# Boy/Girl probability in Monty Hall **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [August 10, 2006, 6:08pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182 "2006-08-10T18:08:19Z")\ **Posts on this page:** 20\ **Page:** 6
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [August 13, 2006, 11:09pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/101 "2006-08-13T23:09:42Z")
[QUOTE=tim314] After all this typing, it would be _really_ nice if we could come to some agreement. [/QUOTE] _ **Ha!** _ > [@wissdok](#): > > The second sentence where we are told of a daughter is, by its very nature, an afterthought. In neither of these sentences does it say that having a daughter was a condition for choosing this family. I will state again, from what we are given, a random family is chosen and then a random child of that family. No child whatsoever is being chosen, for anything. No specific child is having its sex identified, and no specific child is having its sex asked about. It’s not that having a daughter was a condition for choosing the family. It’s that we’re told a number of facts about a family, among them: (1) there are exactly 2 children, and (2) the number of girls is at least 1. There are families in the universe meeting these criteria, and then there are families that don’t. We are asked a probability question about a family known to meet these criteria. We can ignore all the ones that don’t, since if one of them were a possibility, it would mean someone has been lying to us. (And at that point, the math problem evaporates in a puff of logic.) ---
**Author:** ![adbadqc](https://avatars.discourse-cdn.com/v4/letter/a/c4cdca/32.png) [@adbadqc](https://boards.straightdope.com/u/adbadqc)\ **Post date:** [August 14, 2006, 2:23am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/102 "2006-08-14T02:23:22Z")
For the 2/3rds crowd; look at it this way. We have our three groups left after tossing the BB crowd: BG GB GG From this you cry “2/3rds”! Consider these individuals: Albert / Betty Charlotte / Danny Edwina / Florence You have two girls who have brothers (Betty and Charlotte), and two girls who have sisters (Edwina and Florence). For each girl, there is a 50-50 chance her sibling is a boy. You have to “count” each side of the GG pair if you insist on keeping all three “includes a girl” pairings, instead of tossing one as I have suggested earlier. Even in this universe, the odds any given girl has a brother is 50/50. You have to do it this way if you insist “it doesn’t matter whether the girl is the first one in the pair or the second”. If you go down that route, then you have to consider both sides of the GG pairing as separate cases, because you said they are and therefore you have to treat Edwina and Florence as separate cases. Quod Erat Demonstrandem ---
**Author:** ![wissdok](https://avatars.discourse-cdn.com/v4/letter/w/c4cdca/32.png) [@wissdok](https://boards.straightdope.com/u/wissdok)\ **Post date:** [August 14, 2006, 2:25am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/103 "2006-08-14T02:25:37Z")
**Cecil’s original statement:** _There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)_ > [@](#): > > Posted by Bytegeist > No child whatsoever is being chosen, for anything. No specific child is having its sex identified, and no specific child is having its sex asked about. > > It’s not that having a daughter was a condition for choosing the family. It’s that we’re told a number of facts about a family, among them: (1) there are exactly 2 children, and (2) the number of girls is at least 1. There are families in the universe meeting these criteria, and then there are families that don’t. We are asked a probability question about a family known to meet these criteria. We can ignore all the ones that don’t, since if one of them were a possibility, it would mean someone has been lying to us. (And at that point, the math problem evaporates in a puff of logic.) Choice #1 Our source is free to tell of either child in a family. For families that have two girls or two boys our source always give the same answer. For families that have one of each child, our source is free to give information about either child. This example is a natural one; we are at a cocktail party and told about a couple we don’t know. This couple wasn’t checked out by our source to see they had a girl, he just told us what he knew. Our source could have just as likely known of a boy instead. The knowledge of the child is random. If our source tells us about other families, some he might know of a girl, some of a boy. There is no indication in the version that our source will know only of families with at least one girl. (1/2) Choice #2 Our source tells us only of families with girls. In this scenario our source has done his homework and questioned the families beforehand, so we will always be given a family with girls. In this version all the families are of a selected group. This family isn’t just some family we meet on the street, this family was predetermined for this question. The families with both a boy and a girl must alway provide us with the clue “at least one girl”. (2/3) > [@](#): > > Posted by Xema > Note that in post #22 you say that there is just one problem, not 2. It would have saved a lot of trouble if at that time you’d acknowledged that 2/3 is the correct answer to one problem - the one I’m claiming is the “mainstream” interpretation of what Cecil posed. I still say there is only one problem here, because the other problem wasn’t presented. I clearly see a period between the family with two children and the statement about the daughter. A period normally ends a single complete statement of thought. As for the mainstream interpretation, I didn’t start this thread. I didn’t start any of the half dozen threads that have come up about this subject in the last few months. I would have to say not many people respond in the forum, so I don’t know what is mainstream’s opinion is. I can tell you that this has been debated on almost every forum you can think of. I also know from Marilyn vos Savant(who had a boy version), that this puzzle was the second most challenged topic she had ever covered (the first being the Monty Hall). As I posted earlier, \*Scientific America \*also got enough responses that they took another look at the problem. If you Google or Yahoo this problem, you will find that most scholars that even post anything on logic, will proclaim this problem ambiguous and point out how it must be stated correctly to get the desired result. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [August 14, 2006, 2:55am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/104 "2006-08-14T02:55:21Z")
[QUOTE=adbadqc] You have to “count” each side of the GG pair if you insist on keeping all three “includes a girl” pairings [/QUOTE] I think we’ve made it clear that it’s wrong to “count” each side of the GG pair. Why? Because they are in one family, and it’s a family that was selected, not a child. Let’s (yet one again) stipulate that _if_ the problem called for the random selection of a child, the correct answer would be a probability of 1/2. But for that to be true, it’s necessary to believe that “You have been told this family has a daughter” is identical to saying “A child is selected at random and proves to be female”. Do you believe this? ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [August 14, 2006, 3:07am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/105 "2006-08-14T03:07:27Z")
[QUOTE=wissdok] I still say there is only one problem here, because the other problem wasn’t presented. [/QUOTE] To the contrary, the two possible problems were clearly described quite early in the thread. See JR Brown’s post #11 or my post #15. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [August 14, 2006, 5:34am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/106 "2006-08-14T05:34:18Z")
[QUOTE=wissdok] If you Google or Yahoo this problem, you will find that most scholars that even post anything on logic, will proclaim this problem ambiguous and point out how it must be stated correctly to get the desired result. [/QUOTE] Please cite with some google search words, because I’m not finding it. And working, as I do, in the world inhabited by these scholars who publish on logic, I feel some confidence in making the claim that your statement here is incorrect. I do not think most scholars would call this puzzle as originally formulated ambiguous, unless they meant it is ambiguous in the way any sentence in any language can be made “ambiguous” by asking just exactly what the speaker “really meant” when s/he uttered it. -FrL- ---
**Author:** ![tim314](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tim314/32/3468_2.png) [@tim314](https://boards.straightdope.com/u/tim314)\ **Post date:** [August 14, 2006, 9:53am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/107 "2006-08-14T09:53:28Z")
[QUOTE=wissdok] May I ask, where have you concluded we were given a restricted family? The first sentence where we are given a family, neither mentions a son or daughter. From this we can conclude that the family was chosen first from all available families. [/QUOTE] It’s a conditional probability, **wissdok**. The point of the question is to determine the probability that they have a son, given that we _know_ they have at least one daughter. So yes, the family was selected at random to _begin_ with, but then we have been given further information that narrowed it down. Here’s a simpler example: Suppose I draw a card at random from a deck. At this point, it has an equal probability of being red or black. However, I then _tell you_ this card is not black. Surely you must admit that the probability of the card being red is now 100%, not 50%, even though it was drawn from the full population. My point is, even though it was drawn from the full population, some of that population has since been ruled out by the information you were given. The same is true for the boy-girl problem. ---
**Author:** ![tim314](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tim314/32/3468_2.png) [@tim314](https://boards.straightdope.com/u/tim314)\ **Post date:** [August 14, 2006, 10:08am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/108 "2006-08-14T10:08:53Z")
[QUOTE=adbadqc] For the 2/3rds crowd; look at it this way. We have our three groups left after tossing the BB crowd: BG GB GG From this you cry “2/3rds”! Consider these individuals: Albert / Betty Charlotte / Danny Edwina / Florence You have two girls who have brothers (Betty and Charlotte), and two girls who have sisters (Edwina and Florence). For each girl, there is a 50-50 chance her sibling is a boy. You have to “count” each side of the GG pair if you insist on keeping all three “includes a girl” pairings, instead of tossing one as I have suggested earlier. Even in this universe, the odds any given girl has a brother is 50/50. You have to do it this way if you insist “it doesn’t matter whether the girl is the first one in the pair or the second”. If you go down that route, then you have to consider both sides of the GG pairing as separate cases, because you said they are and therefore you have to treat Edwina and Florence as separate cases. Quod Erat Demonstrandem [/QUOTE] If you’re going to persist in being wrong, you could at least do it in English. :rolleyes: You’re still talking about the odds of a particular girl having a male sibling. That’s not the same as the odds of a family having a boy given that they have a girl. See my numerous previous explanations. (Honestly, I think I’ve run out of ways to say it.) It’s really as simple as this: 50% of the total two-child families have a boy and a girl. Only 25% have two girls. All others have no girls. If I rule out “no girls”, 2/3 of the remaining cases have a son. Talking about the odds of an _individual_ girl having a brother is totally misunderstanding the problem. You aren’t asked to calculate the odds of a particular girl having a brother – you’re asked to calculate the odds of a family that has at least one girl having a boy. If you think these are the same, please re-read my previous posts. ---
**Author:** ![misterdls](https://avatars.discourse-cdn.com/v4/letter/m/f4b2a3/32.png) [@misterdls](https://boards.straightdope.com/u/misterdls)\ **Post date:** [August 14, 2006, 1:51pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/109 "2006-08-14T13:51:06Z")
Perhaps this can be made more clear if we substitute coins for children. You are shown two coins, a nickel and a dime. They are tossed in the air, and land on a table behind a screen where you cannot see them. We consider two questions: 1. Someone looks at the coins and says “The nickel came up heads. What is the probability that the dime came up tails?”. 2. Someone looks at the coins and says “One of the coins came up heads. What is the probability the other coin came up tails?”. The four possible outcomes of tossing the two coins are nickel heads/dime heads (NH-DH), nickel heads/dime tails, etc as follows: 1. NH-DH 2. NH-DT 3. NT-DH 4. NT-DT Two coins times two sides = 4 coin/sides. In question 1 the probability the dime shows tails is obviously 50%. We are considering only the two coin/sides of the dime. The sample space here is two coin/sides. Question two is entirely different. When we are only told that one of the coins came up heads, we have eliminated only coin/side 4 above (NT-DT), leaving a sample space of three coin sides. Two of those three coin/sides (2 & 3) would show a tails; only coin/side 1 (NH-DH) would not. Thus the answer to question 2 is 2/3. ---
**Author:** ![adbadqc](https://avatars.discourse-cdn.com/v4/letter/a/c4cdca/32.png) [@adbadqc](https://boards.straightdope.com/u/adbadqc)\ **Post date:** [August 14, 2006, 3:41pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/110 "2006-08-14T15:41:48Z")
I see it now. The correct answer is 2/3rds. ---
**Author:** ![JR\_Brown](https://avatars.discourse-cdn.com/v4/letter/j/d6d6ee/32.png) [@JR\_Brown](https://boards.straightdope.com/u/JR_Brown)\ **Post date:** [August 14, 2006, 3:43pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/111 "2006-08-14T15:43:20Z")
[QUOTE=wissdok] As I posted earlier, \*Scientific America \*also got enough responses that they took another look at the problem. If you Google or Yahoo this problem, you will find that most scholars that even post anything on logic, will proclaim this problem ambiguous and point out how it must be stated correctly to get the desired result. [/QUOTE] I didn’t want to bring this up earlier, since my collection of 1959 _Scientific American_s is unfortunately incomplete 😉 and I therefore can’t look at Martin Gardner’s original problem, but the follow-up article you quote contains the sentence _“Many readers correctly pointed out that the answer depends on the procedure by which the information “at least one is a boy” is obtained.”_ Therefore I _assume_ the original problem contained the phrase “at least one is a boy”. I submit that since Mr Gardner’s column was titled “Mathematical Games”, we can assume that any mathematical jargon used in his problems is being used in its agreed-upon mathematical sense. The phrase “at least one X” has a defined meaning in math, statistics and logic: it is the inverse of the case in which there are no X. Formally: Probability of at least one X = 1 - (probability of no X) Saying the family contains “at least one boy” in this context means that all possible families are being divided into exactly two groups: A) those that have one or more male children, and B) those that have NO male children. In this context, saying “we checked one kid and it was a girl, so we gave up” is NOT a valid answer. Therefore, Mr Gardener’s statement that the problem is ambiguous is wrong (unless of course the original problem did not include the phrase “at least one”), and the only correct answer is 2/3. Unfortunately, Cecil’s rendition of the problem is not unambiguously unambiguous 🙂 , but I stand by the assertion that the most likely interpretation of “You have been told this family has a daughter” is that the family was asked “Do you have any daughters?”. In which case the answer is 2/3. JRB ---
**Author:** ![JR\_Brown](https://avatars.discourse-cdn.com/v4/letter/j/d6d6ee/32.png) [@JR\_Brown](https://boards.straightdope.com/u/JR_Brown)\ **Post date:** [August 14, 2006, 3:47pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/112 "2006-08-14T15:47:29Z")
[QUOTE=adbadqc] Yep; I wuz wrong I see it now. The correct answer is 2/3rds. [/QUOTE] Hallelujah, a convert! Welcome to the fold 🙂 Consider the fatted calf slaughtered. Unless you would prefer a tofu burger? JRB ---
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [August 14, 2006, 4:09pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/113 "2006-08-14T16:09:32Z")
[QUOTE=adbadqc] For the 2/3rds crowd; look at it this way. We have our three groups left after tossing the BB crowd: BG, GB, GG From this you cry “2/3rds”! [/quote] I don’t know about the others, but I’ve actually been singing it. [QUOTE=misterdls] You are shown two coins, a nickel and a dime. They are tossed in the air, and land on a table behind a screen where you cannot see them. We consider two questions: 1. Someone looks at the coins and says “The nickel came up heads. What is the probability that the dime came up tails?”. 2. Someone looks at the coins and says “One of the coins came up heads. What is the probability the other coin came up tails?”. [/quote] I like what you’re trying to do here, but I’m afraid the answer to **both** of these questions is 1/2. Once you say “the other coin”, you’re implying that the first coin mentioned is specific — and then, so is the one being asked about. And this no longer resembles the original Boy-Girl Problem, in which there are no specific children at all. ---
**Author:** ![wissdok](https://avatars.discourse-cdn.com/v4/letter/w/c4cdca/32.png) [@wissdok](https://boards.straightdope.com/u/wissdok)\ **Post date:** [August 14, 2006, 4:23pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/114 "2006-08-14T16:23:05Z")
> [@](#): > > Posted by **Frylock** > Please cite with some google search words, because I’m not finding it. > > And working, as I do, in the world inhabited by these scholars who publish on logic, I feel some confidence in making the claim that your statement here is incorrect. I do not think most scholars would call this puzzle as originally formulated ambiguous, unless they meant it is ambiguous in the way any sentence in any language can be made “ambiguous” by asking just exactly what the speaker “really meant” when s/he uttered it. Google words…Two Children Puzzle, Other Child Puzzle, Ask Marilyn Child Puzzle, etc. Last spring when this subject first came up, the first bunch of Google results were from colleges talking about this puzzle. They don’t appear now high on the list so they very well could have been course related. Wait a few weeks and they may appear again as we are at the start of the school year. But even today I found these in less than 15 minutes. Ask Dr. Math(answered by Dr. TWE?) Question: _From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?_ Answer: _½_ Question: _From the set of all families with two children, a family is selected at random and is found to have a boy. What is the probability that the other child of the family is a girl?_ Answer: _2/3_ \*Why the difference? As in other probability problems, how information is obtained is as important as the information itself. Without knowledge of the data gathering process, ambiguity can result. How do we know that one child is a boy? \* [Ask Dr. Math](http://64.233.161.104/search?q=cache:b2FuesRPPL8J:mathforum.org/library/drmath/view/52186.html+as+1/3+for+it+being+a+boy+and+2/3+for+being+a+girl.+&hl=en&lr=lang_en&ie=UTF-8)) From Tanya Khovanova PhD in Mathematics Her question: _A man says, “I have two children; at least one of them is a boy.” What is the probability that the other one is a boy?_ Her answer: \*The correct answer is that the problem is undefined. For example, there is no indication in the problem that all fathers with at least one boy will tell you, “I have at least one boy.” \* [TanyaKhovanova PhD](http://www.tanyakhovanova.com/Puzzles/solchildren.html) From Robert McMullen, author of **Stranger and Stranger** His question: _Professor Saunders has two children. At least one of them is a boy. What is the probability that both his children are boys?_ His answer: _The answer is that the problem is not answerable without more information._ [Robert McMullen](http://www.curiouser.co.uk/puzzles/kids.htm) Not truly scholarly but…. Wikipedia….Brain Teaser Question: \*If someone has two children, and one of them is a son, what is the probability that the other is also a son? ….but that would be (more) ambiguous, since it could mean that we chose a person at random, and learnt that at least one of their two children was a son (in which case we get 1/3), or it could mean that we chose a person at random, and met one of their children, which turned out to be a son. This would then be a particular child, so the probability of the other being a son is 1/2.\* [Wiki-Brain Teaser](http://en.wikipedia.org/wiki/Brain_teaser) As I have posted before, Scientific American and Martin Gardner have had almost 50 years to retract their statements, but as of yet they haven’t. And lastly, check Ask Marilyn’s website. She once said that between the two child puzzle and the Monty Hall puzzle, she believed every mathematician in America had written her. Do you really think they all wanted to pat her on the back? For **JR Brown** , _“Many readers correctly pointed out that the answer depends on the procedure by which the information “at least one is a boy” is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says “at least one is a boy.” If both are girls, he says “at least one is a girl.” And if both sexes are represented, he picks a child at random and says “at least one is a …” naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases – BB, BG, GB, GG – and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.”_ \*\*Scientific American, October, 1959 \*\* I beleive that the magazine made it clear that if we don’t know how the information was obtain, the puzzle is ambigious. It is my opinion that we have to use a natural approach to this puzzle… we meet a couple (or are told of a couple) and that through conversation find out they have a two child and one of which is a girl. In the nature approach the family is random and the child is random. This version is easily repeatable as the only condition that exist is that the family has to have two children. In the 2/3 version we must always have a family with girl, which would require us to alway ask “do you at least one girl”, to screen the families. You cannot allow the family with both to say “I have a boy.” ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [August 14, 2006, 4:32pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/115 "2006-08-14T16:32:06Z")
[QUOTE=Bytegeist] I don’t know about the others, but I’ve actually been singing it. I like what you’re trying to do here, but I’m afraid the answer to **both** of these questions is 1/2. Once you say “the other coin”, you’re implying that the first coin mentioned is specific — and then, so is the one being asked about. And this no longer resembles the original Boy-Girl Problem, in which there are no specific children at all. [/QUOTE] I don’t think that’s right. “One of the coins” names no specific coin, and it follows from this that “the other coin” names no specific coin either. “At least one of the children is a girl” doesn’t name a specific child any more than does “One of the children is a girl.” And in either case, we can ask, “what’s the probability the other child is a boy?” without asking about a specific child. (Isn’t that how the original is phrased, after all? Namely, in terms of “the other child?”) “At least one” is probably better wording than "one of the, though.’ -FrL- ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [August 14, 2006, 4:37pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/116 "2006-08-14T16:37:09Z")
[QUOTE=wissdok] Google words…Two Children Puzzle, Other Child Puzzle, Ask Marilyn Child Puzzle, etc. Last spring when this subject first came up, the first bunch of Google results were from colleges talking about this puzzle. They don’t appear now high on the list so they very well could have been course related. Wait a few weeks and they may appear again as we are at the start of the school year. But even today I found these in less than 15 minutes. Ask Dr. Math(answered by Dr. TWE?) Question: _From the set of all families with two children, a child is selected at random and is found to be a boy. What is the probability that the other child of the family is a girl?_ Answer: _½_ Question: _From the set of all families with two children, a family is selected at random and is found to have a boy. What is the probability that the other child of the family is a girl?_ Answer: _2/3_ \*Why the difference? As in other probability problems, how information is obtained is as important as the information itself. Without knowledge of the data gathering process, ambiguity can result. How do we know that one child is a boy? \* [Ask Dr. Math](http://64.233.161.104/search?q=cache:b2FuesRPPL8J:mathforum.org/library/drmath/view/52186.html+as+1/3+for+it+being+a+boy+and+2/3+for+being+a+girl.+&hl=en&lr=lang_en&ie=UTF-8)) From Tanya Khovanova PhD in Mathematics Her question: _A man says, “I have two children; at least one of them is a boy.” What is the probability that the other one is a boy?_ Her answer: \*The correct answer is that the problem is undefined. For example, there is no indication in the problem that all fathers with at least one boy will tell you, “I have at least one boy.” \* [TanyaKhovanova PhD](http://www.tanyakhovanova.com/Puzzles/solchildren.html) From Robert McMullen, author of **Stranger and Stranger** His question: _Professor Saunders has two children. At least one of them is a boy. What is the probability that both his children are boys?_ His answer: _The answer is that the problem is not answerable without more information._ [Robert McMullen](http://www.curiouser.co.uk/puzzles/kids.htm) Not truly scholarly but…. Wikipedia….Brain Teaser Question: \*If someone has two children, and one of them is a son, what is the probability that the other is also a son? ….but that would be (more) ambiguous, since it could mean that we chose a person at random, and learnt that at least one of their two children was a son (in which case we get 1/3), or it could mean that we chose a person at random, and met one of their children, which turned out to be a son. This would then be a particular child, so the probability of the other being a son is 1/2.\* [Wiki-Brain Teaser](http://en.wikipedia.org/wiki/Brain_teaser) As I have posted before, Scientific American and Martin Gardner have had almost 50 years to retract their statements, but as of yet they haven’t. And lastly, check Ask Marilyn’s website. She once said that between the two child puzzle and the Monty Hall puzzle, she believed every mathematician in America had written her. Do you really think they all wanted to pat her on the back? For **JR Brown** , _“Many readers correctly pointed out that the answer depends on the procedure by which the information “at least one is a boy” is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says “at least one is a boy.” If both are girls, he says “at least one is a girl.” And if both sexes are represented, he picks a child at random and says “at least one is a …” naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases – BB, BG, GB, GG – and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.”_ \*\*Scientific American, October, 1959 \*\* I beleive that the magazine made it clear that if we don’t know how the information was obtain, the puzzle is ambigious. It is my opinion that we have to use a natural approach to this puzzle… we meet a couple (or are told of a couple) and that through conversation find out they have a two child and one of which is a girl. In the nature approach the family is random and the child is random. This version is easily repeatable as the only condition that exist is that the family has to have two children. In the 2/3 version we must always have a family with girl, which would require us to alway ask “do you at least one girl”, to screen the families. You cannot allow the family with both to say “I have a boy.” [/QUOTE] Okay, thanks for the links. I think they’re wrong, but you’re right that these are experts and what they say has weight. You are right that _if_ we don’t know how the information was obtained, then the puzzle is ambiguous. But I maintain we _do_ know how it was obtained. We don’t need to be told, because the phrasing “at least one child is a girl” implies a method by which the information was obtained. Absent specification, there is a “default” method which any native speaker of English would understand. “At least one child is a girl” means someone looked at the pair, determined whether at least one child is a girl, and reported to me that, yes, at least one is a girl. To wonder, as Gardner et al do, how the information was obtained, is unnecessary, in the context of a puzzle like this. -FrL- ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [August 14, 2006, 4:41pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/117 "2006-08-14T16:41:07Z")
[QUOTE=misterdls] 2. Someone looks at the coins and says “One of the coins came up heads. What is the probability the other coin came up tails?” [/QUOTE] That is seriously ambiguous. Others have suggested the answer might be 2/3 and 1/2. On my reading, the best answer is 1. It depends on what question is being implicitly answered, but I think it’s in answer to the question such as: “Exactly how many coins came up heads?” In that case if one came up heads, it’s the only one showing heads, and the other coin necessarily came up tails. ---
**Author:** ![JR\_Brown](https://avatars.discourse-cdn.com/v4/letter/j/d6d6ee/32.png) [@JR\_Brown](https://boards.straightdope.com/u/JR_Brown)\ **Post date:** [August 14, 2006, 5:11pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/118 "2006-08-14T17:11:19Z")
[QUOTE=Giles] "Quote: Originally Posted by misterdls 2. Someone looks at the coins and says “One of the coins came up heads. What is the probability the other coin came up tails?” " That is seriously ambiguous. Others have suggested the answer might be 2/3 and 1/2. On my reading, the best answer is 1. It depends on what question is being implicitly answered, but I think it’s in answer to the question such as: “Exactly how many coins came up heads?” In that case if one came up heads, it’s the only one showing heads, and the other coin necessarily came up tails. [/QUOTE] You’re only doing this to annoy everybody, aren’t you? :mad: 😉 JRB ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [August 14, 2006, 5:14pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/119 "2006-08-14T17:14:06Z")
[QUOTE=JR Brown] You’re only doing this to annoy everybody, aren’t you? :mad: 😉 [/QUOTE] Of course: you’ve just made my day, you know 🆒 ---
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [August 14, 2006, 5:14pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/120 "2006-08-14T17:14:15Z")
[QUOTE=Frylock] I don’t think that’s right. “One of the coins” names no specific coin, and it follows from this that “the other coin” names no specific coin either. [/quote] This might set off yet another language debate. To me, yes, the phrase “one of the coins . . .” doesn’t specify a particular coin. Or anyway, it shouldn’t be taken to specify one, all by itself. But when you get to “the **other** coin” in the next sentence, that forces an adjustment. There is now a particular coin being asked about, and therefore (since there are only two) a particular coin being described in the previous sentence. [QUOTE=Frylock] “At least one of the children is a girl” doesn’t name a specific child any more than does “One of the children is a girl.” [/quote] Agreed with that, if you look at those sentences in isolation. [QUOTE=Frylock] And in either case, we can ask, “what’s the probability the other child is a boy?” without asking about a specific child. (Isn’t that how the original is phrased, after all? Namely, in terms of “the other child?”) [/quote] I can’t agree with this. To me, your first question there strongly implies that you’re indicating a particular child — which becomes **the** child whose sex you’re being asked about, in contrast with **the** child you know to be a girl. The original problem uses different language: _You have been told this family has a daughter. What are the odds they also have a son . . . ?_ [Previous page](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182.md?page=5) [Next page](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182.md?page=7)