The key is that the host is quite constrained in which box he can show you: He must always show you a box with a goat and he can never show you the box you initially picked. So the choice the host makes is less random than your (initial) choice and can give you new information.
In particular, only two different things can happen:
a) You initially picked the car. In that case, the host is free to pick any of the two remaining boxes since both have goats in it and neither were picked by you. In this case, it would be obviously unwise to switch.
b) You initially picked a goat. In that case, the host has no choice at all: They must pick the one box with the second goat, which leaves the last remaining box as the one with the car. So you should absolutely switch here.
If you knew whether you're in situation a) or situation b), you'd already have won the game: You could switch as needed and would always get the car.
Obviously though, you don't know. However you know that a) happens when you initially picked the car and b) when you didn't. You also know, you initially picked the car with 1/3 probability. So you know that with 1/3 probability you're in situation a) and with 2/3 probability you're in situation b).
Therefore, if you're just always pretend you're in situation b) and switch, you'll win the car with 2/3 probability.
If your initial guess is correct, then switching always loses.
If your initial guess is wrong, then switching always wins.
Everyone who understands the game can agree on those facts without controversy. If they don’t then there’s no point proceeding because they do not understand the rules of the game.
Therefore, the odds of winning when you switch are the inverse of the odds of winning from your initial guess.
If your initial guess wins 1/3 of the time, that means always switching loses 1/3 of the time and conversely always switching wins 2/3 of the time.
A slightly different reframing of this idea makes it even more clear to me:
When you make a choice, you divide the doors into two sets: the set containing your choice, and the set containing the other doors.
Obviously your set has a 1/3 chance of containing the prize, and the other set has a 2/3 chance.
When the host reveals a door in the second set, it has no effect on those initial probabilities. The set as a whole still has a 2/3 chance of containing the door, and now you have the choice of selecting the only item in that set which you know isn't wrong.
This reasoning is just as obvious when you scale up the doors too. With 100 doors, your set has a 1% chance of containing the prize, and the set of unchosen doors a 99% chance. All but one door in that set are revealed to be wrong, then you get a "50/50" choice of swapping to that last remaining item... of course that's what you want to do!
The one which made it clear for me was: imagine you have 4 doors instead. You choose one at random. The host then chooses two of the remaining doors he knows has goats. Is it in your best interest to switch to the one he didn't open?
Suppose you have 100 doors. You choose one at random. The host then chooses 98 of the remaining doors he knows has goats. Is it in your best interest to switch?
In all of these situations, the basic question is: what is more likely - that you chose the one with the car, or that you didn't, and the door remaining is only remaining because the host knew it was the one with the car?
Yeah, this is exactly the easiest way to think about the problem. In summary, by switching you win when you initially picked a goat, which happens two thirds of the time.
This is the first thing to get, the second is that Monty won't pick the car. If you do not understand these two, then you will say 1/2.
I've heard the analogy with 1 000 doors and opening 998 of them before but that doesn't explain anything at all. If you think that makes you understand then you're mistaken; that made you understand something else.
The key is that you have 1/3 of being right, 2/3 of being wrong. So in 2/3 of the cases, you will be offered to pick the car. So 2/3 switching gives you the car and 1/3 loses you the car. So always switch, it has better odds.
Any explanation that doesn't involve scaling up the number of doors to a arbitrarily large amount where the probability becomes obvious is really lacking if you ask me. The only reason this is so confusing in the first place is because it's only 3 doors.
With an infinite amount of doors, the chance of picking the right one at first is zero. When the host opens all doors except one other, it's almost certain that it's the door with the car.
It’s good for the aha intuition moment for some people. It’s the thing that made me get it as a kid. There’s something unsatisfying about it though. You can have 100 doors, and open 98 to leave the initial guess and 1 more. But this is scaling up the rule as “all but one”. I don’t see an obvious reason that the rule shouldn’t scale up as “open one more door”. Then you have 99 closed and 1 open and the result isn’t nearly as obvious, even though switching is the better move.
Like a lot of things in math, I think Monty Hall becomes easier to see when you look at some extreme cases. 3 doors is the lowest amount of doors where the paradox appears, ie no way to go smaller. So try a much bigger problem. I like this version:
After Sheherazade has told the calif 1001 stories, she tells him that (only) one of them is true, not fiction. She has him guess which one, and he picks one (say story #500). Then Sheherazade tells him that the true story is either the one he picked, or another one, say #312.
I think in this formulation, it is much easier to see that the calif would be well advised to switch his guess to #312, as his initial guess only had a 1/1001 chance of being correct. Monty Hall is the same problem with 3 instead of 1001, but the same principle holds.
Shameless plug, I am fascinated by Monty Hall and I had a hard time proving it to my friends so I made this: https://kr1stjans.github.io/monty-hall/ :D feedback appriciated :)
I haven't done the analysis, but one major assumption that the whole result is based on is whether the rules of the game are pre-established or just revealed after the first pick.
With an adversarial game host who has the option to reveal or not, maybe the result changes, but more importantly, it explains the intuitionistic refusal for some to buy the argument.
In an adversarial setup you basically get 1/3 wins like you started with because the optimal host move is to provide no information whether they reveal or not
In an adversarial setup, the optimal host move is to permit you to switch only if you picked the car. You still get 1/3 wins with optimal play (never switching), but it's better for the host because it allows people who think it's the traditional game to get zero wins.
If the host strategy isn't explicitly specified, it's reasonable to assume an adversarial setup, because seeing the player lose after switching would be more entertaining for the audience.
One thing to note is that 33% chance is still pretty good odds. Stats like that only really work with large numbers, and when you've got a single chance, neither of the outcomes is really surprising.
IOW, if every participant switched, the show would be handing out prizes 2/3rds of the time, but that 1/3rd of participants would still be going home empty handed. Do you want to be that one? If participants randomly chose whether to switch or stick, how often would the show hand prizes out?
Now we've got multiple conditional probabilities.
And that choice is what makes casinos, lotteries and games like these work: we want to succeed despite the odds.
My favourite casino game is craps. I find it amazing they created a game that has two basic strategies you can play and the one that has better odds is the "dark side" and is frowned upon to play that strategy.
Sometimes I'll lose a ton of money and a superstitious pass line bettor will make a quip about how I'm playing the dark side and deserve it for betting against the table. My usual reply is along the lines of: we're all playing against the casino and the dark side beats the casino more often than the pass line ;)
Craps is a social game, probably more so than any other table game in a casino. I find the minuscule odds advantage of playing Don’t Pass Line bets to be not worth “being the bad guy” at a table full of superstitious gamblers out having fun, but that’s just me. I’m sure the casinos love this artificial stigma.
I always wondered: how juicy would they have to make Dont Pass to overcome the social stigma at the table? Would it be enough to pay instead of push on 12? How much house edge would cause everyone to switch to the dark side? I have no doubt casinos have spent millions of man-hours of research to understand this.
EDIT: In addition, most craps games offer "odds" bets once a point is established which have zero house edge, so by taking the maximum odds, you reduce the house edge on Pass and Don't Pass to the point where it's not really probabilistically significant which side you play.
I like the variant:
You pick a door. An earthquake hits and one door opens that is a goat. The earthquake of course had know knowledge of that it's just what happened.
Should you switch?
I really thought I understood the problem until I got that variant wrong.
There's a good video with a rant about the problem I will link to it when I find it.
Yep, this is the key. The host knows where the prize is, and the rules of the game force him to use and reveal some of that knowledge when he chooses the door to open.
That rule is fundamentally what changes the odds, but it's worth noting that Monty only revealing goats was not one of the rules of the actual Monty Hall gameshow, nor was it explicitly stated in the original framing of this problem (though arguably it was implied). The somewhat ambiguous description of the problem in the newspaper column that popularized it may have contributed to the confusion.
Say the game has a million boxes. You pick one at random, the host opens the other 999,998 boxes (which all contain goats), and asks if you should switch.
At that point, it's very obvious you really should switch, as the original box has a one in a million chance of being correct, but the second box that is now unopened is wink wink nudge nudge maybe the one you want.
That's really a good way to visualise it. The host opens all the other boxes except your pick and one specific other box somewhere in the middle. It's clear that something special must be going on with that box.
The million box analogy isn't very clear IMO, because it assumes the host always opens all boxes except one. In the original scenario, "all except one" is identical to "exactly one", so "exactly one" might be the correct host behavior.
I see this posted a lot but it doesn't click with me. It only feels obvious once I know how the probabilities work - meaning I have to get the base case first.
Picture, in your mind or maybe draw on paper, 20 brightly lit doors in a row where 19 are labeled L and one is labeled W. Out of those 20 doors, one door gets randomly selected. It's almost certainly going to be an L. After choosing it, the lights go out above 18 of the other L doors so only the W door, and the L door you probably chose, remained lit.
If you stayed with your existing selection— one of two doors that remains lit— you've still almost certainly got an L door. If you switch to the one other door, with all of the other L doors eliminated, the only way it's not the W door is if you chose the W to begin with.
Here's a way to think of it that some people fine helpful. Assume N boxes, the host knows where the prize is, always reveals N-2 goats after the content picks leaving just the contestant's box and one other box unopened, and always gives the contestant a chance to switch after revealing the N-2 goats.
A person who does not switch wins if and only if their initial pick was correct. The probability of that is 1/N where N is the number of boxes.
A person who switches wins if and only if their initial pick was not correct. The probability of that is 1-1/N.
Switching is (1-1/N)/(1/N) = N-1 times as likely to win.
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The key is that the host is quite constrained in which box he can show you: He must always show you a box with a goat and he can never show you the box you initially picked. So the choice the host makes is less random than your (initial) choice and can give you new information.
In particular, only two different things can happen:
a) You initially picked the car. In that case, the host is free to pick any of the two remaining boxes since both have goats in it and neither were picked by you. In this case, it would be obviously unwise to switch.
b) You initially picked a goat. In that case, the host has no choice at all: They must pick the one box with the second goat, which leaves the last remaining box as the one with the car. So you should absolutely switch here.
If you knew whether you're in situation a) or situation b), you'd already have won the game: You could switch as needed and would always get the car.
Obviously though, you don't know. However you know that a) happens when you initially picked the car and b) when you didn't. You also know, you initially picked the car with 1/3 probability. So you know that with 1/3 probability you're in situation a) and with 2/3 probability you're in situation b).
Therefore, if you're just always pretend you're in situation b) and switch, you'll win the car with 2/3 probability.
If your initial guess is correct, then switching always loses.
If your initial guess is wrong, then switching always wins.
Everyone who understands the game can agree on those facts without controversy. If they don’t then there’s no point proceeding because they do not understand the rules of the game.
Therefore, the odds of winning when you switch are the inverse of the odds of winning from your initial guess.
If your initial guess wins 1/3 of the time, that means always switching loses 1/3 of the time and conversely always switching wins 2/3 of the time.
When you make a choice, you divide the doors into two sets: the set containing your choice, and the set containing the other doors. Obviously your set has a 1/3 chance of containing the prize, and the other set has a 2/3 chance.
When the host reveals a door in the second set, it has no effect on those initial probabilities. The set as a whole still has a 2/3 chance of containing the door, and now you have the choice of selecting the only item in that set which you know isn't wrong.
This reasoning is just as obvious when you scale up the doors too. With 100 doors, your set has a 1% chance of containing the prize, and the set of unchosen doors a 99% chance. All but one door in that set are revealed to be wrong, then you get a "50/50" choice of swapping to that last remaining item... of course that's what you want to do!
Suppose you have 100 doors. You choose one at random. The host then chooses 98 of the remaining doors he knows has goats. Is it in your best interest to switch?
In all of these situations, the basic question is: what is more likely - that you chose the one with the car, or that you didn't, and the door remaining is only remaining because the host knew it was the one with the car?
It's really easy to get people arguing about this one, even though the correct answer is so well documented.
I've heard the analogy with 1 000 doors and opening 998 of them before but that doesn't explain anything at all. If you think that makes you understand then you're mistaken; that made you understand something else.
The key is that you have 1/3 of being right, 2/3 of being wrong. So in 2/3 of the cases, you will be offered to pick the car. So 2/3 switching gives you the car and 1/3 loses you the car. So always switch, it has better odds.
Deleted Comment
With an infinite amount of doors, the chance of picking the right one at first is zero. When the host opens all doors except one other, it's almost certain that it's the door with the car.
After Sheherazade has told the calif 1001 stories, she tells him that (only) one of them is true, not fiction. She has him guess which one, and he picks one (say story #500). Then Sheherazade tells him that the true story is either the one he picked, or another one, say #312.
I think in this formulation, it is much easier to see that the calif would be well advised to switch his guess to #312, as his initial guess only had a 1/1001 chance of being correct. Monty Hall is the same problem with 3 instead of 1001, but the same principle holds.
If the host strategy isn't explicitly specified, it's reasonable to assume an adversarial setup, because seeing the player lose after switching would be more entertaining for the audience.
Dead Comment
IOW, if every participant switched, the show would be handing out prizes 2/3rds of the time, but that 1/3rd of participants would still be going home empty handed. Do you want to be that one? If participants randomly chose whether to switch or stick, how often would the show hand prizes out?
Now we've got multiple conditional probabilities.
And that choice is what makes casinos, lotteries and games like these work: we want to succeed despite the odds.
Sometimes I'll lose a ton of money and a superstitious pass line bettor will make a quip about how I'm playing the dark side and deserve it for betting against the table. My usual reply is along the lines of: we're all playing against the casino and the dark side beats the casino more often than the pass line ;)
I always wondered: how juicy would they have to make Dont Pass to overcome the social stigma at the table? Would it be enough to pay instead of push on 12? How much house edge would cause everyone to switch to the dark side? I have no doubt casinos have spent millions of man-hours of research to understand this.
EDIT: In addition, most craps games offer "odds" bets once a point is established which have zero house edge, so by taking the maximum odds, you reduce the house edge on Pass and Don't Pass to the point where it's not really probabilistically significant which side you play.
Should you switch?
I really thought I understood the problem until I got that variant wrong.
There's a good video with a rant about the problem I will link to it when I find it.
Why not? There is no harm in doing so if the probabilities are the same anyway...
Say the game has a million boxes. You pick one at random, the host opens the other 999,998 boxes (which all contain goats), and asks if you should switch.
At that point, it's very obvious you really should switch, as the original box has a one in a million chance of being correct, but the second box that is now unopened is wink wink nudge nudge maybe the one you want.
If you stayed with your existing selection— one of two doors that remains lit— you've still almost certainly got an L door. If you switch to the one other door, with all of the other L doors eliminated, the only way it's not the W door is if you chose the W to begin with.
A person who does not switch wins if and only if their initial pick was correct. The probability of that is 1/N where N is the number of boxes.
A person who switches wins if and only if their initial pick was not correct. The probability of that is 1-1/N.
Switching is (1-1/N)/(1/N) = N-1 times as likely to win.