# Monty Hall Question **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [February 3, 2006, 8:47pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897 "2006-02-03T20:47:02Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Kamino\_Neko](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kamino_neko/32/34_2.png) [@Kamino\_Neko](https://boards.straightdope.com/u/Kamino_Neko)\ **Post date:** [February 7, 2006, 2:07pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/21 "2006-02-07T14:07:59Z")
I find it slightly ironic that math education has done so good a job eliminating the ‘gambler’s fallacy’, they’ve apparently managed to install a new fallacy - the idea that cumulative probability doesn’t exist. There’s a difference between the odds for any given trial in a sequence yielding a particular result, and the odds of the whole sequence yielding a particular set of results. A single flip of a coin, or a single birth has a 50/50 chance of being either result. However, with each subsequent flip, or each subsequent birth, the chances of every result being the same become smaller (they half each time, in this case). So, the first flip and the second both have a 50% chance of coming up heads. But, BECAUSE the first result doesn’t bias the second, the chances of both flips being heads are only 25%. Half the first flips will be heads - and half the flips following those heads will also be heads - or _one quarter_ of the total trials. As all the diagrams demonstrate. ---
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [February 7, 2006, 6:01pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/22 "2006-02-07T18:01:15Z")
**Zut** already gave a good rebuttal, but I’m in the mood to keep arguing… > [@SeeSaw](#): > > The only combinations possible at this point: > A = Male B = Female > A = Female B = Male > A = Female B = Female > these are absolutely the only combinations possible, and ALL the combinations possible, any more and any less is incorrect. Right. The fourth combination, A = Male and B = Male, is excluded by a stated condition of the problem. > [@](#): > > Its tempting to say that since theres only 3 possibilities, and two of them have a male, that the chance is 2/3 66%, however no one ever said THESE outcomes have equal chances. It’s more than tempting; it’s correct. The problem doesn’t say _explicitly_ that these three outcomes are equally likely, because that would destroy all the challenge. It’s not just some hunch pulled out of the air however. The four sex combinations given above are equally likely in the population of all two-child families. (Or anyway, in the “idealized” population which assumes boy and girl births are like coin-flips.) If we exclude any one of those combinations — or even more than one — the remaining combinations will still be equal in size to each other within the reduced population. > [@](#): > > Again, we ONLY know that one child is an F, not whether it was the first born or 2nd born. > 1st child 50% M 2nd child 0% M [impossible due to one child being F no matter what] > 2nd child 100% F [since M is 0% in this branch, F must be 100%] > 1M 2M = 50%_0% = 0.5_0.0 = 0.0 = 0% > 1M FM = 50%_100% = 0.5_1.0 = 0.5 = 50% > > If the first child born is a male, the other absolutely must be female 100%, agreed? In the restricted set of two-child families that we’re encountering, this last statement is certainly true (because all M-M families have been excluded). However, within that restricted set of families, the probability of the first child being a boy is not 50%. I think it might be a distracting mistake to consider the children by order: whether it’s birth order, or alphabetical order, or any other kind. This problem would be essentially the same if it were a guy dealing himself two playing cards at random, telling you he has at least one black card, and asking you the probability that the other card is red. Deal yourself pairs of cards, over and over and over (re-shuffling each time), and you’ll find that you get two blacks about 25% of the time, two reds 25% of the time, and a mixed pair 50% of the time. Whether you think of the cards as ordered relative to each other doesn’t matter; that will still be the breakdown. If the all-red hands are rejected as invalid, then mixed hands will occur in 2/3 of the remaining, valid hands. > [@](#): > > Answer = 75% chance that a family will have a male and female child combination out of all the child gender combinations possible in a family with 2 children. I see you’re the first person to arrive at an answer of 3/4, instead of 1/2 or 2/3. > [@](#): > > The key is, regardless of the fact that we know one child is a female, the first born has a 50% chance of being M or F. We didn’t know that one out of two children would be female at the time of birth of the first child. What we knew or didn’t know after the birth of the first child isn’t helpful. Think of the children as being fraternal twins, born by Caesarian section, both named “Pat”. They don’t have a sequence to them. Not one that’s relevant anyway to this problem. > [@](#): > > At this point the only thing left to argue is interpretation of the problem … _I would argue with that._ ---
**Author:** ![SeeSaw](https://avatars.discourse-cdn.com/v4/letter/s/a183cd/32.png) [@SeeSaw](https://boards.straightdope.com/u/SeeSaw)\ **Post date:** [February 10, 2006, 9:08am UTC](https://boards.straightdope.com/t/monty-hall-question/342897/23 "2006-02-10T09:08:21Z")
> [@zut](#): > > Long and involved, but incorrect. An easy way to see this is to note that the original problem has no information about “first” or “second” children. Thus, I could retype your entire post, substituting “first” for “second” (and vice versa) and have the argument be just as valid. However, I’d some up with the probability that 2M 1F = 50%_100% = 0.5_1.0 = 0.5 = 50%, which is clearly at odds with your post, which give the probability as 25%. Since these are conflicting, the underlying logic must be wrong. except… what youre doing, besides skipping the assumptions, and having your own interpretation of what the problem is asking for, is adding a redundant combination which is exactly the same as 1M 2F = 50%\*100%=50% And my interpretation assumes, a first and second children is assumed, seeing how in normal birth situations, theres is a child born first, and a child born second. Not something i believe can be flipped around at will > [@Tengu](#): > > I find it slightly ironic that math education has done so good a job eliminating the ‘gambler’s fallacy’, they’ve apparently managed to install a new fallacy - the idea that cumulative probability doesn’t exist. > > There’s a difference between the odds for any given trial in a sequence yielding a particular result, and the odds of the whole sequence yielding a particular set of results. > > A single flip of a coin, or a single birth has a 50/50 chance of being either result. > > However, with each subsequent flip, or each subsequent birth, the chances of every result being the same become smaller (they half each time, in this case). > > So, the first flip and the second both have a 50% chance of coming up heads. > > But, BECAUSE the first result doesn’t bias the second, the chances of both flips being heads are only 25%. > > …etc if “1F 2F = 50%_50% = 0.5_0.5 = .25 = 25%” isnt cumulative probability, i dunno what is. > [@Bytegeist](#): > > It’s more than tempting; it’s correct. The problem doesn’t say explicitly that these three outcomes are equally likely, because that would destroy all the challenge. It’s not just some hunch pulled out of the air however. > > The four sex combinations given above are equally likely in the population of all two-child families. (Or anyway, in the “idealized” population which assumes boy and girl births are like coin-flips.) If we exclude any one of those combinations — or even more than one — the remaining combinations will still be equal in size to each other within the reduced population. Difference in interpretation. Im not interpreting this as a situation where 4th outcome is eliminated because families cant have the combination. Im interpreting it as an elimination based on the the fact that we can assume 1 child is a girl, meaning it is technically impossible to gain that combination, which is still there but with a 0%, moving the 50% to the other combination. As i see it, two open doors, 50% chance of going through door A and 50% chance of going through door B, in every case where door A is entered, it leads to a room with 2 doors C and D and a 50% chance of going into either. Except C is locked. In every case where door A is entered, there is a 0% chance of going through locked door C and a 100% chance of going through the only other choice D > [@Bytegeist](#): > > What we knew or didn’t know after the birth of the first child isn’t helpful. Think of the children as being fraternal twins, born by Caesarian section, both named “Pat”. They don’t have a sequence to them. Not one that’s relevant anyway to this problem. Difference in interpretations, i THOUGHT i said that im assuming strange births and twins and etc are not calculated into my answer, but apparently i forgot to. As far as i can tell, the point of this question is to confuse people with its potential to be interpreted in many ways. Since the answer clearly changes depending on the interpretation ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [February 10, 2006, 1:19pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/24 "2006-02-10T13:19:57Z")
> [@SeeSaw](#): > > except… what youre doing, besides skipping the assumptions, and having your own interpretation of what the problem is asking for, is adding a redundant combination which is exactly the same as 1M 2F = 50%\*100%=50% Uh, no. Not really. What I was doing was _exactly the same thing_ that you were doing. I just substituted the word “first” for “second” and vice versa. My point, in fact, was that the original problem diesn’t include any information at all about “first” or “second” children, and there’s nothing special about either order. ---
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [February 10, 2006, 3:13pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/25 "2006-02-10T15:13:52Z")
> [@zut](#): > > Uh, no. Not really. What I was doing was _exactly the same thing_ that you were doing. I just substituted the word “first” for “second” and vice versa. My point, in fact, was that the original problem diesn’t include any information at all about “first” or “second” children, and there’s nothing special about either order. \*Precisely.\*It _doesn’t_ contain any information about the order. It is when it _does_ contain information about the order (or the relative weight, or hair color, or any other difference that will serve to distinguish the two) that the chance changes from 2/3 to 1/2. ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [February 10, 2006, 3:29pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/26 "2006-02-10T15:29:51Z")
There was another discussion somewhere with a slight twist to the birth order thing - you run into an old acquaintance at the grocery store, and she has a boy with her. She informs you that she now has two kids, but you have no indication whether the one with her is the older or younger one. Now what is the probability that the other is a boy? Of course it’s 0.5. This wording might be helpful for people who get stuck on the birth order aspect, thinking that’s what makes the difference, when it’s instead the difference between saying that a specific kid is the boy vs. the non-specific statement that there is at least one boy out of the two kids. ---
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [February 10, 2006, 4:45pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/27 "2006-02-10T16:45:12Z")
> [@Bytegeist](#): > > The four sex combinations given above are equally likely in the population of all two-child families. […] If we exclude any one of those combinations — or even more than one — the remaining combinations will still be equal in size to each other within the reduced population. > [@SeeSaw](#): > > Difference in interpretation. Im not interpreting this as a situation where 4th outcome is eliminated because families cant have the combination. Nor was I, nor the others here who say the answer is 2/3. I don’t think there’s a difference between us in our interpretation of the problem’s text. The male-male combination is eliminated for the reason you describe next, which we agree on… > [@](#): > > Im interpreting it as an **elimination based on the the fact that we can assume 1 child is a girl** , meaning it is technically impossible to gain that combination, which is still there but with a 0%, moving the 50% to the other combination. But, you just can’t “move” the 50% around like that. The problem disqualifies 25% of the entire two-child population from consideration. If any percentages are getting redistributed, it’s the 25% that’s being chopped up and added to the other three categories, making the final sample population 33/33/33. However, this is muddy thinking, I’d say. I don’t think it’s a helpful way to think about probability — which is not really like redistributing water among a bunch of glasses. > [@](#): > > As i see it, two open doors, 50% chance of going through door A and 50% chance of going through door B […] That problem just isn’t analogous to the original one. > [@](#): > > Difference in interpretations, i THOUGHT i said that im assuming strange births and twins and etc are not calculated into my answer, but apparently i forgot to. There was no need to. Nobody’s trying to shake any tabasco sauce into the stew. For example, I mentioned “fraternal” twins in my previous post, not “identical”. I was still adhering to the given that the kids’ sexes are independent of each other. In any case, there are no strange families being snuck in. No hermaphrodites or clones or anything. No time travel. The problem essentially tells us to assume that the kids are like coin flips. (From the text: “… assuming the biological odds of having a male or female child are equal …”) The sexes are equally likely, and independent of each other. It’s under these very assumptions that the probability asked for turns out to be 2/3. > [@](#): > > As far as i can tell, the point of this question is to confuse people with its potential to be interpreted in many ways. Since the answer clearly changes depending on the interpretation I can’t agree with that. The only thing I find mildly, and briefly, vague about the problem’s text is the sentence, “You have been told this family has a daughter.” In ordinary English this could mean either (1) they have _at least_ one daughter, or (2) they have _exactly_ one daughter. But since the problem would be silly and trivial under the second interpretation, we can pretty safely assume we’re supposed to take on the first one. Otherwise, I don’t see what the different interpretations could be. What differences are there in our interpretations of this problem? ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [February 10, 2006, 10:22pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/28 "2006-02-10T22:22:25Z")
I’ll try to explain the logic behind the 2/3 answer for the son/daughter question step by step. Stop me when you think I’ve done something wrong. This will help us all pinpoint what people are misunderstanding. I suspect many who disagree w/ the 2/3 answer will disagree with my sentence number 1. But we’ll see. 1. The situation described can be represented semi-formally as follows: events P and Q are both of probability 1/2. You are told “Either P or Q (or both) obtains.” You are asked, in the light of this information, what the probability is that _both_ P and Q obtain. (P in this problem is “The family’s first child is a daughter.” Q is “The family’s second child is a daughter.” The problem does not specifically reference “first” or “second” children, but look at what it does say: “(At least) one of the children is a daughter.” At least one of the C’s is a D. What can this mean but “Either C1 is a D, or C2 is a D, or C3 is a D,…, or CN is a D, or some combination of the foregoing are all true.” with N==the number of C’s. In the problem at hand, N==2.) This is equivalent to asking, by the way, “What are the chances that both are girls” rather than “what are the chances that the other is a boy.” But the math is easier this way, and the last few steps of this argument will make it clear that answering either question gives you the answer to the other question. 2. The probability that both P and Q obtain is 1/4. 3. For all X and Y, where the probabilities of X and Y are not dependent on each other, where X has probability a/b and Y has probability c/d “Either X or Y or both” has a probability equal to ((ad+bc)-ac)/bd (i.e. Prob(X) + Prob(Y) - (Prob(X and Y)) 4. From 1 and 2 it follows that the probability of either P or Q or both obtaining (prior to your being told that it in fact does) is 3/4. 5. Any event in which both P and Q obtain is also an event in which either P or Q or both obtain, but not vice versa. 6. When two events X and Y are such that any event in which X obtains is an event in which Y obtains, then the probability of X _given_ that Y obtains is prob(X)/prob(Y). 7. From 2, 4, 5, and 6 it follows that the probability that P and Q both obtain, given that either P or Q or both obtain, is (1/4)/(3/4), which simplifies to a probability of 1/3. 8. Translating back from 7–given that one child is a girl, the probability that both children are girls is 1/3. 9. From 8 it follows that, given that one child is a girl, the probability that the other child is also a girl is 1/3. 10. From 9 it follows that, given that one child is a girl, the probability that the ohter child is not a girl is 2/3. In other words, given that one child is a girl, the probability that the other child is a boy is 2/3. I know I didn’t make that as easy to follow as it possibly could have been, but tell us what you think is wrong with the argument, or what you don’t understand about it, and this will help us see what people are misunderstanding in the discussion. -FrL- ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [February 11, 2006, 5:02am UTC](https://boards.straightdope.com/t/monty-hall-question/342897/29 "2006-02-11T05:02:10Z")
Frylock, my eyes rolled back in my head before I ever got to point #2, and I’m pretty geeky. Let me take a stab at an intuitive explanation. Take a typical cross section of two-kid families. Of these, about 50 will have a boy and a girl, about 25 will have two boys, and about 25 will have two girls. Agreed so far? By the statement that this family has a daughter, you’re simply ruling out the 25 who have two boys, and getting no more information. Out of those 75 families, 25 will have two girls: 1/3 of them. That’s pretty simple, right? ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [February 11, 2006, 7:46am UTC](https://boards.straightdope.com/t/monty-hall-question/342897/30 "2006-02-11T07:46:35Z")
> [@CurtC](#): > > Frylock, my eyes rolled back in my head before I ever got to point #2, and I’m pretty geeky. Let me take a stab at an intuitive explanation. Take a typical cross section of two-kid families. Of these, about 50 will have a boy and a girl, about 25 will have two boys, and about 25 will have two girls. Agreed so far? > > By the statement that this family has a daughter, you’re simply ruling out the 25 who have two boys, and getting no more information. Out of those 75 families, 25 will have two girls: 1/3 of them. > > That’s pretty simple, right? I wasn’t trying to offer an intuitive explanation. Plenty of good intuitive explanations have already been posted here and they’re not convincing anyone. Instead, I was trying to give a (readable but still pretty) rigorous step-by-step demonstration of the conclusion. Then whatever people are not understanding can be pinpointed with some exactness. -FrL- ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [February 12, 2006, 9:22pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/31 "2006-02-12T21:22:18Z")
While some people are having difficulties with the calculations involved, the statement of the problem still has an epistemological difficulty. This was alluded to in the alternate problem given by **CurtC** and Len Ragozin. It may also be behind some of the problems with understanding why probability changes based on additional information. I think it’s easiest to understand in terms of objects. ‘Children’ can be easily substituted, subject to some assumptions below. Let there be two types of ball, pink and blue, that are picked from a pile with equal probability (i.e. 50%). Two are chosen at random. I pick them up and put them in a hat. So I know what they are, you don’t. Case 1 : I tell you, “One of the balls in the hat is pink.” _or_ Case 2 : I say nothing. You reach in the hat, and pull one ball out of the hat without looking (at the other). It’s pink. Now in both cases, the same statement could be made : **(At least) one of the balls is pink.** It should be obvious, however, that in Case 1 there is more information available. You are learning about it from someone who does know more than you. This is what changes the problem, leading to different answers to the question, “What is the probability the other ball is blue?” [sup]\*[/sup] Going back to the child problem, this shows the difficulty. The assumptions for the mathematical calculation depend on how you know about the children. It is implicit in this example that I am like a parent - of course they know their own children, so if they tell you, then you can determine better (i.e. what to rule out). In the situation where you, say, meet the child, even with the parent present, it’s like Case 2. You’ve randomly picked a particular child, and you haven’t learned anything more. Even I could have pulled the ball out of the hat, as long as I did so without preference. Examining the wording, all we have is, “You have been told this family has a daughter.” Unfortunately, we do not know **who** told you or how they found this out. If it is a parent, it’s apparent they do know more information. But it could conceivably be someone whose kid goes to school with their daughter, and they don’t know anything about the other child. [sup]\*[/sup]This is essentially the basis of what I consider the best interpretation of what probability means (Bayesian interpretation). I know with certainty (100%) what is in the hat; the probability you have is less, because you have less information. This also underlies the Monty Hall problem. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [February 13, 2006, 1:55am UTC](https://boards.straightdope.com/t/monty-hall-question/342897/32 "2006-02-13T01:55:09Z")
> [@](#): > > Going back to the child problem, this shows the difficulty. The assumptions for the mathematical calculation depend on how you know about the children. Exactly the point I was going to make. Depending on how you came by the information, the answer could go either way. For instance (I think I’ve posted this example before, but the search function disagrees): Suppose my old (all-male) high school holds an appreciation dinner for parents of alumni. My mother goes, and strikes up a conversation with the fellow sitting next to her. Over the course of the conversation, she asks him how many kids he has, and he says “Two”. At this point, my mom knows that this fellow has two kids, since he said so. She also knows that at least one of his kids is male, since he’s a parent of an alumnus, and it’s an all-boys school. From my mom’s point of view, what is the probability that both the fellow’s children are male? 2/3, as argued above. Of course, if she gets additional information, the probabilities can change. Suppose, for instance, that later in the conversation, the man says “I’m sorry I’m so inattentive; my toddler cried all night last night, and I didn’t get any sleep.”. What’s the probability now, from my mom’s point of view, that the man has two boys? Now, the problem is completely different. She knows that he has two kids. She knows that at least one is a high school graduate. She also knows that at least one is a toddler. She knows that the graduate is male, and it’s a 50-50 chance that the toddler is male. The difference between the two situations is that now, she has an identifier to attach to the known male. She couldn’t use “the high-school grad” as an identifier, before, since for all she knew, both kids graduated from high school. Heck, they could both have gone to my alma mater. Likewise, she can’t use “The boy” as a label, since they might both be boys. With the new information that one’s a toddler, though, she now knows that one and only one kid is a high school grad, so that can be a label. ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [February 13, 2006, 3:45pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/33 "2006-02-13T15:45:39Z")
> [@panamajack](#): > > It should be obvious, however, that in Case 1 there is more information available. I disagree with that - in Case 2, the one where you have a pink ball in your hand, there is more information available. In our method of listing out the four equally-likely possible outcomes, and seeing which ones we can rule out, then in Case 1 (you’re told there is at least one pink ball) you have enough information to rule out the two blue ball scenario. But in Case 2, you have enough information to rule out two scenarios: the two that would result in your holding a pink ball. > [@Chronos](#): > > At this point, my mom knows that this fellow has two kids, since he said so. She also knows that at least one of his kids is male, since he’s a parent of an alumnus, and it’s an all-boys school. From my mom’s point of view, what is the probability that both the fellow’s children are male? 2/3, as argued above. I believe this statement is wrong. Since your mom knows _which_ kid (the classmate of yours) is male, the probability that the other is male is 1/2. This is exactly analogous to my scenario above where you see an old acquaintance in the grocery store with a boy. The probability of her other being a boy is 1/2. ---
**Author:** ![wissdok](https://avatars.discourse-cdn.com/v4/letter/w/c4cdca/32.png) [@wissdok](https://boards.straightdope.com/u/wissdok)\ **Post date:** [February 13, 2006, 4:11pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/34 "2006-02-13T16:11:10Z")
I haven’t read every post, so I’m sorry if I am covering old ground. Cecil did a good job of explaining the Monty Haul question, but he was wrong in the girl and the sibling question. He undervalued possibility of number #2, while doubling the normal value of #1 and #3 by splitting them up. Cecil wrote : 1. Child A is female and Child B is male. 2. Child A is female and Child B is female. 3. Child A is male and Child B is female. 4. Child A is male and Child B is male. Even though we know one is a girl (and that eliminates #4), we don’t know which female is the girl we already know. A correct way of writing this is 1. Child A (Girl) is female and Child B is male. 2. Child A (Girl) is female and Child B is female. 3. Child A is female and Child B (Girl) is female. 4. Child A is male and Child B (Girl) is female. OR The GIRL is the oldest with a younger brother. The GIRL is the oldest with a younger sister. The GIRL is the youngest with an older brother. The GIRL is the youngest with an older sister. Either way we are looking at a 50/50 chance boy or girl. ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [February 13, 2006, 5:01pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/35 "2006-02-13T17:01:46Z")
> [@wissdok](#): > > A correct way of writing this is > > 1. Child A (Girl) is female and Child B is male. > 2. Child A (Girl) is female and Child B is female. > 3. Child A is female and Child B (Girl) is female. > 4. Child A is male and Child B (Girl) is female. > > OR > > The GIRL is the oldest with a younger brother. > The GIRL is the oldest with a younger sister. > The GIRL is the youngest with an older brother. > The GIRL is the youngest with an older sister. > > Either way we are looking at a 50/50 chance boy or girl. Now I’m confused, **wissdok**! I had just come to terms with the probabilities being 1/3 and 2/3, but I do believe you’re right! ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [February 13, 2006, 6:04pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/36 "2006-02-13T18:04:07Z")
> [@wissdok](#): > > I haven’t read every post, so I’m sorry if I am covering old ground. > > Cecil did a good job of explaining the Monty Haul question, but he was wrong in the girl and the sibling question. He undervalued possibility of number #2, while doubling the normal value of #1 and #3 by splitting them up. > > Cecil wrote : > > 1. Child A is female and Child B is male. > 2. Child A is female and Child B is female. > 3. Child A is male and Child B is female. > 4. Child A is male and Child B is male. > > Even though we know one is a girl (and that eliminates #4), we don’t know which female is the girl we already know. A correct way of writing this is > > 1. Child A (Girl) is female and Child B is male. > 2. Child A (Girl) is female and Child B is female. > 3. Child A is female and Child B (Girl) is female. > 4. Child A is male and Child B (Girl) is female. > > OR > > The GIRL is the oldest with a younger brother. > The GIRL is the oldest with a younger sister. > The GIRL is the youngest with an older brother. > The GIRL is the youngest with an older sister. > > Either way we are looking at a 50/50 chance boy or girl. Something has to be wrong here. The probability of “P and Q” given “P or Q or both”, where prob(P) and prob(Q) are both 1/2, (and given a couple of other givens I won’t go into,) is 1/3. That’s just the way the math works, straightforwardly. See my previous post, and tell me where you think it goes wrong. Something must be wrong with the way you’re formulating the problem, but I can’t figure it out right now. My intuition is that it is illicit for you to formulate it in a way such that anything depends on _which_ of the two siblings we’re being told about when we’re told “one is a girl” since we’re _not_ being told this about either sibling in particular, but rather, just about the pair. But I’m having trouble formulating this intuition more clearly. Maybe someone else can step in and do so. -FrL- ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [February 13, 2006, 6:08pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/37 "2006-02-13T18:08:22Z")
> [@wissdok](#): > > Even though we know one is a girl (and that eliminates #4), we don’t know which female is the girl we already know. A correct way of writing this is > > 1. Child A (Girl) is female and Child B is male. > 2. Child A (Girl) is female and Child B is female. > 3. Child A is female and Child B (Girl) is female. > 4. Child A is male and Child B (Girl) is female. Assuming that we’re told which child is female at random, scenarios 2) and 3) are only half as likely as scenarios 1) and 4). ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [February 13, 2006, 6:27pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/38 "2006-02-13T18:27:39Z")
> [@Frylock](#): > > Something has to be wrong here. The probability of “P and Q” given “P or Q or both”, where prob(P) and prob(Q) are both 1/2, (and given a couple of other givens I won’t go into,) is 1/3. That’s just the way the math works, straightforwardly. See my previous post, and tell me where you think it goes wrong. > > Something must be wrong with the way you’re formulating the problem, but I can’t figure it out right now. My intuition is that it is illicit for you to formulate it in a way such that anything depends on _which_ of the two siblings we’re being told about when we’re told “one is a girl” since we’re _not_ being told this about either sibling in particular, but rather, just about the pair. But I’m having trouble formulating this intuition more clearly. > > Maybe someone else can step in and do so. > > -FrL- Okay, here it is: We’re not being told anything about any individual sibling. There is no such thing as “The girl sibling” in this problem. We are not dealing with individuals, but with pairs. There are three kinds of pair-entities. Call them doubleG’s, singleG’s, and doubleB’s. SingleG’s occur with probability 1/2. DoubleG’s and DoubleB’s each occur with probability 1/4. We have that information, and we have the information that the entity we’re talking about is not a doubleB. That leaves us with a clear 2/3 chance that what we’re dealing with is a singleG. The move some people are making is tantamount to saying “But there are two kinds of doubleG’s–the kind where we know about the first sibling’s gender, and the kind where we know about the second sibling’s gender.” But A. It’s not true that we know the gender of any particular person. All we know is that the _pair_ is such that it is not a doubleB. B. Even if we allow for an observation that “In a doubleG, it could be that it’s the younger sibling that’s the girl, and it could be that the older sibling that’s the girl,” it’s clear that neither of these possibilities is relevant to the calculation of the probability that there are two girls (and therefore that the “other sibling” is not a boy). When we list the following as two separate possibilities: 1. THE GIRL has a younger female sibling 2. THE GIRL has an older female sibling we fail to mark out two different situations with respect to the problem. In the problem at hand, both of these descriptions point to exactly the same possibility–that “the other sibling” is a girl. It is illicit to tease them apart and assign them separate probabilities. Strictly speaking, 1. THE GIRL has a younger male sibling 2. THE GIRL has an older male sibling _also_ describe exactly the same situation with respect to the problem. It is, strictly speaking, illicit to tease them apart as well–they each describe the situation in which “the other sibling” is a boy–a situation which demonstrably has a probability of 2/3 given that it is known that the pair is not a doubleB. With all of that said, we can nevertheless allow for _this_ way of listing the possibilities: 1. The pair is a doubleG 1a. We were told about the older sibling 1b. We were told about the younger sibling 2. The pair is a singleG 2a. We were told about the older sibling 2b. We were told about the younger sibling. Given that the pair is not a doubleB, the list exhausts the possiblities. Furthermore, 1a and 1b exhaust the possibilities given 1, and are equally likely, and 2a and 2b exhaust the possibilities given 2, and are equally likely. Given that the probability of a doubleG in general is 1/4, of a singleG in general is 1/2, and of a doubleB in general is 1/4, then it follows that: The probability of 1 is 1/3. The probabilities of 1a and 1b are each 1/6. The probability of 2 is 2/3. The probabilities of 2a and 2b are each 1/3. That’s the best I can do right now. -FrL- ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [February 13, 2006, 6:31pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/39 "2006-02-13T18:31:50Z")
> [@ultrafilter](#): > > Assuming that we’re told which child is female at random, scenarios 2) and 3) are only half as likely as scenarios 1) and 4). Thanks. ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [February 13, 2006, 6:35pm UTC](https://boards.straightdope.com/t/monty-hall-question/342897/40 "2006-02-13T18:35:59Z")
[QUOTE=CurtC] I disagree with that - in Case 2, the one where you have a pink ball in your hand, there is more information available. ] I misspoke … what I should have said is, “more information about the unknown ball”, or “more information to help you solve the problem.” In **Chronos** ’s first example, the identification isn’t enough to distinguish. Both children could be alumni. The dinner is not necessarily honoring a particular class (so the child is not necessarily a classmate of **Chronos**. [Previous page](https://boards.straightdope.com/t/monty-hall-question/342897.md?page=1) [Next page](https://boards.straightdope.com/t/monty-hall-question/342897.md?page=3)