9 ms·

Tricking Monty Hall

50 points by zdw 3y ago

Orlan 3y ago
And here’s an explanation from a theater show by Derren Brown @ 5:43 - https://youtu.be/GPoPSNxV1D4?t=342 https://youtu.be/GPoPSNxV1D4?t=342
andy800 3y ago
To the best of my knowledge, this "question" was first presented to the general public around 1990 via a popular syndicated newspaper column, and quickly "went viral" the way things did back then, by being repeatedly discussed in newspaper and magazine columns all across America. It's fascinating that 33 years later, there is still no simple explanation of the optimal strategy, such that most people would say "a-ha" when they hear it. In fact, most people actually express skepticism or disbelief when presented with the "correct" strategy.
tzs 3y ago
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.
johnfn 3y ago
The reasoning I've always used to understand this problem is to imagine that you and the host both choose at exactly the same time. Then 2/3 of the time you've both simultaneously selected both boxes with goats, and therefore 2/3 of the time the unselected box is the car.
patrickthebold 3y ago
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.
- lisper 3y ago
  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.
  - slavik81 3y ago
    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.
    - lisper 3y ago
      Yes, definitely. However, it's a pretty reasonable inference, because Monty would not open a door without being sure that it would not contain the prize because that would just destroy the dramatic tension. "Oops, sorry, guess you lost. Didn't see that coming. Better luck next time."
- kgwgk 3y ago
  > Should you switch? Why not? There is no harm in doing so if the probabilities are the same anyway...
voyager1 3y ago
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/ https://kr1stjans.github.io/monty-hall/ :D feedback appriciated :)
atum47 3y ago
I've made this "intuition" for the Monty Hall into a game [1] so I can test it for myself. I remember being intrigued by the notion that you're better off switching. 1 - https://victorribeiro.com/montyhall/ https://victorribeiro.com/montyhall/ Keep in my that this was made a long time ago, I wasn't even a programmer back then, I was more a webmaster.
silisili 3y ago
I wonder: has anyone went through the old tapes and confirmed that the people who stayed won about a third of the time, and the switchers 2/3 of the time? I agree with the math, but wonder how unscrupulous the show was.
activitypea 3y ago
I feel like most explanations are overcomplicating it. By the rules of the game, the option of switching is asking you the question "Do you think you chose the correct door out of the three?". If you think you did, you don't switch. If you think you didn't, by rules of the game, switching will get you to the right door. Since you had a 33.3% chance to choose correctly and 66.6% to choose incorrectly, betting against yourself and switching is probably the correct choice.
jamjamjamjamjam 3y ago
I honestly wouldn’t mind a goat.
fsckboy 3y ago
if you play this game repeatedly, 2/3 of the time your first guess will be a goat, and Monty will then always show you the other goat, and if you switch you will see the car behind the remaining door. if you play the game once, you should follow the same strategy. don't explain more than that, the person just needs to grok that. a restatement of the main idea might help: 2/3 of the time your first choice is a goat therefore 2/3 of the time Monty is showing you the other goat therefore 2/3 of the time if you switch you will be switching to the car.
c7b 3y ago
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.
lamontcg 3y ago
Bifurcate the world into the 3x3=9 chances of which door the prize is behind and which door the contestant guessed and write down the permutations. These are two independent random events and the permutations are equally likely with a 1/9th chance each. Now for the cases where the contestant guessed the door correctly -- "11", "22", and "33" the right move is to stay and it doesn't matter which door Monty Hall reveals. For the cases where the contestant didn't guess the door correctly -- "12", "21", "13", "31", "23", "32" the right move is always to switch (Monty Hall will have always revealed the third door in each permutation). The probabilities of each are 1/9th so they add up to 1/3 chance of failing if you switch and 2/3 chance of succeeding if you switch. So always switch. If you want to complicate things you can add the door that Monty Hall shows as a third option, but you need to be careful about the probabilities: "112", "113" - these have to add to 1/9th so each has 1/18th chance (50-50 chance MH picked one door or the other) and both fail if you stay, similarly for "221", "223", "331", "332", 6*1/18 = 1/3rd. "123" -- MH is forced to pick "3" so this is just 1/9th chance of success, same for "213", "132", "312", "231" and "321". So this is 6*1/9 = 2/3rd. One error would be writing down all the permutations and assuming they were equally likely, but "112" and "113" must sum to the odds of "11" which is 1/9th, while "123" is the same odds as "12" which is 1/9th.
DoctorOetker 3y ago
When people pose this "puzzle" verbally to another, there usually is a back and forth questions and answers about the exact problem statement. This is where most "puzzle" posers mess it up. In fact vos Savant messed this up herself. A proof is a sequence of steps, which are to be justified individually. Justifications may be general theorems from the field of mathematics, its axioms, or otherwise givens from the problem statement. Hiding withheld "given(s)" in the "solution" instead of providing them in the problem statement means the literal problem statement is ill-posed. Check the section "standard assumptions" on Wikipedia: https://en.wikipedia.org/wiki/Monty_Hall_problem#Standard_assumptions https://en.wikipedia.org/wiki/Monty_Hall_problem#Standard_as... "Under the standard assumptions, the probability of winning the car after switching is 2 / 3. The key to this solution is the behavior of the host. Ambiguities in the Parade version do not explicitly define the protocol of the host. However, Marilyn vos Savant's solution[3] printed alongside Whitaker's question implies, and both Selvin[1] and Savant[5] explicitly define, the role of the host as follows: [...]" When I was first posed the problem, I told the person there was not enough information to solve the puzzle, I saw the Bayesian catapult under a specific assumption, but I realized the problem statement was underspecified and unposed. So while inquiring the acquaintance to explain "how it happened that there was a goat behind the door" the person just reiterated the already communicated sequence of events as a fairy tale, and this went on until he provided the solution. Of course his explanation was the Bayesian catapult, and I responded that the problem statement did not contain the requisite facts and he must have mistold it. So I looked up the problem on Wikipedia only to discover that vos Savant herself ill-posed her problem. Its perfectly fine for a step in a proof of solution to refer to a given. But that given must either be directly present in the problem statement or be rigorously derivable from it. This is simply not the case. If it isn't given it's not a given.
- kgwgk 3y ago
  > In fact vos Savant messed this up herself. I don’t think it’s _that_ bad. She writes that “the host, who knows what's behind the doors, opens another door which has a goat.” Why would the problem statement say that the host knows what's behind the doors (devoting 10% of the description to making this clarification) if not to indicate that he knowingly opens another door which has a goat?
11235 3y ago
Here are a couple of different explanations using Bayes' Theorem as well as a more informal one. https://dev.to/santisbon/the-monty-hall-problem-28om https://dev.to/santisbon/the-monty-hall-problem-28om
necovek 3y ago
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.
- evandale 3y ago
  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 ;)
  - ryandrake 3y ago
    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.
    - 3y ago
stavros 3y ago
What helped me visualize is this: 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.
- xg15 3y ago
  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.
- IanCal 3y ago
  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.
  - chefandy 3y ago
    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.
    - IanCal 3y ago
      Two things about this 1. It requires that I already get the point about the probabilities. 2. You don't get a benefit in Monty hall if the host picks randomly - the description requires that they deliberately leave the winning door available. Any situation that gives you the same intuition but you just picture as happening without intent has given you the wrong feeling. I don't find that increasing the number of doors makes it feel any different.
deeg 3y ago
Here's my attempt to explain the Monty Hall with code: https://github.com/DeegC/monty_hall_paradox https://github.com/DeegC/monty_hall_paradox
evandale 3y ago
This is a good way of putting it and I think would resonate with some people! I've not been able to get my dad to understand why you should switch with 1,000,000 boxes because he refuses to believe the 3 box and 1,000,000 box case is the same. I can't even convince him a 3 and 5 box problem are the same. I actually think I could convince him switching is correct using this strategy because I suspect if he thinks he's tricking or cheating at the game he'll be more open to the explanation.
- arithma 3y ago
  Are we siblings
- Swizec 3y ago
  My problem with Monty Hall is that it’s not a repeated game. Average probabilities don’t help me. This is the point Taleb makes in a lot of his books. You are not the average probability. You only get one play through [life] and the only thing that matters is how the dice fall for you. Even a 90% average chance of getting $prize, is still a 10% chance of walking away with nothing and you better be ready to get that one because you very well might. The strategy to increase your average probabilities works in poker. Because you play multiple hands and even multiple games. In Monty Hall, you’re just making a guess and no matter how clever you are about it, it’s still just a guess. There is no strategy because the game doesn’t last long enough for strategy to matter. Now if you had an iterated monty hall, then yes, bring out the maths.
  - LukasMathis 3y ago
    If I tell you that I'm rolling a die, and ask you whether I rolled a six or not a six, offering you 100$ if you guess right, you very obviously should say "not a six", even if we only play the game once.
    - kgwgk 3y ago
      No, reading Taleb I realized that if it's not repeated it’s just a guess and winning or losing is a 50/50 thing. There is no strategy because the game doesn’t last long enough for strategy to matter. [/s]
Ekaros 3y ago
What this problem always fails for me is that the show host doesn't pick randomly. What if there was option of opening the car? Also, does the goat come butchered or do you need to do it on stage?
- DoctorOetker 3y ago
  Where I live most cars have 4 doors, but since we are expected to pretend this is a well-posed problem we must assume there are literally 3 doors at the game host location. Evidently the participant entered through one of them and already knows whats behind one door. If he saw the car during entry, the correct choice would be the door of entry. The host can open (and close, and open, ...) doors with goats all day long, the participant would know the car is where he just saw it. If the participant did not see the car during entry it must be behind one of the remaining doors: either behind the back door, or behind the emergency exit door. Clearly it would be highly illegal to obstruct the emergency exit door with a car, so under Standard Assumptions (see assumption 42, the last one in the code) we can assume the host did not park the car behind this emergency exit door. Especially since emergency exit doors open outwards. Hence the car was parked behind the remaining backdoor, and since this follows from the Standard Assumptions we can conclude the participant made the same (by definition valid assumption) as described here, since its a Standard Assumption, and assumption 1 of the Standard Assumptions is that in the absence of necessary knowledge, everyone makes the Standard Assumptions. Given that the participant (using the standard assumptions) knew immediately which door hides the car and given that this door was his initial "guess", the correct solution (under Standard Assumptions) is to stay with the initial choice.
- drexlspivey 3y ago
  If the host picked randomly then there would be a 1/6 chance that the show would end rather anticlimactically and you would be indifferent in switching. Now this 1/6 becomes your edge turning your 1/3 to 1/2 after the goat is revealed. EDIT: 1/3 not 1/6
  - IanCal 3y ago
    1/3 not 1/6. There are six outcomes of you picking a door and the host opening one of the remaining doors. In two of them you pick a goat the first time and then the host reveals the car.
rain1 3y ago
This is incorrect, the goats and car are behind doors. They are not inside cardboard boxes.
arithma 3y ago
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.
- jncfhnb 3y ago
  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
  - mrob 3y ago
    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.
    - listenallyall 3y ago
      > seeing the player lose after switching would be more entertaining for the audience Have you ever watched game shows? Because it is far more satisfying and fun for the audience when players win, not lose. If you are out there rooting for people to lose, you're a miserable bastard.
      - mrob 3y ago
        Now that I think about it, I have watched a few episodes of game shows, and I think you're right. They are always set up so the audience identifies with the players and the hosts are the adversaries, so you want the players to win.
    - 3y ago
mikrl 3y ago
I did a presentation on this for a discrete math course in undergrad. IIRC I did the N box case and took the limit to infinity and, unsurprisingly, got the result that you should always switch.
xg15 3y ago
The reasoning that made me get it was basically: 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.
- nkrisc 3y ago
  You can distill it all to this: 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.
  - diputsmonro 3y ago
    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!
jncfhnb 3y ago
I think this approach shown in the article is not great. It adds additional layers of thinking where none are needed. You don’t lie. It also doesn’t feel like it would help explain variants like 5 doors, and only one goat is revealed.
- LukasMathis 3y ago
  I would argue that if you start out knowing that you will switch, you essentially do lie, because you initially pretend to pick a box (or door) that you know you will not end up opening. It is true that this approach adds additional layers of thinking, but the problem is that without those layers, the solution to the problem is simply not intuitive for the vast majority of people. Adding these additional layers helps at least some people gain a more intuitive understanding of why switching helps.
  - jncfhnb 3y ago
    Ehhh. Idk. I think it is not great. A much easier framing on a similar level is simply saying you pick a door, and then the host says you can either open the chosen door, or both other doors one at a time. It’s the same thing. But if you don’t get why the state is unchanged even after opening one of the doors I don’t think you get it with this whole lying thing
rahimnathwani 3y ago
I like this explanation. Another equivalent one that I wrote about a while back[0]: You can choose to have: - The most valuable prize that’s behind door A, or - The most valuable prize that’s behind doors B or C When you look at it this way, it’s obvious that you would rather have the most valuable prize from the ‘other’ doors, and the way to do that is to switch. [0] https://www.encona.com/posts/monty-hall-problem https://www.encona.com/posts/monty-hall-problem

50 points by zdw 3y ago

And here’s an explanation from a theater show by Derren Brown @ 5:43 - https://youtu.be/GPoPSNxV1D4?t=342 https://youtu.be/GPoPSNxV1D4?t=342

andy800 3y ago

To the best of my knowledge, this "question" was first presented to the general public around 1990 via a popular syndicated newspaper column, and quickly "went viral" the way things did back then, by being repeatedly discussed in newspaper and magazine columns all across America. It's fascinating that 33 years later, there is still no simple explanation of the optimal strategy, such that most people would say "a-ha" when they hear it. In fact, most people actually express skepticism or disbelief when presented with the "correct" strategy.

tzs 3y ago

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.

johnfn 3y ago

The reasoning I've always used to understand this problem is to imagine that you and the host both choose at exactly the same time. Then 2/3 of the time you've both simultaneously selected both boxes with goats, and therefore 2/3 of the time the unselected box is the car.

patrickthebold 3y ago

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.

lisper 3y ago
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.
- slavik81 3y ago
  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.
  - lisper 3y ago
    Yes, definitely. However, it's a pretty reasonable inference, because Monty would not open a door without being sure that it would not contain the prize because that would just destroy the dramatic tension. "Oops, sorry, guess you lost. Didn't see that coming. Better luck next time."
kgwgk 3y ago
> Should you switch? Why not? There is no harm in doing so if the probabilities are the same anyway...

voyager1 3y ago

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/ https://kr1stjans.github.io/monty-hall/ :D feedback appriciated :)

atum47 3y ago

I've made this "intuition" for the Monty Hall into a game [1] so I can test it for myself. I remember being intrigued by the notion that you're better off switching. 1 - https://victorribeiro.com/montyhall/ https://victorribeiro.com/montyhall/ Keep in my that this was made a long time ago, I wasn't even a programmer back then, I was more a webmaster.

silisili 3y ago

I wonder: has anyone went through the old tapes and confirmed that the people who stayed won about a third of the time, and the switchers 2/3 of the time? I agree with the math, but wonder how unscrupulous the show was.

activitypea 3y ago

I feel like most explanations are overcomplicating it. By the rules of the game, the option of switching is asking you the question "Do you think you chose the correct door out of the three?". If you think you did, you don't switch. If you think you didn't, by rules of the game, switching will get you to the right door. Since you had a 33.3% chance to choose correctly and 66.6% to choose incorrectly, betting against yourself and switching is probably the correct choice.

jamjamjamjamjam 3y ago

I honestly wouldn’t mind a goat.

fsckboy 3y ago

if you play this game repeatedly, 2/3 of the time your first guess will be a goat, and Monty will then always show you the other goat, and if you switch you will see the car behind the remaining door. if you play the game once, you should follow the same strategy. don't explain more than that, the person just needs to grok that. a restatement of the main idea might help: 2/3 of the time your first choice is a goat therefore 2/3 of the time Monty is showing you the other goat therefore 2/3 of the time if you switch you will be switching to the car.

c7b 3y ago

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.

lamontcg 3y ago

Bifurcate the world into the 3x3=9 chances of which door the prize is behind and which door the contestant guessed and write down the permutations. These are two independent random events and the permutations are equally likely with a 1/9th chance each. Now for the cases where the contestant guessed the door correctly -- "11", "22", and "33" the right move is to stay and it doesn't matter which door Monty Hall reveals. For the cases where the contestant didn't guess the door correctly -- "12", "21", "13", "31", "23", "32" the right move is always to switch (Monty Hall will have always revealed the third door in each permutation). The probabilities of each are 1/9th so they add up to 1/3 chance of failing if you switch and 2/3 chance of succeeding if you switch. So always switch. If you want to complicate things you can add the door that Monty Hall shows as a third option, but you need to be careful about the probabilities: "112", "113" - these have to add to 1/9th so each has 1/18th chance (50-50 chance MH picked one door or the other) and both fail if you stay, similarly for "221", "223", "331", "332", 6*1/18 = 1/3rd. "123" -- MH is forced to pick "3" so this is just 1/9th chance of success, same for "213", "132", "312", "231" and "321". So this is 6*1/9 = 2/3rd. One error would be writing down all the permutations and assuming they were equally likely, but "112" and "113" must sum to the odds of "11" which is 1/9th, while "123" is the same odds as "12" which is 1/9th.

DoctorOetker 3y ago

When people pose this "puzzle" verbally to another, there usually is a back and forth questions and answers about the exact problem statement. This is where most "puzzle" posers mess it up. In fact vos Savant messed this up herself. A proof is a sequence of steps, which are to be justified individually. Justifications may be general theorems from the field of mathematics, its axioms, or otherwise givens from the problem statement. Hiding withheld "given(s)" in the "solution" instead of providing them in the problem statement means the literal problem statement is ill-posed. Check the section "standard assumptions" on Wikipedia: https://en.wikipedia.org/wiki/Monty_Hall_problem#Standard_assumptions https://en.wikipedia.org/wiki/Monty_Hall_problem#Standard_as... "Under the standard assumptions, the probability of winning the car after switching is 2 / 3. The key to this solution is the behavior of the host. Ambiguities in the Parade version do not explicitly define the protocol of the host. However, Marilyn vos Savant's solution[3] printed alongside Whitaker's question implies, and both Selvin[1] and Savant[5] explicitly define, the role of the host as follows: [...]" When I was first posed the problem, I told the person there was not enough information to solve the puzzle, I saw the Bayesian catapult under a specific assumption, but I realized the problem statement was underspecified and unposed. So while inquiring the acquaintance to explain "how it happened that there was a goat behind the door" the person just reiterated the already communicated sequence of events as a fairy tale, and this went on until he provided the solution. Of course his explanation was the Bayesian catapult, and I responded that the problem statement did not contain the requisite facts and he must have mistold it. So I looked up the problem on Wikipedia only to discover that vos Savant herself ill-posed her problem. Its perfectly fine for a step in a proof of solution to refer to a given. But that given must either be directly present in the problem statement or be rigorously derivable from it. This is simply not the case. If it isn't given it's not a given.

kgwgk 3y ago
> In fact vos Savant messed this up herself. I don’t think it’s _that_ bad. She writes that “the host, who knows what's behind the doors, opens another door which has a goat.” Why would the problem statement say that the host knows what's behind the doors (devoting 10% of the description to making this clarification) if not to indicate that he knowingly opens another door which has a goat?

11235 3y ago

Here are a couple of different explanations using Bayes' Theorem as well as a more informal one. https://dev.to/santisbon/the-monty-hall-problem-28om https://dev.to/santisbon/the-monty-hall-problem-28om

necovek 3y ago

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.

evandale 3y ago
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 ;)
- ryandrake 3y ago
  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.
  - 3y ago

stavros 3y ago

What helped me visualize is this: 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.

xg15 3y ago
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.
IanCal 3y ago
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.
- chefandy 3y ago
  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.
  - IanCal 3y ago
    Two things about this 1. It requires that I already get the point about the probabilities. 2. You don't get a benefit in Monty hall if the host picks randomly - the description requires that they deliberately leave the winning door available. Any situation that gives you the same intuition but you just picture as happening without intent has given you the wrong feeling. I don't find that increasing the number of doors makes it feel any different.

deeg 3y ago

Here's my attempt to explain the Monty Hall with code: https://github.com/DeegC/monty_hall_paradox https://github.com/DeegC/monty_hall_paradox

evandale 3y ago

This is a good way of putting it and I think would resonate with some people! I've not been able to get my dad to understand why you should switch with 1,000,000 boxes because he refuses to believe the 3 box and 1,000,000 box case is the same. I can't even convince him a 3 and 5 box problem are the same. I actually think I could convince him switching is correct using this strategy because I suspect if he thinks he's tricking or cheating at the game he'll be more open to the explanation.

arithma 3y ago
Are we siblings
Swizec 3y ago
My problem with Monty Hall is that it’s not a repeated game. Average probabilities don’t help me. This is the point Taleb makes in a lot of his books. You are not the average probability. You only get one play through [life] and the only thing that matters is how the dice fall for you. Even a 90% average chance of getting $prize, is still a 10% chance of walking away with nothing and you better be ready to get that one because you very well might. The strategy to increase your average probabilities works in poker. Because you play multiple hands and even multiple games. In Monty Hall, you’re just making a guess and no matter how clever you are about it, it’s still just a guess. There is no strategy because the game doesn’t last long enough for strategy to matter. Now if you had an iterated monty hall, then yes, bring out the maths.
- LukasMathis 3y ago
  If I tell you that I'm rolling a die, and ask you whether I rolled a six or not a six, offering you 100$ if you guess right, you very obviously should say "not a six", even if we only play the game once.
  - kgwgk 3y ago
    No, reading Taleb I realized that if it's not repeated it’s just a guess and winning or losing is a 50/50 thing. There is no strategy because the game doesn’t last long enough for strategy to matter. [/s]

Ekaros 3y ago

What this problem always fails for me is that the show host doesn't pick randomly. What if there was option of opening the car? Also, does the goat come butchered or do you need to do it on stage?

DoctorOetker 3y ago
Where I live most cars have 4 doors, but since we are expected to pretend this is a well-posed problem we must assume there are literally 3 doors at the game host location. Evidently the participant entered through one of them and already knows whats behind one door. If he saw the car during entry, the correct choice would be the door of entry. The host can open (and close, and open, ...) doors with goats all day long, the participant would know the car is where he just saw it. If the participant did not see the car during entry it must be behind one of the remaining doors: either behind the back door, or behind the emergency exit door. Clearly it would be highly illegal to obstruct the emergency exit door with a car, so under Standard Assumptions (see assumption 42, the last one in the code) we can assume the host did not park the car behind this emergency exit door. Especially since emergency exit doors open outwards. Hence the car was parked behind the remaining backdoor, and since this follows from the Standard Assumptions we can conclude the participant made the same (by definition valid assumption) as described here, since its a Standard Assumption, and assumption 1 of the Standard Assumptions is that in the absence of necessary knowledge, everyone makes the Standard Assumptions. Given that the participant (using the standard assumptions) knew immediately which door hides the car and given that this door was his initial "guess", the correct solution (under Standard Assumptions) is to stay with the initial choice.
drexlspivey 3y ago
If the host picked randomly then there would be a 1/6 chance that the show would end rather anticlimactically and you would be indifferent in switching. Now this 1/6 becomes your edge turning your 1/3 to 1/2 after the goat is revealed. EDIT: 1/3 not 1/6
- IanCal 3y ago
  1/3 not 1/6. There are six outcomes of you picking a door and the host opening one of the remaining doors. In two of them you pick a goat the first time and then the host reveals the car.

rain1 3y ago

This is incorrect, the goats and car are behind doors. They are not inside cardboard boxes.

arithma 3y ago

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.

jncfhnb 3y ago
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
- mrob 3y ago
  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.
  - listenallyall 3y ago
    > seeing the player lose after switching would be more entertaining for the audience Have you ever watched game shows? Because it is far more satisfying and fun for the audience when players win, not lose. If you are out there rooting for people to lose, you're a miserable bastard.
    - mrob 3y ago
      Now that I think about it, I have watched a few episodes of game shows, and I think you're right. They are always set up so the audience identifies with the players and the hosts are the adversaries, so you want the players to win.
  - 3y ago

mikrl 3y ago

I did a presentation on this for a discrete math course in undergrad. IIRC I did the N box case and took the limit to infinity and, unsurprisingly, got the result that you should always switch.

xg15 3y ago

The reasoning that made me get it was basically: 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.

nkrisc 3y ago
You can distill it all to this: 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.
- diputsmonro 3y ago
  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!

jncfhnb 3y ago

I think this approach shown in the article is not great. It adds additional layers of thinking where none are needed. You don’t lie. It also doesn’t feel like it would help explain variants like 5 doors, and only one goat is revealed.

LukasMathis 3y ago
I would argue that if you start out knowing that you will switch, you essentially do lie, because you initially pretend to pick a box (or door) that you know you will not end up opening. It is true that this approach adds additional layers of thinking, but the problem is that without those layers, the solution to the problem is simply not intuitive for the vast majority of people. Adding these additional layers helps at least some people gain a more intuitive understanding of why switching helps.
- jncfhnb 3y ago
  Ehhh. Idk. I think it is not great. A much easier framing on a similar level is simply saying you pick a door, and then the host says you can either open the chosen door, or both other doors one at a time. It’s the same thing. But if you don’t get why the state is unchanged even after opening one of the doors I don’t think you get it with this whole lying thing

rahimnathwani 3y ago

I like this explanation. Another equivalent one that I wrote about a while back[0]: You can choose to have: - The most valuable prize that’s behind door A, or - The most valuable prize that’s behind doors B or C When you look at it this way, it’s obvious that you would rather have the most valuable prize from the ‘other’ doors, and the way to do that is to switch. [0] https://www.encona.com/posts/monty-hall-problem https://www.encona.com/posts/monty-hall-problem