10 ms·

Marilyn vos Savant and the Monty Hall Problem

coreyp_1 12y ago
This was fascinating. I can't believe so many academics got it wrong, and then attacked her gender over it!
- bsdetector 12y ago
  I'm not surprised. At the time, almost every question every week in Parade magazine was just pulp. Questions of the type "what's the world's smartest woman's favorite color?". You could take anybody off the street and get the same answers. Parade magazine was a gossip magazine. Furthermore, at the time she wasn't listed in the current Guinness Book so anybody that looked it up would see the column as a fraud unless they checked a specific old edition. Claiming somebody "is listed" when they had been listed, for one year many years ago, is a lie. So there were a lot of readers who thought of the column as a joke. If you open 99 doors that are full of mindless drivel, what are the odds the 100 is the one that has something interesting? This is why so many people were fooled by it.
  - mcphage 12y ago
    > If you open 99 doors that are full of mindless drivel, what are the odds the 100 is the one that has something interesting? This is why so many people were fooled by it. So they thought "ha, this is foolishly wrong, I'll write in and tell them that—of course, without doing the 5 minute math check to make sure, first!"
asdfologist 12y ago
I felt pretty stupid when it took me a long time to convince myself of the correct answer. But I felt better when I read that even "Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result"! http://en.wikipedia.org/wiki/Monty_Hall_problem http://en.wikipedia.org/wiki/Monty_Hall_problem
- bmelton 12y ago
  The only explanation that convinced me that my original guess was incorrect was this one (which is alluded to in the article): If there are three doors, the odds that the car is behind any of them is 1 in 3, or 1/3. Once you select a door, the odds are 1/3 for your door, 1/3 for each of the other, unselected doors. Put more simply, 1/3 for your door, or 2/3 that it is not in your door. Once another door is opened that does not have the car, then the odds of it being your door are 1/3, and the odds of it not being your door are still 2/3, so it makes sense to switch.
  - mjevans 12y ago
    Thank you. Adding the temporal domain and system states is what makes this easy to understand. At the time of selection you freeze the probability of the sets (as this is the macroscopic world and the implications of the state in the other rooms must have collapsed due to interactions with all sorts of radiation from the rest of the universe). Thus what you say is exactly correct. * T0 - {1/3} {1/3} {1/3} * T1 - {1/3 Your pick} {2/3 = {1/3} + {1/3}} * T2 - {1/3 Your pick} {2/3 = {0/3 Host Pick} {2/3 Remaining door}}
- mariodiana 12y ago
  I read about this problem in the book "The Drunkard's Walk." I rejected the answer viscerally until I coded up a simple simulation (in Python) and ran it 10 million times. And then it clicked. Human beings are really bad at intuiting probability.
rhino42 12y ago
One interesting variant, is the case where the Host opens a door without knowing if that door had a goat. Expressing relief that he opened a goat door, he asks you to consider switching doors. IIRC, you get less information in this case, since there was a 1/3 chance that the host would accidentally pick the "car" door, but you still have a 1/2 chance of winning by switching.
- rando3826 12y ago
  > still have a 1/2 chance of winning by switching It seems that you are referring to the other variant, in which case this is false, because that one has a 2/3 chance of winning by switching.
  - taeric 12y ago
    No, you have a 1/2 chance of winning after a door has been opened. However, you also have a chance of losing on that reveal. Something you don't have in the normal variants. (That is, in normal variants, you only lose at the end of the game. This one, you could effectively lose when the host opens the actual winning door.) So, your overall odds aren't as different as it seems. Just you have a much better odds of winning by switching, if you are still in the game.
MaysonL 12y ago
I still think that the problem is wrongly cast as a probability problem, rather than the game-theoretic problem that it is.
- Terr_ 12y ago
  A fair point, the hosts' choice of door is not random at all. He could open the door you'd already chosen, but that would make your choice to switch-or-not a no-brainer.
JohnTHaller 12y ago
It always amazed me that so many academics got it wrong. Once you understand the way the actual game works, it's pretty obvious that you gain a statistical advantage if you switch.
- savanaly 12y ago
  I know right? I don't blame them for getting it wrong at first glance-- I'm pretty sure 999 out of 1000 humans would. But the mathematics of the probabilities can be written down on a napkin they are so simple, so how was no one able to convince these scientists with a clear, concise technical proof? They really needed a computer simulation, as I've seen it claimed?
  - function_seven 12y ago
    > But the mathematics of the probabilities can be written down on a napkin they are so simple, so how was no one able to convince these scientists with a clear, concise technical proof? I'm even more appalled that no one tried, you know, actually running the game. Have someone take a quarter and two pennies, mix them up under cups, let you choose one, show you a penny, then switch or not. Repeat several times. Switch places. When you're playing the role of Monty, the truth will reveal itself...
    - MichaelGG 12y ago
      I have a close friend that was incensed about my explanation about how he was misunderstand. He even ran a broken simulation that showed he was right. When I corrected his source code with detailed comments, he further accused me of trying to fool him and make fun of him. Didn't speak to me for a month. Bizarre. But I suppose this kind of thing that goes against instincts can really upset people.
- dllthomas 12y ago
  Note that it can depend on the wording of the question. Monty must know that he's revealing a goat for it to work out 2-to-1. If he picks randomly, your odds are 1-to-1 after he's revealed the goat.
Terr_ 12y ago
I'd re-frame it this way: "The host has zero incentive to reveal what's behind the door you already picked (or the choice of switching would be a no-brainer) so the fact that they left it shut doesn't tell you anything." "However, the fact that they didn't (or couldn't) open the remaining shut-door hints that it may contain the prize." The key is that the host's choice is not random.
- ninkendo 12y ago
  Exactly the way I like to think of it. When he opens the goat door there are now two possibilities: you were right the first time, or the remaining door has the car. By selecting the other door your are now betting against your initial choice, which was a 1/3 chance. Thus your new odds are 2/3.
mediocregopher 12y ago
So I believe the 2/3 answer to be true. I see the evidence in those tables. But intuitively, I still don't get it, and the explanation in the article isn't helping > The short answer is that your initial odds of winning with door #1 (⅓) don’t change simply because the host reveals a goat behind door #3; instead, Hall’s action increases the odds to ⅔ that you’ll win by switching. Well, why don't they change simply because the host reveals a goat? If you show me a three sided coin, I'd say there's a 1/3 chance of getting a particular side. If you then magically change gravity such that one of the other sides will never ever be landed on, would the two remaining sides not now have 50/50? I guess the article asserts that the "probability" doesn't get redistributed amongst the remaining options evenly, and I want to know why it doesn't and why it instead all "goes" to the one I didn't pick. I don't mean to be confrontational in asking, and I don't doubt the answer, I just don't get why. Could anyone explain it better, or point me somewhere that does?
- jnbiche 12y ago
  Because when you ponder this problem, you're considering the probability of two different states. When the host reveals the goat, the state of the problem has changed, and so has the probability of choosing a correct answer (it improved). Your mistake is somehow thinking they are the same state.
- zaksoup 12y ago
  The way I like to think about it is as follows: You pick door 1. Out of three doors you have a 1/3 chance of having a car behind your door. The chance that the car is behind one of either of the other two doors is 2/3. Let's pretend that instead of opening a door and revealing a goat, Monty instead says to you "You can switch to both of the remaining doors. If the car is behind either one of them you get to keep it." Your likelyhood of getting the car by switching is now 2/3. Instead, Monty does the statistical equivalent: He allows you to know for sure which of the other doors definitely has a goat.
  - choffstein 12y ago
    This is a great answer. To answer the question about the 3-sided die: the question is flawed for 2 reasons. First, the way you phrased it, you know the information a priori. It's like me learning that the car isn't behind door #1 before I choose. In that case, I have a 50/50 chance on the other doors. But what if you didn't tell me before? This introduces our second problem. Let's say I roll the die and you hide it under a cup. I tell you I chose side A and you tell me A is physically impossible and ask me if I want to switch. Well of course I want to switch ... but I didn't learn any more information about sides B and C. It's a 50/50 toss-up. In terms of the game show, you basically just opened the constant's door. A critical element of the original problem is when the information is learned and knowing that the contestant's door will not be opened. It is what allows the phrasing from zaksoup above.
savanaly 12y ago
I read about this problem and the fact that multiple educated writers falsely called out vos Savant in an old Martin Gardner column, and it shines in my memory as one of the problems that illuminated math for me and made me see recreational math as worthwhile. Of course back when he wrote about it the topic of sexism was mercifully left out, and I honestly hadn't given a moment's thought to the fact that vos Savant was a woman until I read this new take on it, but there you go. Have standards for what makes something confusing and difficult to understand risen significantly over the decades? When I brought this problem up with my group of (admittedly geeky and mathy) friends no one was left unconvinced by the end of the afternoon, and most even had the intuition as well as the mathematical proof of the answer. Clearly we aren't any smarter than the public figures who couldn't comprehend this puzzle a half a century ago-- so it all in the framing of the problem?
- sukilot 12y ago
  Vos Savant originally posed the problem with ambiguous wording.
  - mcphage 12y ago
    The original problem, from her website: "Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? Craig F. Whitaker Columbia, Maryland" (a) she didn't originally pose the problem, it was a question asked of her, and (b) the wording of the problem is not ambiguous.
    - bztzt 12y ago
      the ambiguity is that it's presented as something the host does in this particular instance, but the validity of the probability calculations depends on it being a protocol the host always follows (in particular, that the host will always reveal one of the goats not behind the contestant's first choice of doors).
bwy 12y ago
Impossible to graduate from Berkeley with a CS degree without hearing this problem :)
Strilanc 12y ago
An intuition pump for the problem, especially if you know about conservation of expected evidence [1], is that the asymmetry between the door you picked and the other door that wasn't opened is that the other door passed a test. The door we choose got a bye; we learned nothing about it's ability to win car-containing contests. The other door was at risk, and survived. We gained information about it. 1: http://lesswrong.com/lw/ii/conservation_of_expected_evidence/ http://lesswrong.com/lw/ii/conservation_of_expected_evidence...
- andresp 12y ago
  This was the clearest explanation in my opinion. I'm still a bit confused regarding how much information have we actually got and what mechanism is behind "passing a test" to influence the amount of information when the test result is not definitive (as in this case). We were just given a complete information about one of the options the user didn't choose, but not any direct new information specific of the door the user didn't choose and that remains closed. Imagine that after opening the door everyone goes home and a new participant who has no knowledge of the previous experiment is called to make a decision between the two remaining doors. Obviously in this case the probably is 1:2, as there is no information regarding the previous test. Actually, if we imagine a similar situation happened before the initial problem with a previous show with 4 doors, we may as well see that the initial probability of 1:3 is also wrong if we had access to the knowledge about the previous experiment. I still agree with the switch suggestion, but the mechanism behind it is still not entirely clear. A more intuitive example would also be to invert the problem: leave one of the non picked doors (C) out of the random opening and use the door the contestant picked (A) as a possible case in the random opening experiment. If we start by opening A, then we already know enough to see if the user won or lost, if we open B and there is the car behind it, we already know everything. If we open B and there is a goat behind the door, consider whether you would switch your choice from A to C. I guess you probably wouldn't and it would feel easier to justify your position, as this doesn't involve the "psychological pain" of changing a previous choice by reasons that you don't entirely understand (associated with possible loss-aversion feelings).
simple1 12y ago
One way to intuitively believe the answer is to imagine there were 100 doors, you picked one, then the host opened up 98 of the other doors, revealing goats. Switching would be the obvious answer in that case.
- cdelsolar 12y ago
  That's still not obvious to me
  - peripitea 12y ago
    I'm a little intoxicated so this may be way over doing it or just not make sense, but... Let's make the number even bigger, so we don't even need to deal in probabilities -- a sextillion (10^21) doors. That's more doors than there are grains of sand on the planet. Say you're given a sextillion doors, and you pick one lonesome door, call it door A. What are the odds that door A is the right one? Essentially zero, right? I'm going to say straight up zero because now is not the time for you to be a dreamer. Ok so now that you've made your futile pick, poor Monty has to go and open every other door, all sextillion of them, except for yours and one other (let's call it door B) which he leaves closed. Every door he opens contains a goat. So now out of this vast universe of doors, there are (10^21-2) doors opened with goats bleating away, and only two doors left unopened: door A, aka yours; and door B, the one that Monty mysteriously left unopened. One of them must contain the car. (I'm guessing at this point it will be obvious for many people that door B contains the car, but I'll elaborate further in case it isn't clear yet.) So we're left with your door A and Monty's door B. At this point there are only two scenarios that matter. Either you picked the right door originally, in which case staying put is the winning move. OR, you picked the wrong door originally, in which case switching is the right move. And we already know you picked the wrong door (remember, no dreamers allowed here), so you should for sure switch to this by-now-extremely-conspicuous-looking door B. Door B isn't just some random door; the fact that it was chosen as the only door to not be opened tells you a tremendous amount of information about it. Think about the game from Monty's perspective. Either you pick the correct door initially and he gets to make door B a random goat door, or you picked the wrong door and he has no choice but to make door B the door with the car behind it. Since we know you picked the wrong door, we know what Monty's gonna have to do.
WillNotDownvote 12y ago
It's a testament to the brilliance of game show producers that they took something like this that both the contestant and the audience will misunderstand (but looks so simple!) and made a long running show out of it, making millions along the way.
elihu 12y ago
One thing that bothers me about the problem as it's stated in the article is that it isn't explicit about whether the game show host is required to open one of the goat doors regardless of which door the contestant picks, or whether the host is allowed to choose whether to open a door or not. If the host were to only open a goat door if the contestant initially picks the correct door, then the contestant would lose every single time if he/she always switched doors.
- sukilot 12y ago
  Yes, this was the source of the controversy and why vos Savant's self-congratulatory tempest was so annoying.
  - hudibras 12y ago
    The original column and response is online and is unambiguous for fluent English readers.
- hudibras 12y ago
  I think you're splitting hairs here. It goes both ways: if the host were to only open a goat door if the contestant initially picks the correct door, then the contestants would quickly learn to stop switching doors.
- deleted 12y ago
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- xom 12y ago
  If the host can choose between doing nothing and revealing a goat, and the host wants the contestant to lose, then the contestant's best strategy is to never switch, in order to avoid being exploited by a host who only reveals a goat when switching would lose. Any reader who wrote in to say the probabilities were 50-50 after a goat was revealed was probably not thinking the host wanted the contestant to lose (or to win).
- mcphage 12y ago
  > whether the host is allowed to choose whether to open a door or not The problem as stated is: "you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3". Why would you think the host has the option of not opening a door? And if you think they could do that, why couldn't the host also do other things, like offer the contestant cash to walk away, or swap what's behind the doors, or add more doors, or open all three doors, or sucker-punch them in the gut?
_pius 12y ago
Quickest way I've seen to understand this result is referenced on Wikipedia: An intuitive explanation is that if the contestant picks a goat (2 of 3 doors) the contestant will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (1 of 3 doors) the contestant will not win the car by switching. The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability. http://en.wikipedia.org/wiki/Monty_Hall_problem http://en.wikipedia.org/wiki/Monty_Hall_problem
bhc3 12y ago
My crack at it... 1/3 chance prize is behind a given door. Pick a door, you've got a 1/3 chance it's behind it. But there's a 2/3 chance it's behind one of the other doors. So you've got two sets of outcomes at this point. Set A has 1/3 probability (the door you chose). Set B has a 2/3 probability (the two doors you didn't choose). You then get this incredibly valuable information. The door in Set B that doesn't have the prize. So now Set B still has a 2/3 probability of having the prize. But you know that higher probability now applies to only the one door in Set B. So you end up with: Set A door = 1/3 chance | Set B door = 2/3 chance Make the switch every time.
- deleted 12y ago
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- dperny 12y ago
  I've known the answer to the Monty Hall problem for a long time, but this particular explanation just so happens to be the first one that's brought me closer to grokking the answer. I dunno what's different about this one, but it makes sense to me. Congrats.
  - BerislavLopac 12y ago
    I think that the key sentence in his explanation is "So now Set B still has a 2/3 probability of having the prize." This really nails the whole thing down.
  - chrismcb 12y ago
    One thing to keep in mind, that most people don't understand, this only works if someone KNOWS which one is empty, and shows the door.
- ntucker 12y ago
  Yeah, I look at it this way: Let's just focus on set B. You're staring at two doors. Some random process decided whether the things behind the doors are a goat and a car (2/3 of the time) or two goats (1/3 of the time). Someone privy to what's behind the doors deliberately opens one to reveal the goat. What changed? Nothing. The random outcome was decided before the door was opened and your odds don't change. That closed door now has a 2/3 chance of having a car behind it.
pieguy 12y ago
What many descriptions of the problem leave out is that it is critical that Monty knowingly picks a door with a goat. If Monty chooses at random, then the winning odds do indeed increase to 50% if he happens to open a door with a goat behind it. The randomized variant, by the way, tends to garner roughly the same degree of "you're completely wrong" responses when given to people who know just enough about the normal Monty-hall problem not to be fooled by it.
UnoriginalGuy 12y ago
I just want to start off by saying that I am not, at all, trying to place blame on Marilyn vos Savant for any of this. When someone gets declared as the smartest/best in the world it first off sets people's expectations so high that they expect perfection, and secondly it places the individual onto a plateau for which others would love to knock them down. The article doesn't discuss this, but I'm sure that played a part in why her detractors jumped on her so fast and so fiercely. They got told that this person was smarter than them, and they were inferior, and now was their chance at proving that that wasn't the case. She was of course right, and she likely didn't declare herself the "world's smartest woman" but unfortunately for her, that connotation was still attached to her character, and no matter how modest she was it would have been difficult for other people to take an ego knock by feeling like this person was "better" than them. Essentially the title of "world's smartest woman" set her up for something like this to happen. Luckily for her her detractors jumped on something she was correct about making themselves sound like morons. However it could have happened when she made her first public mistake (it would have happened eventually). Fortunately this episode may have helped release the pressure somewhat that was placed on her character. She saved face, but her detractors feel like they took her down a peg so didn't continuously stalk her looking for mistakes. PS - Just to reiterate I am in no way trying to blame Marilyn vos Savant or deflect any of it onto her. Someone else likely declared her the "world's smartest woman," that is outside of her control, and she dealt with this as well as anyone could hope to.
- misingnoglic 12y ago
  We could instead blame the egos of the PhDs who waited at the chance to tell her off at the first chance they thought they got.
- mcphage 12y ago
  > The article doesn't discuss this, but I'm sure that played a part in why her detractors jumped on her so fast and so fiercely. They got told that this person was smarter than them, and they were inferior, and now was their chance at proving that that wasn't the case. If only they had stopped to see if she was even wrong first, they wouldn't have proven the claim.
planckscnst 12y ago
The way I see the problem is slightly different and I think quite intuitive. If you've chosen a goat, then the host, by eliminating a door, is telling you exactly where the car is. If you have not chosen a goat, then the host not telling you where the car is. What are the chances that the host is telling you the location of the car? It's the same as the chances of you having chosen a goat the first time around.
Orangeair 12y ago
"You made a mistake, but look at the positive side. If all those Ph.D.’s were wrong, the country would be in some very serious trouble." Best quote in the entire article. And if I had to explain the problem in one sentence: Opening the first door obviously gives you a 1/3 chance, which remains unaffected by the host opening another door.
- danellis 12y ago
  > Opening the first door obviously gives you a 1/3 chance, which remains unaffected by the host opening another door. You had a 1/3 chance if you opened the second door too. Why does the second door's chance change, but the first door's chance is somehow locked by you having chosen it?
  - Orangeair 12y ago
    First door as in the first door to be selected. When you select it, it has a 1 in 3 chance of being the correct door. The door presented to you obviously has a 0 in 3 chance, so the remaining one must be 2 in 3.
  - erroneousfunk 12y ago
    The chance is locked in because, by the rules of the game, the host can't open that one and show you anything -- regardless of what it contains. You know that there's at least one door that the host can open (of the two remaining doors), so them opening a door changes nothing about the door you've chosen.
tmuir 12y ago
None of the explanations in the article really convinced me either. But after looking at the chart of all possible outcomes for a while, it finally clicked. It makes more sense to think of it in reverse from the winning position. Winning by switching doors requires your first choice to be incorrect. You have a 2/3 chance of guessing incorrectly when all three doors are closed. Thus, you have a 2/3 chance of putting yourself in the position from which switching will win.
sago 12y ago
She went on later to talk about the Tuesday's child problem, which is even more interesting and confounding. I have a friend who has two children, he has a son born on a Tuesday, what is the probability his other child is a girl? Compared with this problem I have a friend who has two children, one I know is a boy, what is the probability the other child is a girl? And this one I have a friend who has two children, the oldest is a boy, what is the probability the other child is a girl? Why do these three probabilities differ (in particular, since the last two may be more obviously different, why is the first one equal to neither of the others)? I spent the best part of a week in grad school pondering this until I came up with a deeper intuition. But the results are still surprising, I think.
- javajosh 12y ago
  My intuition is totally failing me here, because I can't get over the 1/2. If I have a friend who has one child, and he's about to have another, what's the probability that it will be a girl? Clearly 1/2. What's the difference between asking any of these questions and the one I just asked? I can't find any difference.
  - sago 12y ago
    Right, that's the last case. The eldest is a boy (or a girl, it doesn't matter), so the younger is equally likely to be either. The middle case is 2/3 likely to be a girl. Think of the possibilities of the sex of two children (sex given in birth order) M then M, M-F, F-M, F-F. I tell you one is a boy, so we have MM, MF, FM, in two of the three cases, the other is a girl. In the first case the answer is 14/27 a girl, so almost 1/2, but not quite. To figure this out you can consider all 196 combinations of two children with birth-days-of-the-week, cross out all those that don't match the phrase "he has a son born on a Tuesday", and you're left with 27 possibilities: 14 of them are when the other is a girl (14 because she could be born first or second, and on any day of the week 2x7=14), and 13 when the other is a boy (6 when the other is an older boy born on a non Tuesday, 6 when it is a younger boy born on a non Tuesday, and 1 when both are both on a Tuesday). So that's not the intuition, as much as the answer. The intuition is that as you add more specific knowledge about which child you mean, that question contributes less to the probability of sex, and is more an irrelevant detail. Being more specific can be anything. Do it with birthday-of-the-year, (more specific than day of the week) and it is even closer to 1/2. Do it with something unique, like their phone number (or their birth order, or 'the taller one is a boy') and it is exactly 1/2. Be less specific, make it so they both toss a coin and it is 'one of the children tossed heads and is a boy' and it is 3/7.
QuantumGood 12y ago
I'm a teacher, so I optimize for the result "understanding in my students." I teach it like this: You don't bring up Monty Hall first. It bends minds too much. I give a parallel game, where basics are similar, but you can choose to reveal 3 doors, 2 doors, or just one door each time. Everyone (after making sure they aren't being tricked) chooses to reveal three doors. Winner every time. Then I change the rules: Can't open all three doors. Have to choose between 1 or two. Everyone picks two. Only then do I give the Monty Hall problem, and show (as @bhc3 put it succinctly) that it's essentially the same game they just learned: your choice is between revealing two doors, or revealing one door. The most interesting part of this, and why I do it this way, is that I specifically optimize for my students to be able to explain what they know. I can teach someone to win an argument with someone who is convinced it's a 50/50 chance this way. Other ways, people sorta, kinda get it, but can't explain it to others who are confused. EDIT: The general principle here IMHO is teach something simple first, and show how it can be applied to something more complicated. When people think something is complicated, "breaking it down" into simpler concepts or making easy analogies doesn't have that great a track record. Their brain seems to have already classified it as "complex" and that has to be overcome —they have to be convinced to drop the tag #complex— in addition to however you teach the right understanding.
ucha 12y ago
The problem is not well-posed. The debate should be about its formulation, not about the math... The key here is to understand that no matter what you do, the game organizer will ALWAYS open one of the two others doors. Now, what if the organizer was to open a door only in some cases? He could open the door only when you choose the correct door. In that case, you have 0 chance of winning when you switch doors when the organizer opens another one. What makes the 1/2-1/2 probability - that so may math PhDs defended - acceptable is this: the organizer is selecting a door at random and opening it. Then if I choose a door and the organizer opens another one, I know that I have a 50% chance of having selected the correct one. Of course, he could also have opened my door in which case the game ends. This is a good example of how an ill-posed problem have different solutions depending on its implicit interpretation.
- mcphage 12y ago
  In the problem as stated, the organizer did open a door, which did contain a goat.
  - dllthomas 12y ago
    How the organizer chose the door matters.
    - mcphage 12y ago
      If you didn't know how the organizer chose the door (which you do, the original problem clearly states it), then your decision needs to be made independently of how the organizer chose it. All you can use are the facts in front of you: you chose a door, the organizer chose a different door, that contained a goat, and offered you the chance to choose. You need to make the best decision you can with the facts that you have.
      - dllthomas 12y ago
        My point was that your listing of things that happened in the past omitted an important piece of it. I actually don't agree that "the original problem clearly states it." Fully cognizent of the issues at hand, and with space for reflection entirely outside of the context of trying to solve the problem, I agree that it is clear that this is what the statement of the problem was trying to get at (particularly with the note that Monty knows which door has what) but it strictly does not say that that informs his choice (or how).
whoopdedo 12y ago
From Monty Hall's interview: > He retained the authority to offer the contestant cash NOT to switch. "I wanted to con [them] into switching there." So he wanted them to win the car. That seems interesting. I suppose it makes sense as this was an entertainment program. Viewers like a happy ending. Or if the guest accepts the cash and stays with the lower probability door, it's okay that they get the goat because they have cash. So win-win just for appearing. Also prevents unhappy guests from saying bad things about the show.
pbreit 12y ago
There's an important rule that isn't always communicated correctly: the participant must know ahead of time that the host is always going to open a door.
transfire 12y ago
There is a flaw in the chart. For the three listed Lose cases, it clumps two distinct choices into one line item and counts them as such, and thus provides an incorrect count of the possibilities. Moreover the basic argument is flawed. The knowledge of Monty has no barring on the knowledge of the contestant. If Monty simply presented you with three doors that you were to pick from, but opened an incorrect door before you choose your first door, then what would your odds be? The fact that he opens one after you choose does not matter. Monty has total knowledge of the setup and you still have zero.
- erroneousfunk 12y ago
  After you choose a door, Monty cannot choose it -- he's stuck choosing one of the two doors that you did not. So, the order of choosing the doors absolutely matters.
  - transfire 12y ago
    OK. You are right. I worked through the whole problem meticulously and figured it out. It's actually rather easy to explain once worked through, as follows: The odds of initially choosing correctly are 1 in 3. In that case if you switch you will loose, but if you stay you will win. The odds of choosing wrong initially are 2 in 3. In which case if you stay you will loose, but if you switch you will win. So it is actually that simple. If you always stay you will win 1/3rd of the time and if you always switch you will win 2/3rds of the time. It only seems like it should be 50/50 at first blush b/c that is what the odds would be if you choose to stay or switch arbitrarily. That is also why if you write down all the possibilities and simply add them up it will appear to be 50/50 too (which is what I did). The only thing I still can't quite figure out is exactly how the exposure of one of the renaming two doors increases one's knowledge. I thought I could take the list of possibilities and eliminate those that no longer apply and then add them up, but that doesn't seem to work.
vladislav 12y ago
I don't buy this at all : "Despite its deceptive simplicity, some of the world’s brightest minds -- MIT professors, renowned mathematicians, and MacArthur “Genius” Fellows -- have had trouble grasping its answer." The intuition is simple. Each time you initially pick a door with a goat, you win the car by switching. This clearly happens 2/3 of the time.
- diginet 12y ago
  That's not true, as Wikipedia says: "Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999)."
  - vladislav 12y ago
    Lol, Paul Erdos was 82 in 1995, which is when Vazsonyi claims they had this conversation, a year before Erdos died. Beyond his advanced age, it's well known that he did amphetamines his whole life and by that point his brain was pretty fried. I guarantee you if you gave the Monty Hall problem to the great Erdos in his primetime he would've thought it trivial.
transfire 12y ago
It didn't seem right to me at first, so I worked through it. The trick was to realize that switching always gives the opposite result of the initial choice. So if the initial choice is correct, switching will make it wrong. On the other hand, if the initial choice is wrong, switching will make it right. (Recall that Monty eliminated one wrong choice, so there is only one other choice from which to choose.) Since you have 1/3 chance of being right and 2/3 chance of being wrong on the initial pick, switching will reverse your odds.

This was fascinating. I can't believe so many academics got it wrong, and then attacked her gender over it!

bsdetector 12y ago
I'm not surprised. At the time, almost every question every week in Parade magazine was just pulp. Questions of the type "what's the world's smartest woman's favorite color?". You could take anybody off the street and get the same answers. Parade magazine was a gossip magazine. Furthermore, at the time she wasn't listed in the current Guinness Book so anybody that looked it up would see the column as a fraud unless they checked a specific old edition. Claiming somebody "is listed" when they had been listed, for one year many years ago, is a lie. So there were a lot of readers who thought of the column as a joke. If you open 99 doors that are full of mindless drivel, what are the odds the 100 is the one that has something interesting? This is why so many people were fooled by it.
- mcphage 12y ago
  > If you open 99 doors that are full of mindless drivel, what are the odds the 100 is the one that has something interesting? This is why so many people were fooled by it. So they thought "ha, this is foolishly wrong, I'll write in and tell them that—of course, without doing the 5 minute math check to make sure, first!"

I felt pretty stupid when it took me a long time to convince myself of the correct answer. But I felt better when I read that even "Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result"! http://en.wikipedia.org/wiki/Monty_Hall_problem http://en.wikipedia.org/wiki/Monty_Hall_problem

bmelton 12y ago
The only explanation that convinced me that my original guess was incorrect was this one (which is alluded to in the article): If there are three doors, the odds that the car is behind any of them is 1 in 3, or 1/3. Once you select a door, the odds are 1/3 for your door, 1/3 for each of the other, unselected doors. Put more simply, 1/3 for your door, or 2/3 that it is not in your door. Once another door is opened that does not have the car, then the odds of it being your door are 1/3, and the odds of it not being your door are still 2/3, so it makes sense to switch.
- mjevans 12y ago
  Thank you. Adding the temporal domain and system states is what makes this easy to understand. At the time of selection you freeze the probability of the sets (as this is the macroscopic world and the implications of the state in the other rooms must have collapsed due to interactions with all sorts of radiation from the rest of the universe). Thus what you say is exactly correct. * T0 - {1/3} {1/3} {1/3} * T1 - {1/3 Your pick} {2/3 = {1/3} + {1/3}} * T2 - {1/3 Your pick} {2/3 = {0/3 Host Pick} {2/3 Remaining door}}
mariodiana 12y ago
I read about this problem in the book "The Drunkard's Walk." I rejected the answer viscerally until I coded up a simple simulation (in Python) and ran it 10 million times. And then it clicked. Human beings are really bad at intuiting probability.

rhino42 12y ago

One interesting variant, is the case where the Host opens a door without knowing if that door had a goat. Expressing relief that he opened a goat door, he asks you to consider switching doors. IIRC, you get less information in this case, since there was a 1/3 chance that the host would accidentally pick the "car" door, but you still have a 1/2 chance of winning by switching.

rando3826 12y ago
> still have a 1/2 chance of winning by switching It seems that you are referring to the other variant, in which case this is false, because that one has a 2/3 chance of winning by switching.
- taeric 12y ago
  No, you have a 1/2 chance of winning after a door has been opened. However, you also have a chance of losing on that reveal. Something you don't have in the normal variants. (That is, in normal variants, you only lose at the end of the game. This one, you could effectively lose when the host opens the actual winning door.) So, your overall odds aren't as different as it seems. Just you have a much better odds of winning by switching, if you are still in the game.

MaysonL 12y ago

I still think that the problem is wrongly cast as a probability problem, rather than the game-theoretic problem that it is.

Terr_ 12y ago
A fair point, the hosts' choice of door is not random at all. He could open the door you'd already chosen, but that would make your choice to switch-or-not a no-brainer.

JohnTHaller 12y ago

It always amazed me that so many academics got it wrong. Once you understand the way the actual game works, it's pretty obvious that you gain a statistical advantage if you switch.

savanaly 12y ago
I know right? I don't blame them for getting it wrong at first glance-- I'm pretty sure 999 out of 1000 humans would. But the mathematics of the probabilities can be written down on a napkin they are so simple, so how was no one able to convince these scientists with a clear, concise technical proof? They really needed a computer simulation, as I've seen it claimed?
- function_seven 12y ago
  > But the mathematics of the probabilities can be written down on a napkin they are so simple, so how was no one able to convince these scientists with a clear, concise technical proof? I'm even more appalled that no one tried, you know, actually running the game. Have someone take a quarter and two pennies, mix them up under cups, let you choose one, show you a penny, then switch or not. Repeat several times. Switch places. When you're playing the role of Monty, the truth will reveal itself...
  - MichaelGG 12y ago
    I have a close friend that was incensed about my explanation about how he was misunderstand. He even ran a broken simulation that showed he was right. When I corrected his source code with detailed comments, he further accused me of trying to fool him and make fun of him. Didn't speak to me for a month. Bizarre. But I suppose this kind of thing that goes against instincts can really upset people.
dllthomas 12y ago
Note that it can depend on the wording of the question. Monty must know that he's revealing a goat for it to work out 2-to-1. If he picks randomly, your odds are 1-to-1 after he's revealed the goat.

Terr_ 12y ago

I'd re-frame it this way: "The host has zero incentive to reveal what's behind the door you already picked (or the choice of switching would be a no-brainer) so the fact that they left it shut doesn't tell you anything." "However, the fact that they didn't (or couldn't) open the remaining shut-door hints that it may contain the prize." The key is that the host's choice is not random.

ninkendo 12y ago
Exactly the way I like to think of it. When he opens the goat door there are now two possibilities: you were right the first time, or the remaining door has the car. By selecting the other door your are now betting against your initial choice, which was a 1/3 chance. Thus your new odds are 2/3.

mediocregopher 12y ago

So I believe the 2/3 answer to be true. I see the evidence in those tables. But intuitively, I still don't get it, and the explanation in the article isn't helping > The short answer is that your initial odds of winning with door #1 (⅓) don’t change simply because the host reveals a goat behind door #3; instead, Hall’s action increases the odds to ⅔ that you’ll win by switching. Well, why don't they change simply because the host reveals a goat? If you show me a three sided coin, I'd say there's a 1/3 chance of getting a particular side. If you then magically change gravity such that one of the other sides will never ever be landed on, would the two remaining sides not now have 50/50? I guess the article asserts that the "probability" doesn't get redistributed amongst the remaining options evenly, and I want to know why it doesn't and why it instead all "goes" to the one I didn't pick. I don't mean to be confrontational in asking, and I don't doubt the answer, I just don't get why. Could anyone explain it better, or point me somewhere that does?

jnbiche 12y ago
Because when you ponder this problem, you're considering the probability of two different states. When the host reveals the goat, the state of the problem has changed, and so has the probability of choosing a correct answer (it improved). Your mistake is somehow thinking they are the same state.
zaksoup 12y ago
The way I like to think about it is as follows: You pick door 1. Out of three doors you have a 1/3 chance of having a car behind your door. The chance that the car is behind one of either of the other two doors is 2/3. Let's pretend that instead of opening a door and revealing a goat, Monty instead says to you "You can switch to both of the remaining doors. If the car is behind either one of them you get to keep it." Your likelyhood of getting the car by switching is now 2/3. Instead, Monty does the statistical equivalent: He allows you to know for sure which of the other doors definitely has a goat.
- choffstein 12y ago
  This is a great answer. To answer the question about the 3-sided die: the question is flawed for 2 reasons. First, the way you phrased it, you know the information a priori. It's like me learning that the car isn't behind door #1 before I choose. In that case, I have a 50/50 chance on the other doors. But what if you didn't tell me before? This introduces our second problem. Let's say I roll the die and you hide it under a cup. I tell you I chose side A and you tell me A is physically impossible and ask me if I want to switch. Well of course I want to switch ... but I didn't learn any more information about sides B and C. It's a 50/50 toss-up. In terms of the game show, you basically just opened the constant's door. A critical element of the original problem is when the information is learned and knowing that the contestant's door will not be opened. It is what allows the phrasing from zaksoup above.

savanaly 12y ago

I read about this problem and the fact that multiple educated writers falsely called out vos Savant in an old Martin Gardner column, and it shines in my memory as one of the problems that illuminated math for me and made me see recreational math as worthwhile. Of course back when he wrote about it the topic of sexism was mercifully left out, and I honestly hadn't given a moment's thought to the fact that vos Savant was a woman until I read this new take on it, but there you go. Have standards for what makes something confusing and difficult to understand risen significantly over the decades? When I brought this problem up with my group of (admittedly geeky and mathy) friends no one was left unconvinced by the end of the afternoon, and most even had the intuition as well as the mathematical proof of the answer. Clearly we aren't any smarter than the public figures who couldn't comprehend this puzzle a half a century ago-- so it all in the framing of the problem?

sukilot 12y ago
Vos Savant originally posed the problem with ambiguous wording.
- mcphage 12y ago
  The original problem, from her website: "Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? Craig F. Whitaker Columbia, Maryland" (a) she didn't originally pose the problem, it was a question asked of her, and (b) the wording of the problem is not ambiguous.
  - bztzt 12y ago
    the ambiguity is that it's presented as something the host does in this particular instance, but the validity of the probability calculations depends on it being a protocol the host always follows (in particular, that the host will always reveal one of the goats not behind the contestant's first choice of doors).

bwy 12y ago

Impossible to graduate from Berkeley with a CS degree without hearing this problem :)

Strilanc 12y ago

An intuition pump for the problem, especially if you know about conservation of expected evidence [1], is that the asymmetry between the door you picked and the other door that wasn't opened is that the other door passed a test. The door we choose got a bye; we learned nothing about it's ability to win car-containing contests. The other door was at risk, and survived. We gained information about it. 1: http://lesswrong.com/lw/ii/conservation_of_expected_evidence/ http://lesswrong.com/lw/ii/conservation_of_expected_evidence...

andresp 12y ago
This was the clearest explanation in my opinion. I'm still a bit confused regarding how much information have we actually got and what mechanism is behind "passing a test" to influence the amount of information when the test result is not definitive (as in this case). We were just given a complete information about one of the options the user didn't choose, but not any direct new information specific of the door the user didn't choose and that remains closed. Imagine that after opening the door everyone goes home and a new participant who has no knowledge of the previous experiment is called to make a decision between the two remaining doors. Obviously in this case the probably is 1:2, as there is no information regarding the previous test. Actually, if we imagine a similar situation happened before the initial problem with a previous show with 4 doors, we may as well see that the initial probability of 1:3 is also wrong if we had access to the knowledge about the previous experiment. I still agree with the switch suggestion, but the mechanism behind it is still not entirely clear. A more intuitive example would also be to invert the problem: leave one of the non picked doors (C) out of the random opening and use the door the contestant picked (A) as a possible case in the random opening experiment. If we start by opening A, then we already know enough to see if the user won or lost, if we open B and there is the car behind it, we already know everything. If we open B and there is a goat behind the door, consider whether you would switch your choice from A to C. I guess you probably wouldn't and it would feel easier to justify your position, as this doesn't involve the "psychological pain" of changing a previous choice by reasons that you don't entirely understand (associated with possible loss-aversion feelings).

simple1 12y ago

One way to intuitively believe the answer is to imagine there were 100 doors, you picked one, then the host opened up 98 of the other doors, revealing goats. Switching would be the obvious answer in that case.

cdelsolar 12y ago
That's still not obvious to me
- peripitea 12y ago
  I'm a little intoxicated so this may be way over doing it or just not make sense, but... Let's make the number even bigger, so we don't even need to deal in probabilities -- a sextillion (10^21) doors. That's more doors than there are grains of sand on the planet. Say you're given a sextillion doors, and you pick one lonesome door, call it door A. What are the odds that door A is the right one? Essentially zero, right? I'm going to say straight up zero because now is not the time for you to be a dreamer. Ok so now that you've made your futile pick, poor Monty has to go and open every other door, all sextillion of them, except for yours and one other (let's call it door B) which he leaves closed. Every door he opens contains a goat. So now out of this vast universe of doors, there are (10^21-2) doors opened with goats bleating away, and only two doors left unopened: door A, aka yours; and door B, the one that Monty mysteriously left unopened. One of them must contain the car. (I'm guessing at this point it will be obvious for many people that door B contains the car, but I'll elaborate further in case it isn't clear yet.) So we're left with your door A and Monty's door B. At this point there are only two scenarios that matter. Either you picked the right door originally, in which case staying put is the winning move. OR, you picked the wrong door originally, in which case switching is the right move. And we already know you picked the wrong door (remember, no dreamers allowed here), so you should for sure switch to this by-now-extremely-conspicuous-looking door B. Door B isn't just some random door; the fact that it was chosen as the only door to not be opened tells you a tremendous amount of information about it. Think about the game from Monty's perspective. Either you pick the correct door initially and he gets to make door B a random goat door, or you picked the wrong door and he has no choice but to make door B the door with the car behind it. Since we know you picked the wrong door, we know what Monty's gonna have to do.

WillNotDownvote 12y ago

It's a testament to the brilliance of game show producers that they took something like this that both the contestant and the audience will misunderstand (but looks so simple!) and made a long running show out of it, making millions along the way.

elihu 12y ago

One thing that bothers me about the problem as it's stated in the article is that it isn't explicit about whether the game show host is required to open one of the goat doors regardless of which door the contestant picks, or whether the host is allowed to choose whether to open a door or not. If the host were to only open a goat door if the contestant initially picks the correct door, then the contestant would lose every single time if he/she always switched doors.

sukilot 12y ago
Yes, this was the source of the controversy and why vos Savant's self-congratulatory tempest was so annoying.
- hudibras 12y ago
  The original column and response is online and is unambiguous for fluent English readers.
hudibras 12y ago
I think you're splitting hairs here. It goes both ways: if the host were to only open a goat door if the contestant initially picks the correct door, then the contestants would quickly learn to stop switching doors.
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xom 12y ago
If the host can choose between doing nothing and revealing a goat, and the host wants the contestant to lose, then the contestant's best strategy is to never switch, in order to avoid being exploited by a host who only reveals a goat when switching would lose. Any reader who wrote in to say the probabilities were 50-50 after a goat was revealed was probably not thinking the host wanted the contestant to lose (or to win).
mcphage 12y ago
> whether the host is allowed to choose whether to open a door or not The problem as stated is: "you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3". Why would you think the host has the option of not opening a door? And if you think they could do that, why couldn't the host also do other things, like offer the contestant cash to walk away, or swap what's behind the doors, or add more doors, or open all three doors, or sucker-punch them in the gut?

_pius 12y ago

Quickest way I've seen to understand this result is referenced on Wikipedia: An intuitive explanation is that if the contestant picks a goat (2 of 3 doors) the contestant will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (1 of 3 doors) the contestant will not win the car by switching. The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability. http://en.wikipedia.org/wiki/Monty_Hall_problem http://en.wikipedia.org/wiki/Monty_Hall_problem

bhc3 12y ago

My crack at it... 1/3 chance prize is behind a given door. Pick a door, you've got a 1/3 chance it's behind it. But there's a 2/3 chance it's behind one of the other doors. So you've got two sets of outcomes at this point. Set A has 1/3 probability (the door you chose). Set B has a 2/3 probability (the two doors you didn't choose). You then get this incredibly valuable information. The door in Set B that doesn't have the prize. So now Set B still has a 2/3 probability of having the prize. But you know that higher probability now applies to only the one door in Set B. So you end up with: Set A door = 1/3 chance | Set B door = 2/3 chance Make the switch every time.

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dperny 12y ago
I've known the answer to the Monty Hall problem for a long time, but this particular explanation just so happens to be the first one that's brought me closer to grokking the answer. I dunno what's different about this one, but it makes sense to me. Congrats.
- BerislavLopac 12y ago
  I think that the key sentence in his explanation is "So now Set B still has a 2/3 probability of having the prize." This really nails the whole thing down.
- chrismcb 12y ago
  One thing to keep in mind, that most people don't understand, this only works if someone KNOWS which one is empty, and shows the door.
ntucker 12y ago
Yeah, I look at it this way: Let's just focus on set B. You're staring at two doors. Some random process decided whether the things behind the doors are a goat and a car (2/3 of the time) or two goats (1/3 of the time). Someone privy to what's behind the doors deliberately opens one to reveal the goat. What changed? Nothing. The random outcome was decided before the door was opened and your odds don't change. That closed door now has a 2/3 chance of having a car behind it.

pieguy 12y ago

What many descriptions of the problem leave out is that it is critical that Monty knowingly picks a door with a goat. If Monty chooses at random, then the winning odds do indeed increase to 50% if he happens to open a door with a goat behind it. The randomized variant, by the way, tends to garner roughly the same degree of "you're completely wrong" responses when given to people who know just enough about the normal Monty-hall problem not to be fooled by it.

UnoriginalGuy 12y ago

I just want to start off by saying that I am not, at all, trying to place blame on Marilyn vos Savant for any of this. When someone gets declared as the smartest/best in the world it first off sets people's expectations so high that they expect perfection, and secondly it places the individual onto a plateau for which others would love to knock them down. The article doesn't discuss this, but I'm sure that played a part in why her detractors jumped on her so fast and so fiercely. They got told that this person was smarter than them, and they were inferior, and now was their chance at proving that that wasn't the case. She was of course right, and she likely didn't declare herself the "world's smartest woman" but unfortunately for her, that connotation was still attached to her character, and no matter how modest she was it would have been difficult for other people to take an ego knock by feeling like this person was "better" than them. Essentially the title of "world's smartest woman" set her up for something like this to happen. Luckily for her her detractors jumped on something she was correct about making themselves sound like morons. However it could have happened when she made her first public mistake (it would have happened eventually). Fortunately this episode may have helped release the pressure somewhat that was placed on her character. She saved face, but her detractors feel like they took her down a peg so didn't continuously stalk her looking for mistakes. PS - Just to reiterate I am in no way trying to blame Marilyn vos Savant or deflect any of it onto her. Someone else likely declared her the "world's smartest woman," that is outside of her control, and she dealt with this as well as anyone could hope to.

misingnoglic 12y ago
We could instead blame the egos of the PhDs who waited at the chance to tell her off at the first chance they thought they got.
mcphage 12y ago
> The article doesn't discuss this, but I'm sure that played a part in why her detractors jumped on her so fast and so fiercely. They got told that this person was smarter than them, and they were inferior, and now was their chance at proving that that wasn't the case. If only they had stopped to see if she was even wrong first, they wouldn't have proven the claim.

planckscnst 12y ago

The way I see the problem is slightly different and I think quite intuitive. If you've chosen a goat, then the host, by eliminating a door, is telling you exactly where the car is. If you have not chosen a goat, then the host not telling you where the car is. What are the chances that the host is telling you the location of the car? It's the same as the chances of you having chosen a goat the first time around.

Orangeair 12y ago

"You made a mistake, but look at the positive side. If all those Ph.D.’s were wrong, the country would be in some very serious trouble." Best quote in the entire article. And if I had to explain the problem in one sentence: Opening the first door obviously gives you a 1/3 chance, which remains unaffected by the host opening another door.

danellis 12y ago
> Opening the first door obviously gives you a 1/3 chance, which remains unaffected by the host opening another door. You had a 1/3 chance if you opened the second door too. Why does the second door's chance change, but the first door's chance is somehow locked by you having chosen it?
- Orangeair 12y ago
  First door as in the first door to be selected. When you select it, it has a 1 in 3 chance of being the correct door. The door presented to you obviously has a 0 in 3 chance, so the remaining one must be 2 in 3.
- erroneousfunk 12y ago
  The chance is locked in because, by the rules of the game, the host can't open that one and show you anything -- regardless of what it contains. You know that there's at least one door that the host can open (of the two remaining doors), so them opening a door changes nothing about the door you've chosen.

tmuir 12y ago

None of the explanations in the article really convinced me either. But after looking at the chart of all possible outcomes for a while, it finally clicked. It makes more sense to think of it in reverse from the winning position. Winning by switching doors requires your first choice to be incorrect. You have a 2/3 chance of guessing incorrectly when all three doors are closed. Thus, you have a 2/3 chance of putting yourself in the position from which switching will win.

sago 12y ago

She went on later to talk about the Tuesday's child problem, which is even more interesting and confounding. I have a friend who has two children, he has a son born on a Tuesday, what is the probability his other child is a girl? Compared with this problem I have a friend who has two children, one I know is a boy, what is the probability the other child is a girl? And this one I have a friend who has two children, the oldest is a boy, what is the probability the other child is a girl? Why do these three probabilities differ (in particular, since the last two may be more obviously different, why is the first one equal to neither of the others)? I spent the best part of a week in grad school pondering this until I came up with a deeper intuition. But the results are still surprising, I think.

javajosh 12y ago
My intuition is totally failing me here, because I can't get over the 1/2. If I have a friend who has one child, and he's about to have another, what's the probability that it will be a girl? Clearly 1/2. What's the difference between asking any of these questions and the one I just asked? I can't find any difference.
- sago 12y ago
  Right, that's the last case. The eldest is a boy (or a girl, it doesn't matter), so the younger is equally likely to be either. The middle case is 2/3 likely to be a girl. Think of the possibilities of the sex of two children (sex given in birth order) M then M, M-F, F-M, F-F. I tell you one is a boy, so we have MM, MF, FM, in two of the three cases, the other is a girl. In the first case the answer is 14/27 a girl, so almost 1/2, but not quite. To figure this out you can consider all 196 combinations of two children with birth-days-of-the-week, cross out all those that don't match the phrase "he has a son born on a Tuesday", and you're left with 27 possibilities: 14 of them are when the other is a girl (14 because she could be born first or second, and on any day of the week 2x7=14), and 13 when the other is a boy (6 when the other is an older boy born on a non Tuesday, 6 when it is a younger boy born on a non Tuesday, and 1 when both are both on a Tuesday). So that's not the intuition, as much as the answer. The intuition is that as you add more specific knowledge about which child you mean, that question contributes less to the probability of sex, and is more an irrelevant detail. Being more specific can be anything. Do it with birthday-of-the-year, (more specific than day of the week) and it is even closer to 1/2. Do it with something unique, like their phone number (or their birth order, or 'the taller one is a boy') and it is exactly 1/2. Be less specific, make it so they both toss a coin and it is 'one of the children tossed heads and is a boy' and it is 3/7.

QuantumGood 12y ago

I'm a teacher, so I optimize for the result "understanding in my students." I teach it like this: You don't bring up Monty Hall first. It bends minds too much. I give a parallel game, where basics are similar, but you can choose to reveal 3 doors, 2 doors, or just one door each time. Everyone (after making sure they aren't being tricked) chooses to reveal three doors. Winner every time. Then I change the rules: Can't open all three doors. Have to choose between 1 or two. Everyone picks two. Only then do I give the Monty Hall problem, and show (as @bhc3 put it succinctly) that it's essentially the same game they just learned: your choice is between revealing two doors, or revealing one door. The most interesting part of this, and why I do it this way, is that I specifically optimize for my students to be able to explain what they know. I can teach someone to win an argument with someone who is convinced it's a 50/50 chance this way. Other ways, people sorta, kinda get it, but can't explain it to others who are confused. EDIT: The general principle here IMHO is teach something simple first, and show how it can be applied to something more complicated. When people think something is complicated, "breaking it down" into simpler concepts or making easy analogies doesn't have that great a track record. Their brain seems to have already classified it as "complex" and that has to be overcome —they have to be convinced to drop the tag #complex— in addition to however you teach the right understanding.

ucha 12y ago

The problem is not well-posed. The debate should be about its formulation, not about the math... The key here is to understand that no matter what you do, the game organizer will ALWAYS open one of the two others doors. Now, what if the organizer was to open a door only in some cases? He could open the door only when you choose the correct door. In that case, you have 0 chance of winning when you switch doors when the organizer opens another one. What makes the 1/2-1/2 probability - that so may math PhDs defended - acceptable is this: the organizer is selecting a door at random and opening it. Then if I choose a door and the organizer opens another one, I know that I have a 50% chance of having selected the correct one. Of course, he could also have opened my door in which case the game ends. This is a good example of how an ill-posed problem have different solutions depending on its implicit interpretation.

mcphage 12y ago
In the problem as stated, the organizer did open a door, which did contain a goat.
- dllthomas 12y ago
  How the organizer chose the door matters.
  - mcphage 12y ago
    If you didn't know how the organizer chose the door (which you do, the original problem clearly states it), then your decision needs to be made independently of how the organizer chose it. All you can use are the facts in front of you: you chose a door, the organizer chose a different door, that contained a goat, and offered you the chance to choose. You need to make the best decision you can with the facts that you have.
    - dllthomas 12y ago
      My point was that your listing of things that happened in the past omitted an important piece of it. I actually don't agree that "the original problem clearly states it." Fully cognizent of the issues at hand, and with space for reflection entirely outside of the context of trying to solve the problem, I agree that it is clear that this is what the statement of the problem was trying to get at (particularly with the note that Monty knows which door has what) but it strictly does not say that that informs his choice (or how).

whoopdedo 12y ago

From Monty Hall's interview: > He retained the authority to offer the contestant cash NOT to switch. "I wanted to con [them] into switching there." So he wanted them to win the car. That seems interesting. I suppose it makes sense as this was an entertainment program. Viewers like a happy ending. Or if the guest accepts the cash and stays with the lower probability door, it's okay that they get the goat because they have cash. So win-win just for appearing. Also prevents unhappy guests from saying bad things about the show.

pbreit 12y ago

There's an important rule that isn't always communicated correctly: the participant must know ahead of time that the host is always going to open a door.

transfire 12y ago

There is a flaw in the chart. For the three listed Lose cases, it clumps two distinct choices into one line item and counts them as such, and thus provides an incorrect count of the possibilities. Moreover the basic argument is flawed. The knowledge of Monty has no barring on the knowledge of the contestant. If Monty simply presented you with three doors that you were to pick from, but opened an incorrect door before you choose your first door, then what would your odds be? The fact that he opens one after you choose does not matter. Monty has total knowledge of the setup and you still have zero.

erroneousfunk 12y ago
After you choose a door, Monty cannot choose it -- he's stuck choosing one of the two doors that you did not. So, the order of choosing the doors absolutely matters.
- transfire 12y ago
  OK. You are right. I worked through the whole problem meticulously and figured it out. It's actually rather easy to explain once worked through, as follows: The odds of initially choosing correctly are 1 in 3. In that case if you switch you will loose, but if you stay you will win. The odds of choosing wrong initially are 2 in 3. In which case if you stay you will loose, but if you switch you will win. So it is actually that simple. If you always stay you will win 1/3rd of the time and if you always switch you will win 2/3rds of the time. It only seems like it should be 50/50 at first blush b/c that is what the odds would be if you choose to stay or switch arbitrarily. That is also why if you write down all the possibilities and simply add them up it will appear to be 50/50 too (which is what I did). The only thing I still can't quite figure out is exactly how the exposure of one of the renaming two doors increases one's knowledge. I thought I could take the list of possibilities and eliminate those that no longer apply and then add them up, but that doesn't seem to work.

vladislav 12y ago

I don't buy this at all : "Despite its deceptive simplicity, some of the world’s brightest minds -- MIT professors, renowned mathematicians, and MacArthur “Genius” Fellows -- have had trouble grasping its answer." The intuition is simple. Each time you initially pick a door with a goat, you win the car by switching. This clearly happens 2/3 of the time.

diginet 12y ago
That's not true, as Wikipedia says: "Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999)."
- vladislav 12y ago
  Lol, Paul Erdos was 82 in 1995, which is when Vazsonyi claims they had this conversation, a year before Erdos died. Beyond his advanced age, it's well known that he did amphetamines his whole life and by that point his brain was pretty fried. I guarantee you if you gave the Monty Hall problem to the great Erdos in his primetime he would've thought it trivial.

transfire 12y ago

It didn't seem right to me at first, so I worked through it. The trick was to realize that switching always gives the opposite result of the initial choice. So if the initial choice is correct, switching will make it wrong. On the other hand, if the initial choice is wrong, switching will make it right. (Recall that Monty eliminated one wrong choice, so there is only one other choice from which to choose.) Since you have 1/3 chance of being right and 2/3 chance of being wrong on the initial pick, switching will reverse your odds.