The Intuitive Monty Hall Problem

Many puzzles are difficult to solve from one perspective but easy from another. A challenge on stackexchange was to find an equivalent version of the Monty Hall problem where the correct solution of switching is obvious. Joshua B. Miller has an excellent answer. To recall, in the original there is a great prize hidden behind one of three doors. You choose a door. Monty Hall then reveals a lousy prize behind one of the other two doors (it’s always a lousy prize). Do you switch doors? Most people see no reason to switch. Even Paul Erdos was a no switcher! Moreover, most of those who do switch get to that conclusion with an unintuitive Bayesian calculation.

Here’s the intuitive version.

There are three boxers. Two of the boxers are evenly matched (no draws!); the other boxer will beat either them, always.

You blindly guess that Boxer A is the best and let the other two fight.

Boxer B beats Boxer C.

Do you want to stick with Boxer A in a match-up with Boxer B, or do you want to switch?

See also Miller’s new piece in the JEP which looks at the Monty Hall problem and the Hot hand puzzle.

Comments

I still prefer a simulation. Something about seeing it happen before your eyes made it click for me. No amount of explaining (Bayes nor not) worked for me but this did.

https://rpsychologist.com/monty-hall-simulation

Once it was explained to me once I found it intuitive. But that just makes me think that I don't actually understand it, since people much smarter than me argued against it for years. (Then again at that point I also had the knowledge that simulations had proved it correct.)

No, you understand it. It's not that hard. It's just that some people aren't as smart as they make themselves out to be.

The simulation is definitely much more convincing. I've learned more statistics from simulating data than probably any other technique, include math.

Simulation is analagous using a calculator rather than doing long division in your head or on paper. Perfectly valid and efficient of the model is correct.

I had to code my own version before I (barely) believed it.

I found most helpful the Straight Dope guy's explanation, per wikipedia, that since he will never reveal the goat Monty is basically giving you not one but two doors when he asks if you want to switch.

Also, Marilyn's changing the problem to a million doors instead of three. (Monty giving you not one but 999,999 doors.)

Having successfully wrapped my mind around it that way, the boxer wording isn't helping me at the moment.

Apologies, never reveal the car. I forgot for a second cars are cooler than goats.

Yep. I never read that answer but did the same thing when explaining it to friends but used a bomb going off in one of a hundred rooms.

Oops. I meant 98 bombs going off in a hotel with 100 rooms.

Something about seeing it happen before your eyes made it click for me.

Same. I put three playing cards on the table and was about to do a bunch of trial runs, but literally the second time I turned over the cards, I *saw* what Devon describes below: when you switch, you trade your original card/door for a pair.

I thought the intuitive Monty Hall was that there were 100 doors and 1 has the car, picked at random. You pick one. You get an offer to switch to the other 99 doors. Do you want to switch?

I think you mean:

You pick one. Then a door with no prize is opened. You get an offer to switch to the other 98 doors. Do you want to switch?

You pick one. Then 98 doors with no prize are opened. You get an offer to switch to the one other remaining door. Do you want to switch?

That's the explanation that works best for me. If someone can see 98 doors that are opened (showing goats), and they see the last two doors that are closed are the one they picked and one other, your mind almost demands that the car is behind the *other* closed door.

I always think of the choice to switch as equivalent to the following: You choose door 1, then Monty gives you the option of keeping your door number 1 choice or the choice to open 2 and 3 and then choose between 2 and 3.

In other words, either 1, or the better of 2 and 3. To me this seems obvious and intuitive.

+1, You're given a choice of a 1/3rd chance of being right. Then you're given a choice of switching to a 2/3rds chance of being right.

Choose...

no, thats not right. No matter what you choose first, the host eliminates a wrong choice, therefore you only have 2 doors, one of which has the prize. so really you go from 1/3rd chance to 1/2 chance. Switching or not is no different than a coin flip.

"so really you go from 1/3rd chance to 1/2 chance."

No, That's the common (intuitive) mistake about the Monty Hall problem.

You have 1 door, Monty has 2 doors. You have a 1/3rd chance of having the correct door, he has a 2/3rds chance of having the correct door.

If he didn't open a door, but offered you a chance of sticking with your door Or choosing to open both his doors then what would you choose? Obviously you would switch.

When he opens one of his doors, he always picks one that doesn't have the prize. But the odds don't change. His pair of doors still has a 2/3rds chance of being the correct choice and since he's shown you the one that can't be correct, the remaining door is the 2/3rds door.

This, this, a thousand times, this.
Always switch. Always.

Look at iit from boxer B point of view. He just whooped on C, and is fighting an untested A for the championship. The betting is on B.

The boxer analogy is not apt and is no more "intuitive" than the original. The point of Monty Hall is that eliminating door C tells you (or seems to tell you) nothing about doors A and B. The boxer analogy as given above tells you something about both boxer B and boxer C which violates the "Let's Make a Deal" rules. A correct analogy to the original would tell you only that boxer C fought one of the other two and lost. Do you switch then? Is that more intuitive?

In the alternative, you pick boxer A initially but the first match is between A and C. A wins, do you switch then? The correct answer is yes but is that intuitive?

Anybody who uses the boxer example clearly doesn't understand. It is not at all analogous to the Monty Hall problem.

Agree. It is not analogous, it is basically saying there are three boxers A,B,C who are ranked 1,2,3 in skill levels.

That's right, because the doors don't fight or engage in any sort of contest where the outcome is uncertain. The goat is behind one, and the car behind the other, with certainty before the game starts, Monty knows which is which, and the only uncertainty is that faced by the contestant.

But if Boxer B beats C he must be very tired and will lose to Boxer A who didn't fight anybody before. I stick with Boxer A. But I'm a cuck what do I know.

I don't find this any more intuitive.

As stated, I think it is more intuitive but only because the analogy cheats the problem. One is pointed toward the correct answer by the apparent information about boxer B which is supplied after the initial choice of A. If you don't know that B won but only that C lost, or if you are told that A beat C the "intuitiveness" disappears.

My effort to make it more intuitive would go like this: you make your choice. Then Monty explains that because of Calif. animal welfare laws the show can't keep a goat on set all day anymore. So he sez, like I used to open one of the remaining doors, always reveal a goat, but now, well, let's not and say we did. Anyway, do you want to switch, from your door to the two doors?

This would also explain why a lot of TV shows that should have a dog around, for verisimilitude, no longer do.

Used to bug me about Justified, an otherwise perfect show.

I agree. Apparently this set-up might nudge someone to the right answer for the wrong reason. So, what's the reason?

It's not clear from the wording whether Monty Hall deliberately opened a door with a lousy prize, or whether he happened to open a door with a lousy prize. The boxing analogy only works if he deliberately opens a door with a lousy prize. If he selected between the two other doors at random, then there'd be no reason to switch, no?

"then there'd be no reason to switch, no?"

Yes, you would switch. You know that there was a 2/3rds chance the prize was in the 2/3rds of the doors. Did he open a door with the prize? If not then there's a 2/3rds chance it's in the one he didn't open.

Sorry, on second thought I'm wrong. You have to assume that Monty Hall knows and is intentionally showing you a door without a prize.

If he's doing it randomly then 1/3rd of the time he'll show a prize, and the other times, the prize will have the same odds of being behind either door and switching doesn't matter.

"You have to assume that Monty Hall knows and is intentionally showing you a door without a prize." That's why, in my experience, most problems that purport to be mathematical problems about probabilities are bogus.

They simply don't provide all the info you need such that it really is purely a maths problem. In the context of problems used in teaching, your examiner is, too often, effectively asking you to guess what extra information he'd prefer you to assume. That's a moronic way to teach the subject.

It's especially moronic if the examination candidate is likely to make a better guess at the omitted data that his examiner is.

On the one hand, yes and that's why so many people had so much difficulty with this puzzle -- the ambiguous wording.

But if you grew up in the US in the 1970s (and probably 1980s, maybe 1990s) then unless you grew up under a rock you knew what Monty Hall's "strategy" was: he would always deliberately pick one of the goat doors and reveal it.

So yes, non-US or non-1970s readers will have a lot more problems with this puzzle, or even say with some validity that it's a poorly worded question. But most Americans of a certain age don't have that excuse. (It's still a difficult problem for most of us to figure out, at least initially, because it does conceal its counter-intuitive nature. As the comments show, people have a wide range of ways of eventually wrapping their heads around the puzzle. )

It's an all-time great brain puzzler, but perhaps it's best for Americans of a certain age. Others will have to have the puzzle laboriously explained to them including Monty Hall's strategy.

No, in a truly equivalent version you don't get to know that Boxer B wins but merely that Boxer C does NOT win and on the premise that ties often occur.

I never understood the fuss about the Monty Hall problem. It's a basic probability problem.

Most people dont know and think in terms of probability.

There are many books and entire college courses on cognitive and behavioral biases in investing. There are long lists and descriptions of logical fallacies that seemingly smart and educated people make every day by the tens of millions.

Yet these biases and fallacies are widely resistant to education. Dunning Krueger effect. You are smarter than you give yourself credit for, and many others are far stupider than they know.

The biases and heuristics literature may have overstated the difficulties in the way of improving decision-making capabilities via training.

See for example:
Gigerenzer, G. (2018). The bias bias in behavioral economics. Review of Behavioral Economics, 5(3-4), 303-336.

Other articles by Gigerenzer are worth a look as well. Many available at google scholar.

It's not exactly a basic probability problem. Or at least, most of the smarter non-switchers get there because they are applying basic probability incorrectly.

Wrong logic: Doors have equal probability. If you open one, the remaining doors still have equal probability, which is now 50/50.

Right logic: The door you picked has a 1/3rd chance of being correct. Opening one of the other doors DOES NOT CHANGE THAT.

Hell, I've understood the right way of thinking about this for years, and finding the right words to talk about it is still hard...

The following is based on the book ‘Das Ziegenproblem‘ by Gero von Randow. The book was written in 1992.

Setup:

- There are 3 doors.
- 1 door has a car behind it.
- 2 doors have a goat behind it.
- The content behind the doors is randomly chosen.
- There are 2 candidates, A and B.
- There is 1 moderator who knows which door has a car behind it.

Actions:

- A and B are asked to choose a door and both choose the same door.
- The moderator chooses one door which has a goat behind it.
- A and B are asked if they would like to switch their choice and pick the remaining door.
- A always stays with his choice, the door that has been initially chosen by both A and B.
- B always changes her choice to the remaining third door.

(Scenario 1) Repeat the actions 999 times:

If you repeat the above list of actions 999 times, given the same setup, what will happen?

Candidate A always stays with his initial choice. Which means that he will on average win 1/3 of all games. He will win 1/3*999, 333 cars.

But who won the remaining 666 cars?

Given the setup of the game, the moderator has to choose a door with a goat behind it. Therefore the moderator does win 0 cars.

Candidate B, who always switched her choice, after the moderator picked a door with a goat behind it, must have won the remaining 666 cars (2/3*999)!

(Scenario 2 ) 1 candidate and 100 doors:

Alter the above setup of the game in the following way:

- There are 100 doors.
- 1 door has a car behind it.
- 99 doors have a goat behind it.
- There is 1 candidate, A.

Alter the above actions in the following way:

- The moderator opens 98 doors with goats behind them.

Now let’s say the candidate picks door number 8. By rule of the game the moderator now has to open 98 of the remaining 99 doors behind which there is no car.

Afterwards there is only one door left besides door 8 that the candidate has chosen.

You would probably switch your choice to the remaining door now. If so, the same should be the case with only 3 doors!

Further explanation:

Your chance of picking the car with your initial choice is 1/3 but your chance of choosing a door with a goat behind it, at the beginning, is 2/3. Thus on average, 2/3 of times that you are playing this game you’ll pick a goat at first go. That also means that 2/3 of times that you are playing this game, and by definition pick a goat, the moderator will have to pick the only remaining goat. Because given the laws of the game the moderator knows where the car is and is only allowed to open a door with a goat in it.

What does that mean?

On average, at first go, you pick a goat 2/3 of the time and hence the moderator is forced to pick the remaining goat 2/3 of the time. That means 2/3 of the time there is no goat left, only the car is left behind the remaining door. Therefore 2/3 of the time the remaining door has the car. Which makes switching the winning strategy.

Even before you pick your door, you KNOW Monty is going to show you a door with a booby prize after you pick.

When you pick your door, you KNOW you have a 1/3 chance of being correct.

So, when Monty shows you the door with a booby prize, this adds NO INFORMATION with respect to the door you chose.

This is correct.

The initial probabilities are partitioned as 1/3 for A and 2/3 for NOT A, or 2/3 for Door B or Door C (NOT A contains Doors B and C). If you choose A and then a door is opened, you learn that it is NOT Door B or Door C, but since the initial partitioning hasnt changed, you now know all 2/3 of NOT Door A 'belong to' the unopened door. You switch.

The best way to illustrate this logic is by using 100 doors and opening all of the remaining doors but one

This question is not actually hard - trivial even - if you think visually as a default.

Monty gets a 2/3rd chance of winning, you get a 1/3rd chance of winning. He offers to switch positions and thus the chance of winning. Do you switch?

That analogy is absolutely awful. The crux of Monty Hall is that you DON'T KNOW if you have Boxer A or not.

By staying (unknowingly) with Boxer B, you dont "lose," you just win less than you could have. You're still guaranteed a win. By sticking, the player might be minimizing maximum regret rather than maximizing expected win. It would be easy enough to ask people whether they want to stick (and why) before and after telling them the expected wins from sticking and switching.

This is exactly right. MR is where horrible analogies go to get celebrated.

The way to explain Monty Hall is just to change it to 1 million doors. Pick one at random, the host opens every other door except one. Do you switch? Of course you switch.

I'm a bit dense on the million/hundred door explanation. Why is it obvious you switch in those scenarios?

You pick 1 door out of 100.

Monty opens up 98 doors, all containing goat, he leaves one door closed. Most people at that point realize that, the door he left close, probably is the one with the Car.

It's the same probability problem, except in this case, there is a 99% chance that Monty's door has the car and only a 1% chance your original door is hiding it.

Because your mistaken intuition that you're just as well off sticking to your first choice in the face of one other unopened door - that your chances seem "equal" - falls apart when there are 98 other doors behind which the car can be and Monty is "giving" you all of them if you switch.

That wasn't the challenge as posed on stackexchange -- the question asked for intuition generally, and it was a creative idea for someone to respond in this way.

I like to explain this in two steps:

Step #1:
* There are three doors.
* There is a goat behind two doors, and a car behind one door.
* You pick one door.
* The announcer asks: Do want to keep your door, or switch to both doors.

Step #2:
* There are three doors.
* There is a goat behind two doors, and a car behind one door.
* You pick one door.
* The announcer shows you that one of the other doors has a goat (this gives you no information!)
* The announcer asks: Do want to keep your door, or switch to both doors.

"* ...(this gives you no information!)"

That's incorrect. This absolutely gives you the critical information you need to make your odds better.

If the announcer didn't show you the goat and offered you a chance to switch blindly to 1 of the 2 other doors, it wouldn't improve your odds. You'd be switching from a 1/3rd chance to another 1/3rd chance.

The two other doors have a combined 2/3rds chance of being correct. When the goat is revealed, you now have narrowed the 2/3rds chance down to 1 door. Thus you should switch from your 1/3rd chance door to the 2/3rds chance door.

No, it doesn't give you any information relative to the scenario I described. Notice that you are still being offered the option of switching from your ONE door to the other TWO doors. You already knew that one of them contained a goat.

I think you are (reasonably enough) extrapolating to

Step #3:
* There are three doors.
* There is a goat behind two doors, and a car behind one door.
* You pick one door.
* The announcer shows you that one of the other doors has a goat.
* The announcer asks: Do want to keep your door, or switch to the other, non-goat door.

In this situation, your comments would be correct.

Ah, I see your point.

You don't gain any useful information in your Step 2, because regardless of the goat door, you get to switch to both. He might identify which door has a goat, but since you get both doors, that's useless information.'

You already knew that one of them contained a goat.

But you didn't know WHICH one. So after the revelation, you DO have new and useful information. Before the revelation, the doors not chosen have a 2/3 chance of containing the prize, but switching doesn't help me because I have no additional information, so between those two doors, which are equal to me in my state of knowledge before the revelation, the probability is 2/3*1/2 = 1/3. The revelation provides new and *important* information because before, the doors not chosen had a 50-50 chance between them of having the prize, but after, one door's probability goes up to 100% and the other drops to 0% between those doors *if the doors not chosen contains the prize*. The fact that I needed the revelation to change the probabilities between the two doors tells me that the information was useful. So IF the prize is behind the doors not chosen, I now know that it is DEFINITELY going to be behind that other door (this knowledge did not exist at the beginning. The probability that the prize is behind the unchosen doors is still 2/3, but there is only one unchosen door left. So the probability is 2/3 that the prize is behind that door.

To beat this dead horse:
Before revelation:
Probability that one of the unchosen doors contains prize = 2/3*1/2 = 1/3
Probability that other of the unchosen doors contains prize = 2/3*1/2 =1/3
Probability that chosen door contains prize = 1/3

After revelation:
Probability that one of the unchosen doors contains prize = 2/3*1 = 2/3
Probability that other of the unchosen doors contains prize, revealed to not contain the prize = 2/3*0 =0
Probability that chosen door contains prize = 1/3

In any case, RatInPutinsMaze's explanation is correct, and the boxing analogy is incorrect for reasons already stated in the comments.

Your comments address probabilities that *each* of the unopened doors might contain a car/goat. These probabilities are completely irrelevant to the second scenario I described, since you switch to *both* of the unopened doors. (This is the point of this different framing.)

How about a no-regret switching strategy?

Always switch.

If its a rigged game (the classic Monty Hall problem), you are given 2/3 odds of winning if you switch and 1/3 if you stay.

If its an honest game (announcer *happens* to open a door with a lousy prize), then you dont care if you switch or stay.

Thus, regardless of whether the game is honest or rigged, switching is *at least as good* as sticking.

i think you are correct, but the odds are wrong. First guess is 1/3rd chance, 2nd guess is 1/2 chance, not 2/3rds chance, i think...

Yes, but the probability of your chosen door has also risen to 50%, so chances are greater that you've won, but switching doesn't help.

A more intuitive version closer to the original would simply be to pick one of three doors then be given the choice to switch to both of the other two. I'm sure most people would switch. For some reason the additional step of first eliminating one of the remaining boxes causes people to lose sight of the basic probabilities involved.

+1, This is the best explanation I've seen.

The probability is your scenario doesn't change because no new information is given. The host needs to open a door or else switching gives the exact same probability as not switching.

That's my point. If you choose 1 of 3 equally likely options then there is a 2/3 chance you are wrong and in the context of MontyHall the only 'new information' that comes along is that one of the non-chosen options is a losing choice. But that does change those 2/3 odds that your initial choice was wrong.

However the way that new information is framed seems to change the likelihood that people will intuitively make the optimal choice.

If one of the non-chosen options is presented as being passively removed by the contest organizers then people seem to intuitively (but wrongly ) think that leaves the remaining odds at 50/50.

But if one of the non-chosen options is presented as remaining because it is superior in some way to the removed option then people will tend to want to back the proven winner and think the odds are greater than 50/50 in its favor. I think they may be getting the right result but for the wrong reason.

I’ve read all the comments and watched a couple of YouTube videos and people still don’t get it.

Because you have the option to switch after a door is revealed, the first “pick” is irrelevant. The contestant is never choosing between 3 doors because the contest never ends after the first pick.

People using probability always assume that there really are 2 goats and 1 car behind the doors. In reality, there is 1 car, 1 goat and 1 door that will be revealed as a goat after the contestant chooses.

Once the “door to be revealed as a goat” is shown, the contestant is then given the real contest of choosing between a car and a goat.

It would be no different than if the contest started with one door with a goat already opened. Or a giant sign on the outside of the door that said “a goat is here”.

In the boxing analogy above, the loser of the first fight can never be selected. Therefore your option is always be between the other two boxers.

If the contest always started with two doors and a half-door with a goat peeking out of it, I don't think it would be the same. How would you get the 2/3 chance of winning if there were no switching step?

You never had nor will you ever have a 2/3 chance of winning because you’re not allowed to choose the goat behind the half door.

At the end of the game show “Ellen’s Game of Games” the contestant is shown ten celebrities and they have to name all ten. The contestant is given 3 options to choose from. It’s usually something like “Adorable Actors, Blonde Bombshells or 18th Century French Poets”. The French Poet topic is not really an option and is just there for giggles.

It is the same in the Monty Hall Problem. You are only choosing between the car or the goat. You are never allowed to select the other goat that will be revealed.

Forget the goats. There are three doors, one has a car. The other two have nothing behind them. You pick one. The door you have chosen therefore has a 1/3 chance of having the car behind it. The other two doors collectively have a 2/3 chance of having the car. With me so far?

Without eliminating any doors, you're asked if you want to switch from your one door, to BOTH other doors. Obviously you should switch, because as we established, there is a 2/3 chance that the car is behind one of those two doors. So you switch and you now have two doors. Since there's only one car, you know for certain that at least one of your two doors has nothing behind it. Monty Hall opens one of your two doors he knows is empty, just for drama. This of course tells you no new information. You already knew at least one of your doors had to be empty! Revealing one empty door tells you nothing! Monty Hall is just trying to confuse you! He asks if you want to switch back to your original choice. You say "Fuck you, Monty Hall! Why would I go back to 1/3 odds when I currently have 2/3 odds?" You punch him in the face and drive home in your brand new two-thirds of a car.

"It would be no different than if the contest started with one door with a goat already opened. Or a giant sign on the outside of the door that said 'a goat is here'."

There are three doors. One of them has car behind it. One of them has nothing behind it. One of them has a sign that says "a goat is here" and a goat peeking out, winking and razzing at you. You are asked to select a door.

You select one of the two non-goat doors, because you are not a deranged maniac. Monty Hall then opens whichever door has nothing behind it and asks if you want to switch. You are now 100% certain where the car is. You drive home in your brand new car, with your new goat bride in the passenger seat. Because actually you are a deranged maniac.

You’re almost there with the deranged maniac thought process. However, no matter which other door you choose Monty ALWAYS opens the one with the goat who was razzing you. You would never select the door with the goat AND Monty was never going to reveal any other door but the goat.

Try creating 3 post cards. Write GoatA, GoatB and Car on the 3 different cards .

Step 1)
Randomize and pick the first 1 of the top.
There are 3 options.

1) Your pick says Car, the other 2 are Goats
2) Your pick says Goat A, the other 2 are Car and Goat B
3) Your pick says Goat B, the other 2 are Car and Goat A

Step 2)
Throw away 1 of the two cards that you didn't pick that says Goat

There are still 3 options

1) Your pick says Car, the remaining card says Goat
2) Your pick says Goat A, the remaining card says Car
3) Your pick says Goat B, the remaining card says Car

How often is your pick the correct answer?

Ok one more question. What is the probability that you will choose the card with the goat AND that card will be the one revealed?

It's not one of the 3 options. So 0/3rds = Zero.

IE,
2) If you pick Goat A, Goat B will be discarded & the remaining card says Car
3) If you pick Goat B, Goat A will be discarded & the remaining card says Car

Instead of a prize placed behind 1 of 3 doors, imagine it is placed behind 1 of 1 million doors. You pick a door at random. Then every single door is opened except for the door you randomly selected and one other door. That is, there are now 99,998 open doors and two closed doors, one of which you chose at random. Do you want to switch? How about the same scenario with 100 doors? 20? 3?

If, going in, you decide that you will switch, then: you'll win the prize if you pick the door with one of the goats; and you'll win a goat if you pick the door with the prize. Two of the doors have goats, one has a prize, therefore the "switch" strategy was a 2/3 chance of winning the prize. Hard to be more compelling, simple or intuitive than this.

Sheesh, that isn't any more intuitive than the original problem! The only improvement I have ever seen on getting people to understand the reasoning behind changing your choice is to expand it well beyond 3 doors to something over 10, and for a lot of people, even that doesn't work (I have had both successes and failures doing so).

For those of you who still think switching gives you an advantage, I will provide an additional scenario.

This one is in honor of the Cuckmeister.

Your wife has taken on a black lover. After you learn of this, she tells you she will stop seeing him if you can guess who it is.

She gives you 3 pictures. One is Mark, another Mike and the third is your neighbor Ted. This seems strange to you. Ted is a slightly balding, pudgy white guy. You’ve never seen Ted talk to women much less act like he’s able to bed your wife. And then you remember that she already told you her lover was black.

So you ask her about Ted. Her response is that she will remove Ted’s picture from the list of options after you choose from Mike or Mark.

Still slightly confused, you make your selection. She then removes Ted’s picture like she said and asks if you would like to keep your original pick or switch.

Anyways your choice is irrelevant because she had been bedding both Mike and Mark and she wasn’t going to stop. She just wanted you to know.

Did you win?

I understand the Monty Hall problem insofar as I can provide the correct answer if asked. But perhaps someone more knowledgeable can enlighten me about a problem I've always had with it: imagine the contestant chooses a door (call it Door A) and is shown the "bad" prize behind Door B. But instead of choosing whether or not to switch, they walk out of the room and an entirely new person walks in. They are then given the choice of Door A or C.

My understanding is that the new contestant now has a 50/50 chance of winning on either A or C... whereas obviously the old contestant should have switched to door C (66% probability of being a prize). I've always been confused by the idea that the probabilities here are situational and depend on the person doing the choosing. Can anyone explain this to me?

I assume that in your story the new person doesn't know that the first person picked door A. The first person, of course, does know that. That is the difference, and it is an important difference.

This isn't a quantum physics problem. The Car doesn't move around on it's own.

There would still be a 2/3rds chance the car is behind Door C.

Sure, we're mostly stupid compared to Marilyn et al, but I think our brains are wired for 50/50. Good/bad.

When I was a child I had a phobia of public restrooms and did not use them. However, there was an exception. My cousin and I were sometimes in tow when my grandmother did her banking at a gold-domed building in OKC that had, as I recall, pictures on the doors of the men's and women's lavatories, rather than words. I believe those pictures were supposed to be silhouettes of a gentleman and a lady riding bicycles in gay nineties clothing. Gilded age, the gold roof, I just got that, ha. Anyway, I had difficulty telling them apart. I didn't feel completely sure. Had I merely followed my usual stubborn inclination, I might have avoided the question - the lady, or the tiger? - entirely. But something about that tantalizing 50/50 chance that I would get it wrong, perhaps egged on by my twin cousin, induced me to go and face my fate.

I’ll try once again because I’m bored.

You and I are going to bet on a coin flip. Heads I win, tails you do. How many consecutive times would you have to lose before you realized I was tossing a double - headed coin?

Every one keeps assuming you are choosing from 3 doors. You’re not. You never, ever get to pick the door that will reveal the goat. Just like tails will never come up on a two-headed coin.

You never ever get to pick ONE OF THE DOORS out of two goat doors out of three total doors that will reveal the goat. A 2 sided coin is not a good analogy.

"You never, ever get to pick the door that will reveal the goat. "
Why are you so caught up on the location of the goat? So they only have one goat and they move it depending on which door you choose so that they can do a dramatic goat reveal without buying two goats for their show. Big deal. Imagine there are no goats, just three doors and one car.

You pick one of the three doors. You can pick any one of the three. There are no mysterious goats forcing your choice. Then, *after* you have chosen, the host eliminates one of the remaining bad doors. He can do this because there will always be at least one remaining bad door after you have made your choice. The particular door he reveals was not predetermined from the start of the game, but rather is based on which door you chose. The host deliberately chooses to reveal one door with the conditions that 1) it has no car behind it, and 2) you did not choose it. So it is not like there was always a door labelled "bad door". You could have picked that particular bad door from the start. Then the host would simply have revealed a different door.

You chose one of three doors. The car has a 1/3 probability of being behind that door you chose. This probability never changes throughout the game.

What about the explanation where there's 100 doors and still only one car? You choose one door (1/100 chance of being right) and then the host eliminates 98 bad doors, leaving only two doors left. Do you really think there's suddenly a 50/50 chance you picked the right door?

What's interesting is that if Monty Haul randomly picks one of the two remaining doors, he has a 1 in 3 chance of picking the car (0% chance if you already picked the car, 50% chance if you have not, the former happens 1/3 of the time, the latter, 2/3 of the time). If he picks the car, obviously you lose. But if he doesn't, then you truly have a 50-50 shot at the car, no matter which remaining door you choose to go with.

The key to the problem is that Monty Haul always reveals a known negative result. The fact that he will never pick the car is what prevents your initial odds from ever getting better than 1/3.

I used to think this was a theoretical problem with no real-world significance. Turns out that when you check "Notify me of follow-up comments by email", you get Door #3 which---instead of a goat---contains an endless stream of email notifications with contrived "simplifications" that somehow turn a relatively easy problem into an impenetrable nest of wire-crossed logic. ;-)

Remember, if you start by picking a door with a goat, you'll always get the car if you switch after Monte shows you the door with the other goat. The chance of picking a door with a goat is 2/3. It really is that simple folks.

I like this one (that second paragraph). +5 internet points.

Waitbutwhy has a good explanation, using Jelly Beans.
https://waitbutwhy.com/2016/03/the-jellybean-problem.html

Interestingly intuition is correct if you play the game in reverse:

Monty gives you two boxes. One of them has a car inside, one of them has a goat inside. You pick one, then Monty adds another goat box & shuffles the two boxes you didn't pick.

Do you switch?

Did Monty Hall always offer his contestants the chance to switch doors? Or did he do so sometimes, and sometimes not?

Sometimes he did, usually he didn't, the door-switching option was to raise the excitement and tension level. IIRC he'd do it at least once per show, maybe even a few times. Maybe he gradually did it more often but I stopped watching the show by then.

Geez, the Monty Hall problem has been solved for decades yet AFAICT all of the comments are about it and not about the meat of the paper.

I'm not familiar with the term "principle of restricted choice" but I'm certainly familiar with the phenomenon. My reaction is that this seems like a good phrase to describe a phenomenon that we frequently encounter, sometimes in hidden form as with the Monty Hall problem.

It's perhaps even more hidden in the hot hand and shooting streaks topic. There's a lot more research and development of approaches to testing the hot hand hypothesis that the authors could've described. But I admit they had to make an editorial decision about which topics to describe just briefly.

AFAIK, economists don't use the phrase "principle of restricted choice" but this paper argues that they should, and that seems like a good idea to me. I wonder what statisticians think? And perhaps mathematicians in general, but I think the principle is of more use and interest to people who work with data. Including people who aren't even statisticians or economists, but who simply want to look at sports statistics. (Or who play Let's Make a Deal.)

The problem with the Monty Hall problem is that it is often incompletely expressed, and there are reasonable ways to "fill in the blanks" that can create quite different optimum solutions.

The standard solution works, IF we know for certain that Monty will always offer the switch. But that stipulation is seldom clarified, including in the brief outline of the problem here.

Another possibility is that Monty will ONLY offer the switch, if you've picked the correct curtain to begin with. In that case, switching (if given the option) is bad 100% of the time. And this possibility is quite reasonable - it's analagous to a salesman trying hard to get you to switch from your preferred item to a different, which offers him a bigger profit margin/commission (generally at the expense of less consumer surplus for you on the transaction).

There are other possibilities that can create other solutions.

Good point: you are not normally told that he will always offer a switch.

And, a perhaps more plausible version of the boxer counter-example is a footrace. Three high school athletes of similar builds, ages, etc. will be racing 50M. The first two contestants are randomly chosen. A races against B (and wins). A sufficient time (say, 30-60 minutes) is given for A to recuperate, then A will race C. Without the extraordinary stipulation that we knew in advance that one racer is 100% dominant, it's reasonable to assume A (with rest) will beat C.

In my previous comment, "it's reasonable to assume A (with rest) will beat C.", should be rephrased as "A (with rest) has a >50% likelihood of beating C"

Why I'm having trouble with "switch:"

I pick A, so 1/3 chance
Monte shows the goat behind C
So you're advising I switch to B so my chance to win the car goes up from 1/3 to 2/3

But I could just as easily have picked B to start.
Monte shows the goat behind C
So now I should switch to A to hike my chances from 1/3 to 2/3

But A and B both cannot have 2/3 odds -- that's 4/3.
As I see it, they both now have 1/2 odds, adding up to 1 as they now must. So no particular reason to switch.

I lay out the 3 options above.

Basically, you have a 1/3rd chance of being right with your 1 door, Monty has the other 2 doors and thus has a 2/3rds chance of being correct.

1/3 (your door) vs 1/3 (Monty door A) + 1/3 (Monty door B)

If you stop right there, what are the odds of you winning the car? answer: 1/3rd

What are the odds Monty keeps the car?
answer: 2/3rds

Monty looks behind his two doors and then opens the Door that doesn't have the car, so

1/3 (your door) vs 0/3 (open door) + 2/3 (Monty door B) OR
1/3 (your door) vs 2/3 (Monty door A) + 0/3 (open door)

Assume Monty doesn't offer you a chance to Switch:

What is your chance of winning? Answer: 1/3rd, nothing has changed.
What are the odds Monty keeps the car? Answer: 2/3rds, nothing has changed

Monty offers you the option of Switching doors:

Does that in anyway effect where the car is sitting? Answer: No, the Car doesn't move.

"But A and B both cannot have 2/3 odds -- that's 4/3."

Your math is wrong. One door has a 1/3rd chance, the Other door has a 2/3rds chance, the third door (revealed goat) has a 0/3rds chance -- That's 3/3rds.

You are adding up the probabilities from two different rounds of the game, that's how you're getting impossible numbers. Pick a number between 1-6 and I'll roll a die. You have a 5/6 chance of being wrong. Pick a different number between 1-6 and I'll roll the die again. You have a 5/6 chance of being wrong. So you have a 10/6 chance of being wrong? Impossible! This is the logic you have employed.

I did watch Make a Deal as a kid so I know Monte Hall might be up to no good. So let's take this out to the racetrack:

The 3rd at Belmont draws only three entries, but they're perfectly evenly matched. On a hunch I bet Horse A to win. My chance is 1/3.

Coming out of the gate, Horse C throws its rider and is thus out of the race, it's the goat, as the remaining contestants A and B trot evenly down the backstretch.

A and B each now have a 50-50 shot at the win, no? If a friend with a win ticket on B says, "hey, let's switch," should I? Why would I?

Clever, I think, and that you're pondering it without recourse to the wikipedia entry, which makes clear the difference in the two scenarios - your picking the door on the game show influences what Monty does next, whereas your picking horse A or B does not cause horse C to throw the rider, unless the horse is in on it.

"The 3rd at Belmont draws only three entries, but they're perfectly evenly matched."

Sure, but in the Monty Hall example, the doors aren't equally matched. One door has a parked car behind it. All the others don't. That will never change.

So, let's modify your example. The Mob has fixed the race. There are 7 horses in the race. One horse will definitely win and your buddy knows which one, but won't tell you. You both have enough money to bet on one horse and you both can't bet on the same horse. You bet on one randomly, your buddy bets on one of the other 6 horses.

As the horses leave the gate, the jockies look to the mob enforcer right by the gate and 5 of the horses come to a screeching halt leaving two horses still running.

"A and B each now have a 50-50 shot at the win, no? "

Do you think your odds of winning, a random 1 in 7 chance of winning, are the same a that of your buddies chances who knew which horse was going to win?

While your buddy is looking the other way do you swap tickets with him? Or keep your own?

"... and 5 of the horses come to a screeching halt leaving the two horses you each bet on still running."

Bridge players learn the "principle of restricted choice", a variation of the Monty Hall problem, when learning the game. The problem can be illustrated easily. North, the dummy, holds A K 10 x x of the trump suit while South, declarer, holds 9 8 x x. North plays the Ace, East plays the Jack, and South and West follow low. South leads low toward the dummy, West follows low again. Should South finesse the 10 and play West for the Queen, or should South play the King, hoping to drop East's Queen? The odds are 2:1 in favor of the finesse, according to the principle of restricted choice. Most bridge players--for that matter, teachers--have trouble deriving the principle and simply memorize: play for two touching honors to be split.

Indeed the article cites bridge as the source of the principle of restricted choice, with Reese in _Master Play in Contract Bridge_ (1960) apparently the first to describe the principle.

"Do you want the door you picked? Or do you want all the other doors?"

I have tried this argument before, David, to convince people- it works sometimes, sometimes it doesn't. It usually works better if the example isn't 3 doors, but 10 or more. But I don't think I have ever tried at before "revealing" all the booby prizes first, and maybe that is why it fails so often.

Monty Hall has been clear to me since I first came across it, and over time I also figured why it trips up so many smart people. The real intuition is in the fact that the presenter's choice is not random, in case you select a goat.

If you choose goat, presenter is forced to choose the other goat. If you choose car, presenter can choose either goat.

Most people trip because they mentally model the presenter actions as 'at random', while in 2/3 rd of cases, they are not.

The easies way to see the difference is to look at not 3 boxes but at 1000 boxes.

1. You choose one,
2. The quizmaster opens 998 boxes which are empty and leaves one lonely box closed
3. Now you choose: your original box or the one singled out by the quizmaster.

The odds are now also very easy to calculate: if you stick to your original box you have a 1/1000 chance. If you switch, your chance is 999/1000

It also clears up the quaestion why the odds change: The quizmaster adds knowledge. He has to know which box contains the price and carefully skip it while opening 998 empty boxes.

So he doesn't open randomly 998 boxes, he really has to know which one contains the big price. Therefor the odds can never be 50-50. In that case, a gorilla could also act as the quizmaster.

The "think of 1,000,000 doors instead of 3" does not really help formulate an intuition, because we don't know if, under that regime, Monty would open 999,998 doors or just one.

Intuition: Monty will always open a door with a goat, so his action tells you nothing. You are being asked to choose between one door (1/3 chance of a car) or two doors (2/3 chance of a car). The first of these is "don't switch," the second is "switch." Switch.

Next week: Optimum Blocking Strategies on Hollywood Squares -- The Charles Nelson Reilly Effect

Lots of confusing explanations here. Think the most straightforward way to look at it is the value of information provided when Monty selects which door to open in two cases: the first when your initial choice was right, the second when your initial choice was wrong.

(1) If you are initially right (1/3 prob), then Monty opens at random and gives you no additional information. If you switch you lose 100% of the time.

(2) If you're initially wrong (2/3 prob), then Monty deliberately opens the door that does not have the prize (this is the valuable information). So the third door must have the prize, and if you switch you win the prize 100% of the time.

So a switching strategy results in you winning 1/3 * 0% + 2/3 * 100% = 67% of the time.

Imagine there are 1,000 doors instead of three. You select your door. Then Monty opens 998 of the doors and asks you would like to switch. I think most folks (even if they can't put their finger on it) would intuitively understand that the opening of 998 doors without prizes behind them tells them that there is something "special" about the other door that they didn't originally choose ...

I figured out the obvious case years ago, b/c I got it wrong at first and it bothered me.

As other pointed the really obvious version just uses a larger number of doors. Once you understand the concept, extrapolating it back to 3 doors is just math.

There are two different problems being solved here. They have the same answer, but only one matches the question:

1) If you have to decide before Monty Hall opens a door, should you switch?
2) If you can wait until you see what door Month Hall opened, say D3 after you picked D1, should you switch?

The answer to #1 is that you had a 2/3 chance to originally pick a goat, that can't change since MH can always open a door, and your prize changes if you switch.

The answer to #2 is that there originally was a 1/3 chance of reaching the game state where you picked a goat and MH opens D3, and a Q/3 chance if you picked the car. Here Q is the probability that MH would pick D# when both D2 and D3 have goats. This makes the chance that switching wins (1/3)/(1/3+Q/3)=1/(1+Q), using Bayes Law.

Since you must assume Q=1/2, the answer is still 2/3. But it would change if Q was not 1/2. Th solution to #1 is not a valid solution to #2 without explaining this, and Erdos was right.

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