefit from switching doors.

Indeed, this is what made it click for me. Switching is equivalent to saying "I didn't get it right the first time", which is a good bet it has a probability of 2/3.

Keeping the same door means "I bet I got it right the first time", which is only 1/3 probable!

3 Traits can cause a person to get the Monty Hall problem wrong:

1) Stubborness: "I've made my pick, and I'm sticking to it, options will just make me double down!"

2) Suspiciousness: "I'm faced with a man in a suit, looking like a salesperson, and he's trying to make me change my mind. I think he is up to no good!"

3) Stupidity (relative to those that get it right): "I overestimate my ability at statistics, and I think there is an even chance of a car between each door after one goat has been revealed!"

And for those who get it right, not from luck, but from understanding the statistics, there is always the Two Envelopes problem that is very similar, yet so much harder.

What works for me is scaling the number of doors to 100 (or more). Now the initial guess is correct only 1% of the time, while switching is 99%.

Simulated many more variations at https://simonduff.net/monty_hall/

> It’s important that Monty looked behind the doors before choosing which to open. This is where people’s intuition usually fails. If he had chosen a door at random — in a way that he risked possibly exposing a car, then the situation would be different. (In that case, there’s no advantage or harm in switching.)

This one took a second to rationalize. The reason it works is you have the same chance of winning overall (assuming you can safely choose the car if he opens the car by chance), it's just that the value of winning from switching vs. staying has been shifted into the probability of winning by default. The usual mental trick is to extend to 1,000,000 doors.

If you pick one door, then are told that all of the alternatives except one are the correct answer, you should obviously reason that the door you didn't choose is the correct answer, unless you got the 1/1,000,000 guess. Odds of winning if you switch are 999,999 / 1,000,000

If the host instead opens 999,998 doors randomly, you have a 999,998 / 1,000,000 chance of winning by default. The remaining two doors have equal chance of winning, giving you the same total odds of 999,999 / 1,000,000 no matter which you choose.

Which makes sense, because both situations are, more or less, being given 999,999 chances to guess the lucky door.

This seems overly complicated. Here's an answer on Quora which is a more straight forward deweirdification.

https://www.quora.com/In-the-monty-hall-problem-how-does-ope...

Reprinted:

Q: In the monty hall problem, how does opening the second door skew the probability in favor of the initially unchosen door?

A: The Monty Hall problem is generally poorly described, in order to make the conclusion seem more surprising then it is.

The actual Monty Hall game — as imagined by the people who are asking the question— is set up like this:

  In front of you are 3 doors, there is a goat behind two 
  of them, and a car behind the other one. 

  In *round 1* of the game, you select a *pair* of doors,  
  from which *one* “goat containing door” will be
  *automatically eliminated from*, leaving only *one* door
  of the selected pair of doors in play, (and only *two* of
  initial *three* doors in play).

  In round 2 of the game, you guess which of the two
  remaining doors in play has the car.

The choice is this: should you choose the remaining door from the pair selected in round 1, or should you choose the door which was not part of the selected pair in round 1?

When phrased like this, the answer is fairly obvious: the pair of doors contains a car 2/3 of the time, whereas the non-paired door contains a car 1/3 of the time.

The Monty Hall problem— as normally described— messes this all up by introducing a game show host. This is an agent who— seemingly by their own whim— changes the game you thought you were playing, and introduces round 1 of the game once you have guessed the door you initially think the car is behind. The rules this game show host agent are following are almost never described to a sufficient degree to ensure the game is equivalent to the game laid out above. And yet, the people asking this problem pretend that it is exactly equivalent when they ask you for an answer.

It’s generally a poorly described problem, whose answer depends entirely on what kind of agent the game show host is.

Don’t worry if it doesn’t make sense to you as it’s usually described. If you can understand why— in the two round game I describe above— it’s better to pick the door from the pair of doors, rather than the single door, you understand probability just fine.

If you swap, monty gives you the best outcome from the 2 doors you didn't originally pick. Better 2 doors than 1.

The kind of answer to this problem that I see most frequently uses logical reasoning. I don't think that's the best approach to start with, because it is abstract and disconnected from reality. It has a high chance of error.

I think people would do better to write a computer simulation of the problem, or, for non-programmers, to design a game with dice that simulates the problem. When you execute the simulation, you pretty quickly realize what's going on. Then you can use logic or formal reasoning to put your intuition into words.

A couple people here have said something similar (ctrl-f "simulation" on this page). According to https://www.mwsug.org/proceedings/2010/stats/MWSUG-2010-87.p... , even the great mathematician Paul Erdős wasn't convinced that switching was better until he saw a simulation:

> Vazsonyi ran the program 100,000 times. Erdős watched the results of the simulation. The simulation results indicated that by switching, the odds of winning are indeed two out of three. Finally, he was grudgingly convinced that switching was better. He did not like it but seeing was believing. He could not argue with the results.

What works for me is thinking: if you pick a goat, switching always wins you the game.

How about this?

    import numpy as np
    
    def play_monty_hall(rounds=100):
        car = np.random.randint(low=1,high=4,size=rounds)
        first_door = np.random.randint(low=1,high=4,size=rounds)
        switching_wins = car != first_door
        staying_wins = np.logical_not(switching_wins)
        print(sum(switching_wins))
        print(sum(staying_wins))

Result from 1,000,000 games: Switching wins: 667,299 Staying wins: 332,701

When you pick a door, there's three possible scenarios: A) You picked the car door B) You picked a goat door C) You picked the other goat door. All three are equally likely. But in B and C, the host only has one goat door available to open (because you already picked the other goat door) so since he has to show you a goat, he is effectively forced to tell you where the car is, since that must be the third door that has neither been picked nor opened yet - and switching doors is how you act on that information to get the car. It's only in scenario A that he gets a choice of which of the two doors to open. So there's a 2/3rds chance that he's telling you exactly where the car is, so you're better off acting on that info by switching.

I think the confusion comes from the oracle-like knowledge of Monty Hall - he knows what is behind each door and decides which door to open based on that knowledge. If Monty Hall wasn't an oracle, and just opened a random door after the guess, people's intuitions would be correct.

When explaining to friends, I found simply increasing the number of doors the most helpful way of explaining.

There are 1,000 doors - you are passed to pick one.

Monty removes all the doors except for your door and one other. There is money behind one of the doors. Do you stick, or change?

Of course you stick -- Monte obviously opened the other doors just to trick you into switching because he knows you got the right answer. Of course he can do this; you already know that 999 doors have goats, so whether or not you picked correctly he can open 998 doors to reveal goats.

If you had picked a goat door, he would have just opened the door you picked or opened the door that has the money to show you that you got the wrong answer.

The assumption that Monty Hall will always offer you a chance to switch is what is broken in the problem statement and the reason why so many people think that the correct answer is unintuitive.

> Since you don’t care about goats, this makes no difference.

Says you. I choose the goat!

For anyone that doesn’t get it after this explanation or even the 1000 doors trick:

Try visualizing how you’d pseudocode this game - it literally didn’t click for me until right now, and now it seems much more intuitive.

> But he doesn’t choose the door at random. He deliberately chooses to show you goats. Since this is always possible, it tells you nothing

What happens if Monty does not ever choose at random. Say that Monty always opens the the highest possible door.

If you choose door 1 and Monty reveals door 2, then switching (to 3) is 100% win.

If you choose door 1 and Monty reveals door 3, then switching (to 2) is 50% win.

I think it's a crucial unstated assumption that Monty does choose randomly among available goat doors.

The assumption is that you don’t know how he picks the door when there are two goats. It’s a reasonable assumption when the problem statement doesn’t say anything about it.

It could be random, it could be the highest, it could be the lowest, it could depend on the position of the stars. The point is that you don’t know so the answer cannot depend on it.

Just using more doors (both for the total, and the number he opens) is probably enough to make it clear. Like 1000 doors. If I choose one at random I am pretty unlikely to be right on that first choice. But then he opens 998 of the unchosen doors, knowing he won't open the door with the car, I would think oh if it was originally in any of the 999 unchosen doors it is in the one he didn't open.

If there are infinite number of doors, then does switching guarantee us the car?

Also, Game 5: There are 2 doors. A car is randomly placed behind one, and goats behind the others. You pick one door. Monty looks behind the other doors. He chooses 0 of them with goats behind them, and opens them. You get two options: Option A: You get whatever is behind the door you picked. Option B: You get whatever is behind the other closed door. Should you switch?

If Monty opens all infinity-minus-two doors except for your initial choice and the one with the car, then switching almost surely[1] gives you a car.

1. https://en.wikipedia.org/wiki/Almost_surely

It's counterintuitive, as magically the car appears behind the door that Monty chose. But tbf chances of it being behind the door you've chosen were 0 to begin with.

For the “The host knowledge and deliberate choice are irrelevant, if he opened a random door nothing would change.” crowd:

If he’s going to open a random door he may just leave it to you (because you don’t know anything).

The problem becomes:

0) you’re presented three doors, there is a car behind one of them

1) you pick one door

2) you choose one of the other doors and open it: there is a goat

Now, you have the choice between keeping your initial door and switching to the other unopened door.

Even in this scenario, you should switch too. Failing to switch is equivalent to saying "I chose correctly the first time", which has 1/3 probability of being right.

No, in this sceneario it has a 1/2 probability of being right.

   P(I chose correctly the first time | all doors are closed) = 1/3

but

   P(I chose correctly the first time | I opened one of the other doors at random and there was a goat) = 1/2

Without loss of generality, we can say that I picked door A and opened door B findind a goat.

   P(car@A | goat@B) = P(car@A and goat@B) / P(goat@B) = P(goat@B | car@A) P(car@A) / P(goat@B) = 1 x 1/3 / 2/3 = 1/2

In case it's not clear where the formulas above come from:

  P(car@A and goat@B) = P(car@A) P(goat@B | car@A)

is equal to

  P(goat@B and car@A) = P(goat@B) P(car@A | goat@B)

Hmmm. Makes sense. I iterated a simulation of 10000 cases, discarding every case where the second random door reveals a car (because this breaks the game; at this point there is nothing to guess anymore) and it gives a probability of winning of only 0.42 if you switch. So it's neither 1/2 nor 2/3. Curious!

This problem is good because it illustrates a good way to think through any counter-intuitive problem - increase the numbers to amplify the effect. I remember being taught a similar thing in physics class, to make the weight stupidly heavy or the charge stupidly large, and then see what happens. This kind of thinking can be applied in a lot of different areas.

An even easier way to achieve this is to just say there are 100 doors, you pick one, he opens 98 goat doors and asks you to if you want to switch.

I have yet to meet someone who didn't get it from that. No need to have this complex evolving ruleset and so on.

The only "surprising" concept in Monty Hall is that a game show host would do anything at all to help a contestant. I don't know if this is just an Americanism, but the default assumption is this guy's trying to trick you out of your money.

To explain Monty is helping you is, at all times, a stretch of the imagination.

I had given a very similar answer here: https://stats.stackexchange.com/questions/41208/the-sleeping... in an attempt to draw a parallel to the Sleeping Beauty paradox.

...I am not a very smart man, and after reading this - it still did not click why the correct choice is to switch.

So I went on youtube.. Watching the first 3 mins of this 5 min video made it click: https://www.youtube.com/watch?v=4Lb-6rxZxx0

Assuming you understand the rule (that Monty will always open a door you didn't choose, and that door always has a donkey behind it), then one way to understand the game is:

You can choose to have:

- The most valuable prize that's behind door A, or

- The most valuable prize that's behind doors B or C

When you look at it this way, it's obvious that you would rather have the most valuable prize from the 'other' doors, and the way to do that is to switch.

Then you don't need to think about whether the probabilities change once Monty opens the door with the donkey.

If you need convincing, here's a simple python script based on the above:

https://gist.github.com/rahimnathwani/2b6ca328a74b37b952c75d...

If you're still thinking about whether the probabilities change, consider whether 'Monty opens a door with a donkey behind it' is new information. It's not because he always does that. And the two doors you picked are fungible/identical except for physical position, as both are in the set of doors you didn't pick. So which one he opens is irrelevant.

But that was the beauty of the original MH problem:

It's so everyday and normal seeming -- and while too cryptic or opaque -- just opaque and non-obvious enough to make it, even for many smart and well educated people -- an exquisitely slippery trap to fall in.

Actually, conditional probability is interesting

https://en.m.wikipedia.org/wiki/Doomsday_argument

Just the fact that you’re doing an experiment is already extra information!

If you were correct with your initial choice, switching is a loss. If you were wrong on your initial choice, switching is a win. Your initial choice is only correct 1/3 times, but your initial choice is wrong 2/3 times. It's pretty simple.

This worked for me. Avoiding the "and now you have 2 doors..." moment in your explanation I think made it easier to stay away from seeing it as a 50/50

Does (Did) Monty always make the offer, or did he only make the offer to switch sometimes? Perhaps he could have only made the offer if the contestant had chosen the winning door. This would change the odds considerably.

I don’t know about the game show, but the premise of the classic problem is that he always makes the offer.

I think one way people get it wrong is by arguing from symmetry; if I choose 1 and 3 is shown to be a goat, then that's the same as choosing 2 and being shown 3 is a goat.

So what's the diff?

Which is wrong but seductive.

Has anyone gone through old episodes of Let's Make A Deal, and looked at the success rate of those who change their guess, vs those who don't?

Pretty much every description of the Monty Hall problem has the same flaw, and it is here also. The problem as given does not describe the general rules by which Monty operates. It describes only a single round of playing the game.

Thefefore, Monty could be using the strategy of "if the player chose the car door, open a goat door and give the option to switch. Otherwise don't give the option to switch and the player wins the goat." In that case switching is a losing strategy.

That's no more a flaw than failing to explicitly say the car is valuable. I mean what if the car is a matchbox toy and the goats are worth more? That's not explicit either.

Sometimes you have to use common sense, and I think every instance of the monty hall problem I've seen was sufficiently explicit (without being absurd), and the confusion was always around the math and probability and never around semantics or trickery.

Obviously the goal of the problem is to get the car. Pretending that the argument I have given is like making up something about a toy car just does not do anything.

The fact is that the argument I have presented demonstrates that the problem as given is flawed and does not have a unique answer. Most people don't understand this and substitute the correct version of the problem in their mind, and then proceed to solve that by arguing about the probabilities. Of course the probabilities are what the problem is "supposed" to be about.

I disagree. Every person I've ever talked to who struggles with this problem is confused over the probabilities, even after we discuss the situation in great detail. They perfectly understand the game and the rules.

You seem to be finding flaws where there are none. You are suggesting there is subtle unspoken trickery hidden in the problem (just as in my example) but nothing about the problem suggests that should be the case.

What if the host lets you choose to switch every game but if you make the right choice he says you are wrong without opening the door and the game is just over? What if he says the door you chose is eliminated and the remaining door is your prize if you chose correctly? What if you win and then he says you have to play best 2 out of 3 to really win, and if you win again he says 3 out of 5, and keeps moving the goalposts until you lose?

Suggesting the problem is flawed because you can imagine up fringe scenarios that are not explicitly excluded does not seem like a useful criticism.

There being many of you does not make you any more correct. I'm not making up "fringe scenarios". I have clearly (I think) explained how, given the problem as stated, switching is not necessarily beneficial, and can be harmful. You need to make an additional assumption (that Monty necessarily behaves in a certain way) that is not stated in the problem to get to the "always switch" answer.

By the way, once you make that assumption, those other scenarios you presented are also excluded.

Quantity doesn't affect the correctness of facts but it sure affects questions of perception, and your argument seems to be one of perception.

> By the way, once you make that assumption, those other scenarios you presented are also excluded.

Even more of a reason that "flaw" shouldn't even be considered.

There is no hidden trickery and no reason to assume the game isn't fair.

I'll try one more time. Let's imagine there are two worlds, A and B.

In A, Monty Hall behaves like you think: always opens a goat door, always gives the option to switch.

In B, he behaves like I described: opens a goat door and gives the option to switch, but only when the player has chosen the car door. Otherwise he does not let you switch.

And then we find ourselves in this situation:

> Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

At this point, how do you know you are in world A and not B (or some other world)? What in this problem as given here allows you to determine that?

There is nothing in the problem statement that should reasonably lead you to believe there are any tricks at play. The fact that you could come up with some chicanery that the host might pull which isn't explicitly disallowed doesn't make the original problem flawed in any way.

I bet you could take any problem of similar nature from anywhere and scrutinize hard enough and find some gimmick that lets you claim similar claims about it, but I don't think that's useful or noteworthy.

At the end of the day we all rely on common sense and common assumptions about these kinds of things. It may be true that some don't have the same shared experience to draw upon and lead them to the same understanding as others, but that doesn't make the problem flawed. It just means people understand things differently.

If a significant number of people brought up this issue then my opinion might change, but as I said, this is the only time I've heard of this particular complaint (and I've been enjoying posing this problem to people for decades now), and that means, in my opinion, there is no grounds to claim your misunderstanding as some objective flaw in the wording or presentation of the problem.

You reject my argument without being able to point to any flaw in it. Because you and a lot of other people do not accept or have not thought of it. I can't really help with that.

If it's about popularity instead of logical argument, of course I'm not the only one who thinks this. There was a really good blog post that laid it out but I can't find it currently. Instead, here's a scientific paper I found just now, the introduction contains the same argument I'm making. And it's far from the only place which agrees.

https://link.springer.com/article/10.1186/2195-5468-2-2

> The problem posed in this way may lead to a lot of controversy, mainly because we do not know whether the behavior of the host had anything to do with your first choice or not.

> Perhaps the host would open a door with a goat only when your first choice was right. In this case, it was not a good choice to change doors.

EDIT: Also several people have pointed this same thing out elsewhere in this thread.

I pointed out several reasons why I think your argument is flawed.

In the real world, you know by watching the show. Monte Hall always offers to switch, so that you know that you're in world A.

If it was the first time the show ever aired, you could be in a situation where you didn't know whether it was A or B. But in the problem as normally posed, you have time to watch the show for months to years, and you know how Monty behaves.

I've never watched the show, but I have read that that is not the case; sometimes he gave the option to switch, and sometimes not. Even if that were the case, it is not stated in the problem and therefore you cannot assume he behaves like he might in the real world. Also even if he had allowed the switch every previous time, it still does not logically mean that he does that the one time you are playing.

And once more, I'm talking about the problem as given, not some other problem. It is a self contained math / logic problem.

I would not say it is a flaw. I think it is a reasonable implicit assumption that there is only one round, unless explicitly stated otherwise.

It does not matter if there are many rounds or one, what matters is how Monty behaves. And that is given only for the current round, and not as a general rule.

Some explanations of this problem are far too complicated. This is how I explain it.

1/3 of the time, you guess correctly on your initial guess. If you switch, you'd be wrong.

2/3 of the time, you guess incorrectly on your initial guess. If you switch, you're right.

So when switching, the expecting outcome 1/3 of the time is 0, but the expected outcome 2/3 of the time 1. (1/3)0 + (2/3)1 = 2/3

You have solved a different problem. You didn’t include in your a analysis the “he opens one of the other doors and there is a goat” bit.

The original problem is more complicated.

So the strategy "always choose the other door" boils down to the following simulation, which kind of makes it obvious

   {
      int count = 0;
      int i;

      for (i = 0; i < 1000; i++) {
         int choice1 = rand()%3;
         int actual = rand()%3;
         if (choice1 != actual)
             count++;
   }

I'm just stunned anyone remembers Monte Hall.

Kudos to the author.

What if you really just wanted one goat?

https://xkcd.com/1282/

There's an XKCD for everything

The analysis of the Game 3 in the article is wrong. You should switch.

Your comment would be better if you could explain why you think it is wrong, rather than merely making an empty assertion. I think you are just misinterpreting the analysis. It says "Option B still gets you the car 90% of the time", and Option B entails switching. Do you read it differently?

If someone is struggling to understand it with three initial doors, instead start with a million initial doors and close all but two.

As has been explained in the other comments, Door A has a 1/3 chance of being the right choice and Door C (the unopened door) has a 2/3 chance of being the right choice.

All true... but if you toss a coin to choose whether to switch or not, the odds are 50:50...

Sure, but why is this noteworthy?

Choosing between any two outcomes with a coin flip where one outcome is good X% of the time and the other is good 1-X% (the rest) of the time will always give you a 50/50 good outcome...

When I first read Monty Hall problem, I actually thought it was blindingly obvious to switch. The fact the presenter had 2 doors to chose from and the one he opens is a goat _should_ increase the chance of the other one being not(goat) right?

Only if he's not obliged to open all the goat doors.

That's where the "imagine there are 1000 doors" extrapolations fail. If there are 1000 doors and Monty can choose to open just one, then there's no point switching to one of the other 998. Those comparisons only work if he is obliged to open every goat door.

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