If you phrase it adversarially it makes more sense... 2/3 of the time you select a goat and force Monty to reveal the other goat. For those 2/3, you're guaranteed to get a car by switching. The remaining third you get a goat.
If you take the "stay" strategy... 1/3 you hit the car and keep it. 2/3 you hit a goat and keep it.
In summary... "switch" is 2/3 car, "stay" is 2/3 goat.
In these times you should choose "stay" and hope to get a goat so you can turn grass into food. Cars are overrated.
The easiest explanation I've ever heard and which immediately made me understand it was the following:
Instead of 3 doors, imagine there are 100. 99 of which have a goat and only one of which has a price behind it. Now blindly choose a door and the host opens 98 of the other doors which have a goat behind it. Would you switch your door now, given the choice?
It's easy to see that your probability of choosing a "wrong" door when you had 100 doors to choose from was much higher than choosing the right door when you only have two doors to choose from.
This method of thinking, i.e. increasing or decreasing the problem space by some orders of magintude has helped me a lot in thinking about problems and their solutions in general.
I also do this, among other things to find subtle bugs in code and math. For every unbounded variable, what happens if it is 10^100, or 10^-100? It's a good habit that also makes you view division and logarithms with great suspicion :)
One of my first computer science teachers pronounced "/" (in C) as "divide and throw away the remainder", which sounds awkward at first but turns out to be quite helpful. It's one of those operators which, unfortunately, looks just like one from math class but acts very differently, even in common cases.
Interesting. I was explained this question the exact same way (100 doors) some years ago and it was instantly intuitive to me. Maybe it's how every brain works differently.
I understand that it makes sense intuitively, but how do you justify it to yourself mathematically? As far as I can tell there is nothing in the original formulation that explains why it would scale that way.
I'm the host, and I also happen to be blind. I chose 98 doors and none of them have the car - terrible luck on my part! But probably this lends some credence to your initial guess?
This is correct, but based on specific assumptions. If the host flipped a coin and selected a door and 1/Nth of the time it had the car then you switching would not change your odds.
Similarly, an adversarial host who sometimes opens a door and sometimes just shows you your choice can similarly mess with the odds.
It’s really due to the host both being forced to show a door and knowing which door to choose that you gain from leveraging that knowledge.
That is also a very interesting way to think about it, if I understood you correctly. Seeing the host as another player, any "bad luck" he has, should translate into myself having a higher chance of success if used correctly.
With the host as another blind player, his opening of 98 goat doors only increases the probability you were right from 0.01 to 0.5, so still makes no difference for you to change. But of course the original version of the problem is predicated on the host knowing where the car is and only revealing goats.
One thing that intrigued me about this puzzle is wondering what mental model for probability would make solving it intuitive? The one that works for me was thinking about it in terms of information:
Because Monty can never choose the door you first picked he can't give you any new information about that door.
So when he reveals which of the remaining doors has a goat, he is only giving us new information about those remaining doors. That information reduces the odds on the remaining doors and that is why you should always switch.
If you switch "you" get to "open" two doors in fact, not one. It's just that Monty picks which one for you. And since your choice is random, you might as well let an RNG do it for you.
If you consider it two independent rounds instead of stay/switch,
Stay: you get to roll 1d3, "1" wins
Switch: you get to roll 1d3, "1" loses, "2" or "3" wins
This puzzle has stayed with me for a decades for two reasons. #1 is it was the first time I was embarrassed in public discourse on Usenet. Guy shot me down pretty hard, but when he was done there was no uncertainty- he was right and I was wrong.
#2 is I finally realized the importance of that postulate of Quantum Mechanics: The wavefunction contains everything you can know about the system. This problem pointed out for me that probability is not an unvariable thing. Different people will calculate different probabilities for different events (think if we're playing cards and I have three aces and you have none- you will calculate a different probability of drawing a ace that I will) and we will both be correct. This without this postulate the probabilities of QM would be meaningless.
The main thing that seems to cause the paradox is a lack of specificity around the door opening mechanism. If you make it clear that Monty only opens a door that will definitely be empty, I find that most commentators will agree that Marilyn is correct, you should switch.
The original question that vos Savant responded to contains the phrase "...and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat..." - I don't know how they could be any more specific.
If the host opens the door with the car, the game would be over anyway. It is clearly implied that the host always opens a door with a goat.
Whether the host opens a door with a goat or a car determines whether the game can still be played at all. How that door is selected does not factor into whether the contestant should switch or not in any way.
I think the source of the paradox lies somewhere in the biases (the Endowment effect or Status Quo bias) as discussed in the Wikipedia article about the problem - not with how the question was stated and especially not with the part about how the host selects a door.
Ah, you are wrong (and this is often confusing to people).
If you model this game where the host chooses randomly, and might open the car door (when that hapoens the player instantly loses) , then there is no point switching.
I think the fact this does matter is the source of many people's confusion.
Why not? Your chances of choosing the correct door were 1 in 3 from the start. That doesn't change with the fact that the host opens the doors at random.
Sorry for the late reply, but if you look at the original article on GitHub I explore that scenario in the code. The answer is that if Monty randomly opens a door then the contestant automatically loses 1/3 of the time before given a chance to switch.
That's what the GP said. In reality, the host doesn't open the door at random. They know which door hides the car and avoid opening it. That's why switching the doors doubles your chances of finding it (because the host has already eliminated one of the goat doors).
Your chance was 1 in 3 yes, but if the host can randomly open the car door, there is a chance of 1 in 3 of that happening, leaving only 1 in 3 chwnce of switching being legal and useful.
I don't think I've ever seen this rule in any variants.
Since we're making up rules: Could the player not still have the option to switch, and thus win the game?
I'm not convinced that I am wrong after your explanation.
Either way, my point is that focusing on the problem statement as the "main thing that seems to cause the paradox" allows the reader to stop there and disregard a far more interesting discussion about biases.
How is that rule made up? There's one car, if that door's been opened it's not left for the player to open. Obviously you can't win by switching to the location you know it's in because you've seen it, that's not a game.
If the host knows where the car is, why would it ask the player to change door if he knows the player already choose a goat?
The problem should specify that those are the rules of the game that the host always open a door.
Otherwise, as a host, I would only offer the players to change if they picked the car. (And if they heard about this paradox before, they might even change their choice)
If you only ever offered the players to switch if they picked the car, they'd soon realize this and never switch when they're offered this chance. Also, players that picked a goat at first would have no chance of winning the car.
That's assuming players knows about the other players, which is not specified in the problem description. It is not even specified if the same game is played several time with the same rules.
Given the description from the article:
> 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?
I'd say, no, it is not to my advantage because I'd think the host would only ask me to switch if I had taken the "right" choice and want to make me lose.
Unless I knew that the host always ask if one wants to change, in which case the paradox indeed apply.
Game show regulations would not allow that behavior. The problem assumes a typical game show in western media so the reader should assume the player knows all the rules governing the host’s behavior, which you can see in the other discussions.
The lack of specificity is not that the host might open a door with a car, but whether the host had to open a door at all. This is not clear from Savants formulation, and indeed does not match the actual game show, where the host had discretion in every episode whether to open a door.
If the host does not have to open a door for the candidate (which is quite likely, he's playing his own strategy of making the show exciting, not predictable), then the usual solution (always switch) is not correct.
All this, and probably any variation and every solution of the problem that comes up in this thread (and probably every other Monty Hall forum thread in the post 3 decades) are discussed extensively in the Wikipedia entry (e.g. https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...).
Maybe every time a new Monty Hall thread comes a mod should just stick the Wikipedia entry at the top. I've never seen anything in any of the following discussion that hasn't been covered in Wikipedia already.
I don't think it is clearly implied. There is an old UK quiz show called "strike it lucky" where the host would show future symbols which might be good or bad for the players, rather than explictly showing "good" things.
> If you make it clear that Monty only opens a door that will definitely be empty, I find that most commentators will agree that Marilyn is correct, you should switch.
I can’t argue with what you’ve observed, but my own experience is not that there’s a misunderstanding about the question, but that people simply don’t think properly about the question. Nobody I’ve ever seen confused about the answer was confused by the question, and many numerically literate people I know have struggled with it.
I don't think it's true that most people will agree. The problem is that intuitively it's easy to think that the chance of guessing the car is 1:3. That chance doesn't change just because you later know that another door has a goat. So you think that your choice was originally 1:3, the door that never gets revealed is 1:3 which are both equal. So when the other door is revealed, both chances move to 1:2. It's wrong, of course, but it's very easy to think that way (of course, I never thought that... ever... really... ha ha...)
I think what's confusing, is that if one doesn't switch, the first selection is still 1:3 - since you randomly picked 1 door out of 3 (the fact that one is revealed does not change that).
The strange thing is that "choosing to stay" somehow doesn't behave like a new 50/50 choice, but choosing to change does.
Intuition implies that given a random choice between a, b and c - chances are 1:3. And given a new choice a or b, chances are 50/50. But you only get to act on new information (make a new choice) if you change.
I must admit I'm still confused on this, and wonder if it bears out in simulation (that always change is better than random stay/change when offered to choose again
Ed: paultopia has a nice simulation and explanation below:
Significantly - staying means the player remains at 1:3, but since we now know the chosen door, or the alternate, holds the car - if staying is 1:3 wins, changing must account for the remaining 2:3 - since there are only two options. And random choice is also better than staying (it's 50/50) - but worse than changing.
> Intuition implies that given a random choice between a, b and c - chances are 1:3. And given a new choice a or b, chances are 50/50. But you only get to act on new information (make a new choice) if you change.
This is not the correct interpretation. In fact, the choice of getting the car if you switch is not 50/50, up from 1/3. It's 2/3. The choice of winning if you don't switch is 1/3.
The reason for this is that the second choice is not independent of the first choice, so you shouldn't model them as different choices.
Essentially the problem could be reformulated as: you choose one door at random. Then the host asks you if you would rather keep that door, or choose to get what is behind both of the other doors. The door opening is just a red herring.
The topic of the counter-intuitive nature of probability reminds of Newton's letter to Pepys - "In 1693, Isaac Newton answered a query from Samuel Pepys about a problem involving dice. Newton’s analysis is discussed and attention is drawn to an error he made."
It seems to me that a lot of what trips people up is that they don't realize that the host specifically opens a door which they know does not to contain a car (as opposed to selecting the door to open randomly).
Therefore, it seems to me that many people fall for it not so much because they misunderstand the probability but because the rules of the game are designed to be misleading.
For a more prosaic use of code to help people understand, I once wrote a quick simulation to help get it into the intuitions of my students: https://paul-gowder.com/montyhall/
I always wondered why it's not 50/50 if I enter the room late. How can a past event that now seems irrelevant change the odds.
Basically you watch Monte jump around and see which doors he avoids because they have prices. Now you can't make that observation about your own door because he'd never touch it anyways and he jumps just once but sometimes skipping a door if his random hits the price. The fact that it's just 3 doors so just one is left makes it even more quirky, but doesn't change much.
So you know the other door is a door monty pontentialy avoided not to reveal the price.You don't have that information about your door.
edit oops i should have refreshed before duplicating this thought !!!!
funny how different people grasp things.. this code example didn't really click with me, but someone else explained it by imagining 100 doors, not 3.
After you choose one door, 98 of the remaining 99 doors (all with goats behind them..) are opened.... leaving 1.
This note in particular could be a lesson for many things in life:
> When I believe the answer is 1/3, and I hear someone say the answer is 1/2, my response is not "You're wrong!", rather it is "How interesting! You must have a different interpretation of the problem; I should try to discover what your interpretation is, and why your answer is correct for your interpretation." The first step is to be more precise in my wording of the experiment...
Why do explanations of this problem never mention conditional probability, on which the explanation is based?
There are dozens of videos explaining conditional probability on youtube, but basically, the taking away of a door gives us additional information about the state of the system. It is counter-intuitive, but it's not a mystery.
This principle is used everywhere to optimize real-world problems.
for me it clicked when i realized he will only open the bad door which you have not chosen. Even if you chose the bad door he will never open that one to show you. He’ll always open the other
The simplest solution is you Always switch as it meant 2/3 instead of 1/3 winning chance. the door opening earlier or (clearer and easier to understand) later is just to help you to check your 2/3 pool.
If you take the "stay" strategy... 1/3 you hit the car and keep it. 2/3 you hit a goat and keep it.
In summary... "switch" is 2/3 car, "stay" is 2/3 goat.
In these times you should choose "stay" and hope to get a goat so you can turn grass into food. Cars are overrated.