Hacker News Reader

Comment by lqet | Hacker News Reader

> 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.

View on HN

>> 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.)

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?

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.

> 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.

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

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.

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.

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.

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.

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.

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.

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.

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

>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 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.

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.

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.

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.

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.

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`.)

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.

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 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.

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 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.

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.

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.

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!

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."

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.

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.

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.

> 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)

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.

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?

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

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.

Thank you!

Bravo! This explanation wins in every way.

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?

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.

> 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

Reply on news.ycombinator.com