# 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:** 4
**Author:** ![LIT123](https://avatars.discourse-cdn.com/v4/letter/l/5daacb/32.png) [@LIT123](https://boards.straightdope.com/u/LIT123)\ **Post date:** [August 12, 2006, 6:31pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/61 "2006-08-12T18:31:22Z")
I’m lousy at these things which is bad news for the 50-50 camp since that is where I am. I might understand the 2/3 CROWD if I could grasp why they have both a BG and a GB. Is the order of birth being addressed and if so why not 2 BB and 2 GG which after elimination leaves _G, G_, _B, and B_. That gets me back to 50-50. If there is some “rule of” thing that covers it then post that phrase and I can Google to save all your time. My next questuin is,assuming adequate O2, h20, and food, if the second child is sealed in a steel tube is it G, B, or neither? ---
**Author:** ![wissdok](https://avatars.discourse-cdn.com/v4/letter/w/c4cdca/32.png) [@wissdok](https://boards.straightdope.com/u/wissdok)\ **Post date:** [August 12, 2006, 6:32pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/62 "2006-08-12T18:32:12Z")
> [@](#): > > Posted by **Larry Borgia** > We each flip a nickel. > > If we each show heads, nothing happens. > If you show a head and I show a tail, you give me a dollar. > If you show a tail and I show a head, you give me a dollar. > If we each show tails, I give you a dollar. > > I’ll spot you twenty bucks and we’ll play 400 rounds. I don’t really understand why we throw out two heads. I’ll take your bet if you let me win on two heads too. And to make it worth my while….each time I win you pay me $1.25. I believe in the end you will owe me $50. ---
**Author:** ![Larry\_Borgia](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/larry_borgia/32/156_2.png) [@Larry\_Borgia](https://boards.straightdope.com/u/Larry_Borgia)\ **Post date:** [August 12, 2006, 6:36pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/63 "2006-08-12T18:36:42Z")
[QUOTE=wissdok] I don’t really understand why we throw out two heads. I’ll take your bet if you let me win on two heads too. And to make it worth my while….each time I win you pay me $1.25. I believe in the end you will owe me $50. [/QUOTE] The rules of the game mirror the original boy girl rule. We throw out the families who have two boys, and two thirds of the families will have a boy and a girl. That was the original problem. You can make a new problem in which the answer is 50/50, but it’s not the problem Cecil describes in his column. ---
**Author:** ![wissdok](https://avatars.discourse-cdn.com/v4/letter/w/c4cdca/32.png) [@wissdok](https://boards.straightdope.com/u/wissdok)\ **Post date:** [August 12, 2006, 7:02pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/64 "2006-08-12T19:02:17Z")
> [@](#): > > Posted by **LIT123** > I’m lousy at these things which is bad news for the 50-50 camp since that is where I am. I might understand the 2/3 CROWD if I could grasp why they > have both a BG and a GB. Is the order of birth being addressed and if so > why not 2 BB and 2 GG which after elimination leaves _G, G_, _B, and B_. > That gets me back to 50-50. If there is some “rule of” thing that covers it > then post that phrase and I can Google to save all your time. My next questuin > is,assuming adequate O2, h20, and food, if the second child is sealed in a steel tube is it G, B, or neither? I am also in the 50/50 camp (as you can see from my posts) but it is generally agreed that the two children are variables that are not interchangeable. While for simplicity we can call it “age”, they are independent events that are different. In figuring this problem we assume that if we have a child, it has an equal chance to have a brother or sister. We assume that we have an equal chance for these combinations: Girl with a sister Girl with a brother Boy with a sister Boy with a brother In the above example girl/boy is represents event #1 and the sister/brother represents event #2. This is the same as flipping two coins, a nickel and a dime, we would record their outcomes separately. The possible outcomes are: Nickel…………………Dime Heads…………………Heads Heads…………………Tails Tails…………………..Heads Tails…………………..Tails As to your second question, it is the argument of the 2/3 camp that once we have been given a girl then we can just drop the two boy family as a possibility and the remaining three families are equally likely to by her family. I can’t speak for the other 50/50 “campers” but I believe that this is a fallacy. I believe that if you remove the boy/boy family it is because we now know that either event #1 or event #2 was a girl which means we must also remove one of the GB or BG families as well. ---
**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 12, 2006, 8:35pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/65 "2006-08-12T20:35:04Z")
To **LIT123** : [QUOTE=LIT123] I’m lousy at these things which is bad news for the 50-50 camp since that is where I am. I might understand the 2/3 CROWD if I could grasp why they have both a BG and a GB. Is the order of birth being addressed and if so why not 2 BB and 2 GG which after elimination leaves _G, G_, _B, and B_. That gets me back to 50-50. If there is some “rule of” thing that covers it then post that phrase and I can Google to save all your time. [/QUOTE] The order of birth is not important; listing GB and BG is just to make it clear to all that “family with one boy and one girl” is twice as common as “family with two girls”. So (BG + GB) = (BB + GG), and (BG + GB) / (BG + GB + GG) = 2/3. In general, the class of statistics that covers these problems is called “conditional probability”. To **wissdok** : [QUOTE=wissdok] "Posted by JR Brown You have four people in your kitchen, Adam, Bill, Charlie and Dave. You give them each two cards, like so: Adam: two black cards Bill: two red cards Charlie: one red and then one black Dave: one black and then one red You ask each person who has any red cards to move into the living room. Question: of the people in the living room, what percentage will also have a black card? Answer: 2/3. Bill, Charlie and Dave have at least one red card and will thus be in the living room; of these, Charlie and Dave both have a black card." Here you have chosen “red”. [/QUOTE] It also works if you choose “black”. All those with at least one black card move into the living room: Adam, Charlie and Dave are now in the living room. How many of them have red cards? Two: Charlie and Dave = 2/3. Note that the set {Bill, Charlie and Dave} and the set {Adam, Charlie and Dave} have the same number of people (three) and the same aggregate number of cards (six), but are different sets. [QUOTE=wissdok] Let’s try… you go get a Mountain Dew and find a red card on the floor next to the table? What are the chances this card came from Charlie’s or Dave’s hand? Are they not combined, equal to the chance that it came from Bill’s hand? [/QUOTE] Yes, but that is A STASTISTICALLY DIFFERENT QUESTION! It is the equivalent to the _second_ boy/girl problem in Cecil’s column (“The first child enters–a girl. The second knocks”, etc). Reading your explanations, you consistently phrase the question as if we were discussing a specific girl, and asking if she had a brother. All of your explanations are relevant to what I’m calling the _second_ boy/girl problem, where this is indeed the question. But in the _first_ boy/girl question, we are asking about _families_ (as **gazpacho** and **matt\_mcl** and **borschevsky** previously noted). To reiterate: the problem I am solving is “What is the probability that a pair of siblings, of which at least one is a girl, also includes a boy” NOT “what is the probability that a girl with one sibling has a brother” [QUOTE=wissdok] As to your second question, it is the argument of the 2/3 camp that once we have been given a girl then we can just drop the two boy family as a possibility and the remaining three families are equally likely to by her family. I can’t speak for the other 50/50 “campers” but I believe that this is a fallacy. I believe that if you remove the boy/boy family it is because we now know that either event #1 or event #2 was a girl which means we must also remove one of the GB or BG families as well. [/QUOTE] OK, here’s the nub. “I believe that if you remove the boy/boy family it is because we now know that either event #1 or event #2 was a girl.” The key word in that sentence is EITHER. We don’t know _which one_. As such, BOTH the GB and BG families are still possible, as well as the GG family. Only the BB family has been excluded. Here is an [article](http://www.amstat.org/publications/jse/v14n2/ancker.html) from the _Journal of Statistics Education_ that discusses a related problem (complete with heavy-duty math): “United States life-table data in a certain year shows that the probability of a male birth is 0.512. If two pregnant women are chosen at random from that population, what is the conditional probability that both will give birth to boys, given that we know that at least one did so?” Answer: 34% Why? Math, much simplified: probability of 2 boys = square of 0.512 = 0.262 probability of 2 girls = square of (1 - 0.512) = square of 0.488 = 0.238 probability of one of each = 2 x (0.512 x 0.488) = 0.500 (check: the above sum to 1. Good) SO: probability of at least one boy = probability of 2 boys + probability of one of each = 0.762 NOW: probability of two boys, GIVEN THAT there is at least one = (probability of 2 boys) divided by (probability of at least one boy) = 0.262 / 0.762 = 0.344 QED. (Note that this is the inverse of our problem, and that they are not setting the boy-girl probability equal). But why get bogged down in all that? The simplest explanation of the problem was back in post #9: [QUOTE=gazpacho (slightly edited for typos and clarity)] If we assume that girls and boys are equally likely then the probabilities for these families are: 1/4 two boys 1/4 two girls 1/2 one of each In a population of 1000 families: 250 have two boys 250 have two girls 500 have one of each There are 750 families that have at least one child that is a girl: 250 have two girls 500 have one of each 500 out of these 750 have a boy, or 2/3 of them. [/QUOTE] Your response to this was: [QUOTE=wissdock (edited to change family ID’s for clarity)] \*If the families are equal, then the children are not. If we are given a girl and the families are equal, then each families (GG, BG, GB) are each likely 33.3% of the time… making each girl in family GG worth half the value of the girls in families GB and BG. [/QUOTE] So what? We aren’t being given _a girl_, we are being given _a family that includes at least one girl_. The distinction between these two situations determines how the problem must be addressed; if you can’t see that, then you need to brush up on your conditional probability! JRB ---
**Author:** ![Bytegeist](https://avatars.discourse-cdn.com/v4/letter/b/9dc877/32.png) [@Bytegeist](https://boards.straightdope.com/u/Bytegeist)\ **Post date:** [August 12, 2006, 10:29pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/66 "2006-08-12T22:29:14Z")
[QUOTE=wissdok] We have to conclude that the source that gave us the information about the girl, told us of all he knew. [/quote] There is no “the girl”. There is no specific child we’re being given information for. > [@Bytegeist](#): > > My favorite way of seeing why the answer isn’t 50% is to expand the number of children. Suppose a family has 10 children. You are told that 9 of them are daughters. What’s the probability that they also have a son? [QUOTE=wissdok] This analogy isn’t very relevant. In your question the “lost” boy could have been in any of 10 places….in our problem just the one not occupied by the given girl. [/QUOTE] And in the _two_-child problem, the boy can be in any of _two_ places. Moreover, there is no “given girl”. My analogy is completely relevant because it’s the exact same problem, only expanded to a larger number of children. To make it more general though, restate the problem this way: > [@](#): > > There is a family with **N** children. You have been told this family has N – 1 daughters [at least]. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? For N = 2, this is the exact same problem Cecil gave in his column. In the world at large, the N-child families having exactly 1 boy and N-1 girls always outnumber the ones having all girls — outnumber them by a ratio of N to 1, in fact. So the answer to the generalized problem is **N/(N+1)**. Which of course is 2/3 for N = 2. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [August 12, 2006, 10:56pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/67 "2006-08-12T22:56:44Z")
[QUOTE=matt\_mcl] Okay. Since I have a penny and a certain amount of free time, I just sat down to flip my silly little coin and perform an actual experiment. Here are the results of the coin tosses: TH TH HT TT TT HT HH HT HT HT HT TH TH HT TH HH I tossed the coin a total of 32 times. Of those, exactly 16 are tails and 16 are heads. Overall heads/tails rate: 50:50. Considering the pairs. We will not even look at the order of the pairs. There are two TT pairs and two HH pairs, so each is 2/16 or 1/8 of the time. This is in fact _less_ than the expected 25%. Pairs with one head and one tail make up 7/8ths of the trials. Now, a total of 14 of the 16 trials had a head in them. Of those 14 trials, 12 also had a tail. The odds over this sample were therefore 86% over this trial that a pair containing a head also had a tail. In other words, if you were discussing any of these pairs, and you were informed that one of the members of the pair was a head, you would be right 86% of the time - **not 50%** - that the other member of the pair was a tail. The order has nothing to do with it, as we can see, since we have never discussed which flip was first. But as you can also see, if you were discussing any of these pairs the same way, and you were told that one of the pair was a tail, you would _still_ be right 86% of the time if you were to say that the other member of the pair was a head. [/QUOTE] Damn straight. Here’s a post where I did something like this with a sample size of a hundred pairs: [http://boards.straightdope.com/sdmb/showpost.php?p=7121957&postcount=60](http://boards.straightdope.com/sdmb/showpost.php?p=7121957&postcount=60) To my recollection, Wissdoc was still not convinced. 🙂 -FrL- ---
**Author:** ![wissdok](https://avatars.discourse-cdn.com/v4/letter/w/c4cdca/32.png) [@wissdok](https://boards.straightdope.com/u/wissdok)\ **Post date:** [August 12, 2006, 11:25pm UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/68 "2006-08-12T23:25:03Z")
> [@](#): > > Posted by **JRBrown** > So what? We aren’t being given a girl, we are being given a family that includes at least one girl. The distinction between these two situations determines how the problem must be addressed; if you can’t see that, then you need to brush up on your conditional probability! > [@](#): > > \*\*Cecil’s puzzle > 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.) > \*\* \*“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 \* Did I miss something in Cecil’s puzzle, it didn’t say anything about choosing families that had only girls. The first sentence says, “There is a family with two children.” This sentence by itself doesn’t say it limits itself to just families with girls. In the next sentence we are told of a daughter. This second sentence doesn’t say that all families that have daughters are now in a subgroup, we only know of one possible family that will always say they have girls. Without some safeguard in place those families that have both a boy and a girl, could have just as likely tell us of a boy. We don’t know how they would have answered. An example: Cecil Adams and Marilyn vos Savant climb into a elevator together. They start talking to each other about this puzzle. In Cecil’s version, he used a girl; In Marilyn’s a boy. They decide to test the 2/3 rule. Lucky for them they have 4 other people on the elevator with them and they all have two kids apiece. They ask the first person to tell a clue about his children. I have a least one boy. Cecil disregards the answer, Marilyn uses the 2/3 rule and predicts the other is a girl. The man has a son and a daughter so……….Marilyn is CORRECT. They the repeat the question to the second man. He says he has at least one son. Again Cecil disregards this answer and Marilyn chooses opposite. But this time the man says he has two sons…..Marilyn is INCORRECT. They ask the third person, and she says she has at least one daughter. This time Cecil picks opposite and Marilyn passes on this question. The woman proclaims that she has a son and daughter so…….Cecil is CORRECT. They approach the final person and she says she has a daughter too. Cecil again chooses opposite and Marilyn again passes. This woman has two daughters so….Cecil is INCORRECT. Final score: Cecil has 1 right and 1 wrong. Marilyn has 1 right and 1 wrong. If they had answered on all questions then they both would be with 2 and 2. ---
**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 13, 2006, 1:11am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/69 "2006-08-13T01:11:59Z")
[QUOTE=wissdok] “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 [/QUOTE] This is true, for what it’s worth, and it’s an example of how probabilities will change with different selection criteria that give (superficially) similar outcomes. But from my perspective Mr Gardner’s readers’ objections are like asking someone “Will an unsupported rock fall down?”, and when they say “Yes”, responding “It wouldn’t if it were in geosynchronous orbit, so you’re wrong!” If I were asked about “families with two children, at least one of whom is a boy”, I would _never_ spontaneously imagine the second selection process described. Cecil’s phrasing (“There is a family with two children. You have been told this family has a daughter.”) implies to me that all children have been inspected and an “at least one girl” rule applied. There are certain assumptions built in to any statement, and while those assumptions may in some cases be wrong, it is, in general, up to the questioner to mention any deviations from standard conditions. Otherwise it is _always_ possible to say “ahh, but it could have been otherwise!” For instance, neither Cecil’s nor Gardner’s questions specify a _human_ family, so they could equally well be about a race of aliens who have three sexes (male, female, and hermaphrodite) in a 1:1:1 ratio; would you be satisfied with an answer that worked out the probabilities for that family? JRB ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [August 13, 2006, 1:29am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/70 "2006-08-13T01:29:12Z")
[QUOTE=JR Brown] This is true, for what it’s worth, and it’s an example of how probabilities will change with different selection criteria that give (superficially) similar outcomes. But from my perspective Mr Gardner’s readers’ objections are like asking someone “Will an unsupported rock fall down?”, and when they say “Yes”, responding “It wouldn’t if it were in geosynchronous orbit, so you’re wrong!” If I were asked about “families with two children, at least one of whom is a boy”, I would _never_ spontaneously imagine the second selection process described. Cecil’s phrasing (“There is a family with two children. You have been told this family has a daughter.”) implies to me that all children have been inspected and an “at least one girl” rule applied. There are certain assumptions built in to any statement, and while those assumptions may in some cases be wrong, it is, in general, up to the questioner to mention any deviations from standard conditions. Otherwise it is _always_ possible to say “ahh, but it could have been otherwise!” For instance, neither Cecil’s nor Gardner’s questions specify a _human_ family, so they could equally well be about a race of aliens who have three sexes (male, female, and hermaphrodite) in a 1:1:1 ratio; would you be satisfied with an answer that worked out the probabilities for that family? JRB [/QUOTE] JRB is exactly right. There are literally an infinite number of criteria by which it might have been decided that “At least one child is a boy” should be asserted. We are not told which of these criteria were adapted. But this does not mean we don’t know which criterion was adapted. This is because, as the puzzle is phrased, it is not to be suspected that anything but the _obvious_ criterion was adapted. The obvious criterion is based on the principle that, in general, an honest and unmistaken person will say “X” if and only if X is true. Similarly, since we assume the puzzle is not lying to us, and since it does not specify a criterion for decision as to when to say what about gender, we take it the obvious criterion was used: That we are told “at least one is aboy” _if and only if_ at least one is a boy. Based on an assumption that this was the criterion used, the answer is 2/3. Well, that didn’t turn out nearly as clear as I hoped. Trust me: I’m just saying Jr. Brown is right. -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 13, 2006, 1:55am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/71 "2006-08-13T01:55:50Z")
[QUOTE=adbadqc] Wrong. Here’s why. YOU HAVE TO THROW AWAY HALF OF THE BRs! ??? That’s right. Why? Because I said so. No; just kidding. Because **the selection from that subset of the population results in a R only 50% of the time** [/quote] Here’s where the key is how they’re answering the question. If the family is reporting the sex of a _random_ kid, then yes, the selection of that subset results in them saying “girl” half the time. _However_, if they are answering the question “Do you have any girl children?” then they will say yes **100%** of the time when they have at least one girl. The 2/3 number is based on this second interpretation. Do you see the distinction I’m making? I’m almost entirely sure that this is why we’re getting different answers. ---
**Author:** ![wissdok](https://avatars.discourse-cdn.com/v4/letter/w/c4cdca/32.png) [@wissdok](https://boards.straightdope.com/u/wissdok)\ **Post date:** [August 13, 2006, 2:12am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/72 "2006-08-13T02:12:11Z")
I don’t have time to write a lengthy response, so for the time being, please checkout this website I found the last time we went over this subject. [Ask Marilyn Response](http://www.wiskit.com/marilyn/boys.html#Letter) ---
**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 13, 2006, 2:20am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/73 "2006-08-13T02:20:01Z")
[QUOTE=adbadqc] OK; I think I need to stop trying to explain this and just ask folk a few questions: 1. Based on your (speaking to the audience at large here) understanding of biology and how the world works (put DOWN that statistics textbook and step away from the table), what would you expect the sex ratio of a first child to be? I assume most of you answered 50-50 male/female. **Yes, 50-50. Even if that _might_ not be true in reality, it’s _definitely_ supposed to be assumed for this question.** 1. Based on your understanding of biology etc. etc. etc, what would you expect the sex ratio of the SECOND child in a family to be? I’m guessing (hoping) most of you said 50-50 male/female. **Yes, each child has a 50-50 chance of being either sex, and they’re independent of each other.** 1. Based on your understanding of biology etc. etc. etc., would you expect the sex of the first child in a family to have an impact on the sex of the second child in a family? (If you answered “yes”, please explain the mechanics of this effect). **No, the sex of one _particular child_ won’t affect the sex of the other**. 2. Given the above, and leaving that statistics book lying there; what would you expect the sex ratio of the other sibling in a family to be where there is one daughter? Unless you’re cheating, you said 50-50 male/female. Think about it. [/quote] Number 4 is where we disagree. Most families with one daughter also have one son (assuming they really are independent 50-50 choices). I know this _seems_ like it contradicts the above, but it doesn’t. Let me try to explain one more time: 1. There’s a 25% chance of them having two girls. (50% times 50%) 2. There’s a 25% chance of them having two boys (50% times 50%) 3. There’s a 50% chance of them having one of each (100% minus 25% minus 25%) **Please let me know if you disagree with any of these three.** If you accept these three points, then you can see that _most_ families that have a girl also have a boy: 50% of the total families have one of each, whereas 25% have girls only. The reason you think of it as contradicting your points above is that you’re not drawing a distinction between saying a _particular_ child is a girl and saying _at least one_ of their two children is a girl. But these are distinct. If the first child is a girl, then yes, the second child is equally likely to be a girl or boy. If the second child is a girl, then yes, the first child was equally likely to be a girl or boy. But if I ask “Do you have any girls,” you’ll say yes if the first _or_ the second one is a girl. That’s a different scenario. Again, we have these possibilities: (A) Girl then girl: 25% chance (B) Girl then boy: 25% chance (C) Boy then girl: 25% chance (D) Boy then boy: 25% chance. If they tell you the first child is a girl, you have possibilities (A) and (B), with a 1/2 chance that the second one is a boy. If they tell you the _second_ child is a girl, you have possibilities (A) and (C), with a 1/2 chance that the first was a boy. _But_, if all they say is “Yes, we have a daughter,” then you have (A), (B), and (C), with a 2/3 chance that they had at least one boy. The 2/3 answer is based on the assumption that the family only said “yes, we have a daughter” – i.e., they are telling you they have _at least one_ daughter. This is equivalent to looking at both cards and then saying “yes” if _at least one_ was red. Please let me know if you don’t agree that _with this interpretation of the information given_ the answer is 2/3, and please quote the exact step in my reasoning that you think is wrong and explain why (as I have attempted to do). After all this typing, it would be _really_ nice if we could come to some agreement. ---
**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 13, 2006, 2:40am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/74 "2006-08-13T02:40:36Z")
If you don’t agree with the 2/3 number, please take the following quiz. **True or False:** **1)** There are four **equally likely** possibilities: - The parents had a boy, then a girl - The parents had a girl, then a boy - The parents had a girl, then another girl - The parents had a boy, then another boy. **2)** In _three_ of these four cases, the parents will answer “yes” to the question “Do you have a daughter?” **3)** In **two out of those three** cases, the parents also have a son. So the odds are 2/3, _if_ the parents were asked “Do you have a daughter?” rather than “Was your first child a daughter?” or “Was your second child a daughter?” The key is that “Do you have a daughter?” produces more “yes” answers than either of those other questions, and all the extra “yes” answers come from cases where they had _both_ a girl and 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 13, 2006, 2:58am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/75 "2006-08-13T02:58:59Z")
[QUOTE=wissdok] I don’t have time to write a lengthy response, so for the time being, please checkout this website I found the last time we went over this subject. [Ask Marilyn Response](http://www.wiskit.com/marilyn/boys.html#Letter) [/QUOTE] Right, it looks like that link is reiterating the interpretation of the problem as ambiguous. But I maintain the problem is not ambiguous. There is a single, plainly intended reading of the problem. And on that reading, the answer to the question is 2/3. -FrL- ---
**Author:** ![Alex\_Dubinsky](https://avatars.discourse-cdn.com/v4/letter/a/dbc845/32.png) [@Alex\_Dubinsky](https://boards.straightdope.com/u/Alex_Dubinsky)\ **Post date:** [August 13, 2006, 8:02am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/76 "2006-08-13T08:02:33Z")
Everybody saying 2/3… I am with you. However, let’s take a second to appreciate the weird train of logic that makes 1/2 seem pretty reasonable as well. A family tells you they have a girl. Case 1: the girl was the firstborn Sub-case 1.1: the secondborn was a boy Sub-case 1.2: the secondborn was a girl Case 2: the girl was the secondborn Sub-case 1.1: the firstborn was a boy Sub-case 1.2: the firstborn was a girl At every fork it seems you are making a 50/50 choice. There’s a 50/50 chance the girl was firstborn or secondborn, and a 50/50 chance that the other-born was either boy or girl. Now, which one of you, the 2/3 people, can figure out where in that logic there is a mistake. I would assert it to be a challenging thing to effectively do, and it is very clear where (at least some of) the 1/2 people are coming from. I can’t say I can explain it elegantly, but “the firstborn girl has a second born sister” and “the second born girl has a firstborn sister” are actually the same situation, because you don’t know which girl the parents were actually talking about. Even if they name her Debbie, you can’t separate Debbie having a younger sister and Debbie having an older sister into two cases. The above explanation is flimsy, but the train of logic that leads to the answer “2/3” is impervious to error. Three-quarters of two-child families have at least one girl. One-half of families have one boy one girl. These are observationally proven, verified facts. The one-half with a boy and a girl is fully a subset of the three-quarters which have at least one girl. If you know you have a family which definately belongs to the superset, it’s got a 2/3 chance of belonging to the subset. Thus, now it is simply a matter of finding a good reason that the above logic is wrong. Now, if you meet a particular girl Debbie, then the cases of her having a younger sister or an older sister become distinct. Thus, the probability of a brother is one half. ---
**Author:** ![Uzi](https://avatars.discourse-cdn.com/v4/letter/u/91b2a8/32.png) [@Uzi](https://boards.straightdope.com/u/Uzi)\ **Post date:** [August 13, 2006, 8:11am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/77 "2006-08-13T08:11:54Z")
[QUOTE=tim314] If you don’t agree with the 2/3 number, please take the following quiz. [/QUOTE] Please see my post earlier. ---
**Author:** ![garygnu](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/garygnu/32/2924_2.png) [@garygnu](https://boards.straightdope.com/u/garygnu)\ **Post date:** [August 13, 2006, 8:18am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/78 "2006-08-13T08:18:43Z")
With two children, there are four options: 1. Older brother, younger brother 2. Older brother younger sister 3. Older sister, younger brother 4. Older sister, younger sister By knowing the sex of one child, in this case a sister, one of the four options can be completely eliminated, in this case option 1). In two out of those remaining three cases, a brother is present. Therefore, a sister with one and only one sibling is statistically more likely to have a brother than to have a sister by a ration of 66.67% to 33.33%. Is there something wrong with my logic? ---
**Author:** ![garygnu](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/garygnu/32/2924_2.png) [@garygnu](https://boards.straightdope.com/u/garygnu)\ **Post date:** [August 13, 2006, 8:19am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/79 "2006-08-13T08:19:47Z")
…besides my misspelling? ---
**Author:** ![Uzi](https://avatars.discourse-cdn.com/v4/letter/u/91b2a8/32.png) [@Uzi](https://boards.straightdope.com/u/Uzi)\ **Post date:** [August 13, 2006, 10:57am UTC](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182/80 "2006-08-13T10:57:47Z")
[QUOTE=garygnu] Is there something wrong with my logic? [/QUOTE] This is what your options are according to your scenario: OB YB OB YS OS YB OS YS No, two of the options can be eliminated. OB YB and one of the sister choices. If the sister is younger you have to eliminate OS YB. if the sister is older then you have to eliminate OB YS. If the sister is (O)lder you are left with: [del]OB YB[/del] [del]OB YS[/del] OS YB OS YS Because the sister is older you can’t have OB YS. It is impossible. If the sister is (Y)ounger you are left with: [del]OB YB[/del] OB YS [del]OS YB[/del] OS YS Because the sister is younger you can’t have OS YB. Again, it is impossible. And it doesn’t matter whether you ‘know’ she is older or younger. She can only ever be older or younger. [Previous page](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182.md?page=3) [Next page](https://boards.straightdope.com/t/boy-girl-probability-in-monty-hall/368182.md?page=5)