Paul, you are absolutely right of course. 50% is probably the most acceptable answer because people should be able to assume that all known information has been presented in the problem, and it has not been rigged in some arbitrary way. The rigged problem is essentially similar to the Monty Hall problem http://en.wikipedia.org/wiki/Monty_Hall_problem where motivations and asymmetric information come into play.
Kevin
· 11 months ago
I would agree with that. I saw the same post a couple days ago, and had heard the problem posed before. When I thought about it again, I also came to "I don't know." The given reasoning makes sense, though one has to consider if the parents are twice as likely to say "I have one girl" if they have two because they could be referring to either one.
Mona N.
· 11 months ago
Do you have algorithms for the marshmallow concoctions too, Paul? Just wondering! :)
idnan
· 11 months ago
is the question wrong, really? i would have said 2/3 to "What are the odds that person has a boy and a girl?" and 1/2 to "What are the odds the other child is a boy?" is that wrong? those seem like different questions to me ... of course, i've never been much for puzzles ...
Manolis
· 11 months ago
I think you've made a mistake there.
Step 2 in algorithm 2 is equivalent to step 2 in algorithm 1 in the situation where there are 2 children.
What you are missing is that the probability calculation is being made after you are told more information in step 2.
Let's say, though, that we consider parents with 3 children, meaning there are eight possibilities:
bbb, bgb, bgg, bbg, ggg, gbg, gbb, ggb
If we eliminate the case where parents have three boys, then the probability that there are at least two children of different genders is 6/7.
If, however, we arbitrarily announce the gender of two of the children, there is no telling whether we will be told two girls, one girl and one boy or two boys.
Nevertheless, when we are told that extra information, we make our calculations after we are told the extra information. Obviously, in the case of our being told that there is one girl and one boy, the chances of there being both boys and girls is 100%. If, however, we are told that there are two girls or two boys, the chances of there being both boys and girls is 75%.
llimllib
· 11 months ago
Paul,
That was my comment on the post where he *asked* the question, not where he answered it:
unfortunately, there's no permalinks on the comments there - grep for "bill mill".
Toby DiPasquale
· 11 months ago
Right on, Paul. This is a key factor in the Monty Hall problem that a lot of people don't get, including a lot of medical specialists who really should do.
logicalextremes
· 11 months ago
The simpler, intuitive to some, way to think about the problem is that children ARE like coins... successive children are independent events. It's irrelevant whether one is a girl or not. The odds of any given child being a boy or girl is 50%.
Manolis
· 11 months ago
The difference is that you are told information after the events have taken place.
Let's say we are asked the probability that of three children, at least one is a boy and at least one is a girl.
The probability of that happening is 6/8 or 75%.
Let's say I tell you that of three children, at least one is a boy and at least one is a girl. If I asked after providing that information the probability of there being at least one boy and at least one girl, would you still say it is 75%?
Of course not, because the extra information provided changes the probabilities.
logicalextremes
· 11 months ago
But in your new 3-child problem, the game has ended once you give that information. There are no more unknowns. It's a different class of problem.
In the original problem, we are told only that of two children, one is a girl. The other is still a mystery, and has a 50% chance of being a boy and a 50% chance of being a girl.
anonymous
· 11 months ago
Jeff Atwood isn't very smart. He writes stuff that not very smart people think is very smart and very smart people think is only smart enough to get not smart people into trouble.
And then, after he writes it, lots of other people link to it and comment on it, for no other reason that lots of other people link to it and comment on it.
It's a big waste of time mostly.
breck
· 11 months ago
Thank you! I almost wrote the same thing but didn't want to get into a comment war. Glad someone corrected him.
Manolis
· 11 months ago
"The catch is that announcing the boy also reduced the boy+girl probability by an equal amount, so the result is still the same (it eliminated either BG or GB, I don't know which, but it doesn't matter)."
In fact, announcing the boy does not eliminate either of those two options. You're getting confused between asking whether or not a particular child is a boy or girl and whether or not any child is a boy or girl.
If the question was "is the second (or any particular) child a boy or girl?", then the answer will always be 50%.
That, however, was not the question. The question was "is one of the children a boy or girl?". That means we are not concerned about the order of birth. If we are not concerned about the order of birth, both BG and GB are possibilities, while GG definitely is not.
Manolis
· 11 months ago
Let me reply to my own comment.
I should have said the question was "is the second (or any particular) child a boy?", which always has a 50% chance.
The second question should have read "is one of the children a boy?", which always has a higher probability regardless of whether or not one of the children's gender is revealed.
If we are asking whether or not the second child is a boy, regardless of what is the first child's gender, there will be a 50% it will be a boy.
If one of the children is revealed to be a girl, then the chances of there being a boy is 66%.
If no child's gender is revealed, the chances of having one boy is 75%.
If one of the children is revealed to be a boy, then the chances of having a boy is 100%.
Yann
· 11 months ago
Paul,
The answer to the second algorithm should be 2/3 if you use all the information provided.
WHY?
The parents ALREADY have 2 children in this context.
Each point in the space of probability is a family of two.
First, let's find the weight of all the points of families containing both sexes. ANSWER: P(GB U BG) = 1/2
Second, let's plug our extra information. We know the family in question is PART OF THE POINTS of families containing at least one girl.
In other words, this family belongs to either GB, BG, GG.
ANSWER: P(GB U BG | GB U BG U GG) = 2/3
Read as: "The weight of GB U BG given that we're in the subspace GB U BG U GG".
The magic is in how you plug the information, not the algorithm of the question.
rdickerson
· 11 months ago
I wrote up a couple detailed explanations of this in a (vain) attempt to explain this very thing to the Atwooders:
Jeff Atwood's Bayesian probability question was obviously lacking in certain stated assumptions, but no more than the problems you'd find in a probability textbook. In your update, you mention that the random announcement of gender (say, boy) eliminates both BB and "(it eliminated either BG or GB, I don't know which, but it doesn't matter)." Problem is, it DOES matter which of those it eliminated, and if asked, you wouldn't be able to answer. Probability is inextricably about what is KNOWN about a situation, not the situation itself. As some consolation, your reasoning would be correct if the parent said "my first|second child is a boy|girl," because then we CAN eliminate either BG or GB.
Andi Kalsch
· 11 months ago
Nice discussion. It's all a question about the correct precodition definition. I think people who heard about conditional probability know that the correct answer is "2/3". Most people will just imagine the first precondition "they have two children" and forget the second one.
What about going in the opposite direction and simply ask "What is the probability that a mother has a boy and a girl?" I think many people would answer "1/2", too, simply adding the precodition "she has 2 children" ;)
The origin for this is - I think - an implicit context people are thinking of without real awareness of it. You think quicker with implicit context.
Anatoly Vorobey
· 11 months ago
Paul,
"The catch is that announcing the boy also reduced the boy+girl probability by an equal amount, so the result is still the same (it eliminated either BG or GB, I don't know which, but it doesn't matter)."
That really depends on what you mean by "Arbitrarily announce the gender of one of the children". If by "arbitrarily" you mean "randomly", then the result is that it eliminated half of each from GB and BG, and the result, as you say, is 50%.
Now consider a parent that has a preference for announcing girls, so will announce a girl if at all possible, and only otherwise a boy. In that case [still within your "second algorithm" choice of the problem], hearing the parent say that at least one child is a boy will mean the probability of a boy and girl is 0, while hearing that at least one is a girl means it's 2/3, because no possibilities from the GB/BG pool are eliminated.
In a reasonable set-up where the parent will announce the gender equally likely in the GB/BG case, the answer is 1/2. Of course, the set-up that you call the first algorithm, where the parent's announcement is merely taken to be information to be conditioned to, rather than a separate choice with its own probabilities, is also very reasonable, and leads to 2/3.
P.S. In the discussion at Jeff Atwood's you link to, comments by the commenter VoiceOfUnReason seem to be the most illuminating with respect to ferreting out the difference between the several versions of the problem and their different answers.
P.P.S. Thanks for this entry. I originally wrote a critical comment to it yesterday, trying to claim that the answer is 2/3 in either case, and also that you're wrong in speaking in terms of algorithms, as it doesn't elucidate anything. I deleted it without posting after I double-checked myself and saw I was wrong on the substantive claim. I still think that speaking of algorithms isn't necessarily the best idea here - "elimination" is a metaphor that doesn't lead naturally to an exact understanding of what's going on. It's possible - and easy, once you get the essential point - to write everything up in terms of sample spaces, outcomes, and events. But hey, if thinking in terms of algorithms allowed you to see the real issue here, more power to you. I'm glad I didn't miss this post - I didn't think there was anything new to learn about this old chestnjut.
brlewis
· 11 months ago
Hi Paul. Both your probability and Jeff's are way too low. More likely, something like 99% of the time if an English speaker "told you they had two children, and one of them is a girl," the intended meaning is that exactly one of them is a girl. To precisely answer this question you need to determine the percent of people geeky enough to say "one of them is a girl" without necessarily meaning exactly one of them is a girl. Your analysis of choosing algorithms, while correct, is likely in the noise. Additionally, if the person is that geeky, then "I have two children" might not mean they have exactly two children. They might have three or more.
Pickle Pumpers
· 6 months ago
Not to mention the person said they HAD two children which is clearly past tense but they also said one of them IS a girl so one of the poor person's children is dead but we have no idea of the sex and it would be insensitive to ask so I don't think we'll ever know!
mike
· 11 months ago
Exactly. If you think the answer to the question is exactly one of 1/2 and 2/3, then you're wrong because there is simply not enough information provided to justify the ruling out of the other interpretation.
Just as you said, it comes down to whether the child is chosen first and sex is reported or the sex is chosen first and the # (= 0 or >0) of children is reported.
Ryan Van Slooten
· 11 months ago
This problem has been discussed time and time again and is a staple of statistics class.
Curiously enough, I was watching the NCAA Football Championship game last night (Florida defeated Oklahoma for those outside the US) and a TV commercial came up with Rick Rosner who, according to the commercial has the world's highest IQ. So I looked him up on Wikipedia which, in turn, led to Marilyn vos Savant. One section of her entry deals with this problem: http://en.wikipedia.org/wiki/Marilyn_Vos_Savant... This problem pre-dates her columns as well, but serves to show that the issue has been discussed extensively. The answer usually depends upon the preconditions or the manner in which the question is asked, hence the "it depends" answer as you mention.
Mitch
· 11 months ago
Here is a better question: (it isnt mine)
A final round of a game show presents the contestant with three doors. Behind two there is nothing, but behind one is a new car. They have a 33% chance of winning the car. The contestant picks a door they wish to open.
To add to the suspense, before they open the door the game show host opens one of the other doors which contains nothing. He then puts to the contestant, do you want to stick the door you chose, or do you want to switch to the other unopened door? What would you do?
It may appear there is now a 50/50 chance, switch to the other door or stick with your initial guess.
Its not obvious, but think about it, it isn't 50/50
Reinier Post
· 11 months ago
Mitch, that is the Monty Hall problem (see Wikipedia for both). It has the same "quality"of requiring additional, usually unstated assumptions before we can interpret the instance as an example of some arbitrarily large series of similar instances, which needs to be done before a valid statistical calculation can be made. In this case, the assumptions relate to how the quizmaster will behave.
If the quizmaster *always* opens a door, and the door he opens *always* comtains a goat, we arrive at the ïntended"answer.
If the quizmaster doesn't always open a door, we're in trouble. If the door he opens sometimes hides the car, we're in even more trouble. We now have to assume some consistent strategy behind the quizmaster's behavior before calculating a propability makes sense. The Wikipedia states that Monty Hall himself really didn't appear to have a consistent strategy.
Sajid
· 10 months ago
Paul, you're taking the wording of the puzzle too literally.
The puzzle should be posed as follows:
If someone has 2 children and one of them is a girl. What is the probabilty they have a boy and a girl?
The correct answer is 2/3.
akshay bhat
· 7 months ago
here is another interpretation. given that first child of a couple is boy. The probability that the other child is girl/boy is independent of the gender of of first kid. Now consider the Adam/Eve scenario assume that there is only one couple in universe, there first child is a boy. now when they mate next time, the sperms generated in are in equal proportion, the probability of the sperm either boy or girl is 50%. Thus it is independent of gender of other kid.
Lane
· 5 months ago
I might have found a better way of explaining the issue I think;
Version 1: If the person has at least 1 girl, he must always declare he has one girl, even if he has b+g.
Version 2: The person randomly picks one of the 2 children. That child happens to be a girl.
V1 will yield 66%, V2 will yield 50%.
Because, in V1, 75% of the population will say they have a girl. Which results in the 50/75 solution.
And, in V2, 50% of the population will say they have a girl (since a random child is 50% girl). But half of those (or 25% of total population) will be the g+g people. And what remains, the other half, will be b+g people.
All Comments
Step 2 in algorithm 2 is equivalent to step 2 in algorithm 1 in the situation where there are 2 children.
What you are missing is that the probability calculation is being made after you are told more information in step 2.
Let's say, though, that we consider parents with 3 children, meaning there are eight possibilities:
bbb, bgb, bgg, bbg, ggg, gbg, gbb, ggb
If we eliminate the case where parents have three boys, then the probability that there are at least two children of different genders is 6/7.
If, however, we arbitrarily announce the gender of two of the children, there is no telling whether we will be told two girls, one girl and one boy or two boys.
Nevertheless, when we are told that extra information, we make our calculations after we are told the extra information. Obviously, in the case of our being told that there is one girl and one boy, the chances of there being both boys and girls is 100%. If, however, we are told that there are two girls or two boys, the chances of there being both boys and girls is 75%.
That was my comment on the post where he *asked* the question, not where he answered it:
http://www.codinghorror.com/blog/archives/00120...
unfortunately, there's no permalinks on the comments there - grep for "bill mill".
Let's say we are asked the probability that of three children, at least one is a boy and at least one is a girl.
The probability of that happening is 6/8 or 75%.
Let's say I tell you that of three children, at least one is a boy and at least one is a girl. If I asked after providing that information the probability of there being at least one boy and at least one girl, would you still say it is 75%?
Of course not, because the extra information provided changes the probabilities.
In the original problem, we are told only that of two children, one is a girl. The other is still a mystery, and has a 50% chance of being a boy and a 50% chance of being a girl.
And then, after he writes it, lots of other people link to it and comment on it, for no other reason that lots of other people link to it and comment on it.
It's a big waste of time mostly.
In fact, announcing the boy does not eliminate either of those two options. You're getting confused between asking whether or not a particular child is a boy or girl and whether or not any child is a boy or girl.
If the question was "is the second (or any particular) child a boy or girl?", then the answer will always be 50%.
That, however, was not the question. The question was "is one of the children a boy or girl?". That means we are not concerned about the order of birth. If we are not concerned about the order of birth, both BG and GB are possibilities, while GG definitely is not.
I should have said the question was "is the second (or any particular) child a boy?", which always has a 50% chance.
The second question should have read "is one of the children a boy?", which always has a higher probability regardless of whether or not one of the children's gender is revealed.
If we are asking whether or not the second child is a boy, regardless of what is the first child's gender, there will be a 50% it will be a boy.
If one of the children is revealed to be a girl, then the chances of there being a boy is 66%.
If no child's gender is revealed, the chances of having one boy is 75%.
If one of the children is revealed to be a boy, then the chances of having a boy is 100%.
The answer to the second algorithm should be 2/3 if you use all the information provided.
WHY?
The parents ALREADY have 2 children in this context.
Each point in the space of probability is a family of two.
First, let's find the weight of all the points of families containing both sexes.
ANSWER: P(GB U BG) = 1/2
Second, let's plug our extra information. We know the family in question is PART OF THE POINTS of families containing at least one girl.
In other words, this family belongs to either GB, BG, GG.
ANSWER: P(GB U BG | GB U BG U GG) = 2/3
Read as: "The weight of GB U BG given that we're in the subspace GB U BG U GG".
The magic is in how you plug the information, not the algorithm of the question.
http://www.robdickerson.net/?p=81
http://www.robdickerson.net/?p=91
Nobody really listened :)
What about going in the opposite direction and simply ask "What is the probability that a mother has a boy and a girl?" I think many people would answer "1/2", too, simply adding the precodition "she has 2 children" ;)
The origin for this is - I think - an implicit context people are thinking of without real awareness of it. You think quicker with implicit context.
"The catch is that announcing the boy also reduced the boy+girl probability by an equal amount, so the result is still the same (it eliminated either BG or GB, I don't know which, but it doesn't matter)."
That really depends on what you mean by "Arbitrarily announce the gender of one of the children". If by "arbitrarily" you mean "randomly", then the result is that it eliminated half of each from GB and BG, and the result, as you say, is 50%.
Now consider a parent that has a preference for announcing girls, so will announce a girl if at all possible, and only otherwise a boy. In that case [still within your "second algorithm" choice of the problem], hearing the parent say that at least one child is a boy will mean the probability of a boy and girl is 0, while hearing that at least one is a girl means it's 2/3, because no possibilities from the GB/BG pool are eliminated.
In a reasonable set-up where the parent will announce the gender equally likely in the GB/BG case, the answer is 1/2. Of course, the set-up that you call the first algorithm, where the parent's announcement is merely taken to be information to be conditioned to, rather than a separate choice with its own probabilities, is also very reasonable, and leads to 2/3.
P.S. In the discussion at Jeff Atwood's you link to, comments by the commenter VoiceOfUnReason seem to be the most illuminating with respect to ferreting out the difference between the several versions of the problem and their different answers.
P.P.S. Thanks for this entry. I originally wrote a critical comment to it yesterday, trying to claim that the answer is 2/3 in either case, and also that you're wrong in speaking in terms of algorithms, as it doesn't elucidate anything. I deleted it without posting after I double-checked myself and saw I was wrong on the substantive claim. I still think that speaking of algorithms isn't necessarily the best idea here - "elimination" is a metaphor that doesn't lead naturally to an exact understanding of what's going on. It's possible - and easy, once you get the essential point - to write everything up in terms of sample spaces, outcomes, and events. But hey, if thinking in terms of algorithms allowed you to see the real issue here, more power to you. I'm glad I didn't miss this post - I didn't think there was anything new to learn about this old chestnjut.
Just as you said, it comes down to whether the child is chosen first and sex is reported or the sex is chosen first and the # (= 0 or >0) of children is reported.
Curiously enough, I was watching the NCAA Football Championship game last night (Florida defeated Oklahoma for those outside the US) and a TV commercial came up with Rick Rosner who, according to the commercial has the world's highest IQ. So I looked him up on Wikipedia which, in turn, led to Marilyn vos Savant. One section of her entry deals with this problem: http://en.wikipedia.org/wiki/Marilyn_Vos_Savant...
This problem pre-dates her columns as well, but serves to show that the issue has been discussed extensively. The answer usually depends upon the preconditions or the manner in which the question is asked, hence the "it depends" answer as you mention.
A final round of a game show presents the contestant with three doors. Behind two there is nothing, but behind one is a new car. They have a 33% chance of winning the car. The contestant picks a door they wish to open.
To add to the suspense, before they open the door the game show host opens one of the other doors which contains nothing. He then puts to the contestant, do you want to stick the door you chose, or do you want to switch to the other unopened door? What would you do?
It may appear there is now a 50/50 chance, switch to the other door or stick with your initial guess.
Its not obvious, but think about it, it isn't 50/50
It has the same "quality"of requiring additional, usually unstated assumptions before we can interpret the instance as an example of some arbitrarily large series of similar instances, which needs to be done before a valid statistical calculation can be made. In this case, the assumptions relate to how the quizmaster will behave.
If the quizmaster *always* opens a door, and the door he opens *always* comtains a goat, we arrive at the ïntended"answer.
If the quizmaster doesn't always open a door, we're in trouble. If the door he opens sometimes hides the car, we're in even more trouble. We now have to assume some consistent strategy behind the quizmaster's behavior before calculating a propability makes sense. The Wikipedia states that Monty Hall himself really didn't appear to have a consistent strategy.
The puzzle should be posed as follows:
If someone has 2 children and one of them is a girl. What is the probabilty they have a boy and a girl?
The correct answer is 2/3.
given that first child of a couple is boy.
The probability that the other child is girl/boy is independent of the gender of of first kid.
Now consider the Adam/Eve scenario assume that there is only one couple in universe, there first child is a boy. now when they mate next time, the sperms generated in are in equal proportion, the probability of the sperm either boy or girl is 50%.
Thus it is independent of gender of other kid.
Version 1: If the person has at least 1 girl, he must always declare he has one girl, even if he has b+g.
Version 2: The person randomly picks one of the 2 children. That child happens to be a girl.
V1 will yield 66%, V2 will yield 50%.
Because, in V1, 75% of the population will say they have a girl. Which results in the 50/75 solution.
And, in V2, 50% of the population will say they have a girl (since a random child is 50% girl). But half of those (or 25% of total population) will be the g+g people. And what remains, the other half, will be b+g people.