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 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.
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.
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.
To LIT123:
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:
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.
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”
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 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:
Your response to this was:
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
There is no “the girl”. There is no specific child we’re being given information for.
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:
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.
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
To my recollection, Wissdoc was still not convinced. 
-FrL-
*“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.
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
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-
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.
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.
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:
- There’s a 25% chance of them having two girls. (50% times 50%)
- There’s a 25% chance of them having two boys (50% times 50%)
- 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.
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.
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-
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.
Please see my post earlier.
With two children, there are four options:
- Older brother, younger brother
- Older brother younger sister
- Older sister, younger brother
- 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?
…besides my misspelling?
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.