In the problem with the siblings, there are many levels to it. The problem was: Given 2 children in a family, and you know one of them is female, what are the odds that the other is male. The answer was 2/3, not the initial reaction of 1/2.

However, for real life the answer is still 1/2. It matters how the information on the gender of the one sibling is aquired. If, for example, you ask. “Is at least one of your childer female?”, and the answer is “yes”, then the 2/3rds answer is correct. However, if the knowledge of the gender is just a random piece of information, then the answer is 1/2. A way to see this is to think about a computer program that generates random familys (1/4 - all male, 1/4 all female, 1/4 male older than female, 1/4 female older than male) Then the program randomly picks a child and give you the gender. If it tells you a child is female, one of 4 things has happened. 1. you know the gender of the young child in a mixed family. 2. you know the gender of the older child in a mixed family. 3. you know the gender of the younger of two girls. 4. you know the gender of the older of two girls. In 2 of the 4 cases the sibling is male, so the odds are back to the intuitive 1/2.

The answer definitely depends on the exact situation. Mind providing a link to the situation mentioned, so that others can comment on which answer is correct?

-Ryan

" ‘Ideas on Earth were badges of friendship or enmity. Their content did not matter.’ " -Kurt Vonnegut, * Breakfast of Champions *

And yet… Having kids is like flipping coins. Each time you do it (no pun intended), you have a 1/2 probability of having a boy or a girl. This seems suspiciously like the Monty Hall scandal from a while back…

Do the Punnet squares, though. 1/2 chance!

I think Ennius is right. It seems to depend heavily of the phrasing of the problem, just like the Monty Hall thing. Imagine the following bit of conversation:

“So, you have two kids? What do they do?”

“Well, Jane’s an accountant, and…”

“Tickets, please!”

So, what do you know at the point of interruption? That either the younger *or* the older child is female. I’m going with GentDave on this one.

It doesn’t have to be “older” or “younger”; it can be any distinction – “taller” and “shorter”, “smarter” and “stupider”, whatever.

John W. Kennedy

“Compact is becoming contract; man only earns and pays.”

– Charles Williams

Yep, in the example TheIncredibleHolg makes, the odds on the other child are 1/2. If the information is random the odds are 1/2. However, if the information is specificly requested, i.e. “Is at least one of your children female?” the odds on the other child being male are 2/3. The difference is because a person is more likely so say something about a daughter, if there are 2 daughters, than if there is just 1.

I’ve got to disagree here. The probability of GentDave’s 4 cases are not equal; the mixed ones are twice as likely. This problem is similar to the Monty Hall problem & a similar analysis in that thread leading to a 1/2 answer.

The 4 cases are:

B B

B G

G B

G G

The math is Bayes’ Theorem: The probability of A, given B - written Pr(A|B) - is: the probability of (A and B), divided by the probability of B.

A = one kid is a boy

B = one kid is a girl

Pr(A and B) = 0.5

Pr (B) = 0.75

Pr(A|B) = 0.5/0.75 = 2/3

Daniel Moore is correct, but only for the specific case given. In the case solved here the information is simply that there are two children, and one is female. In that case, the odds are indeed 2/3rds. The extra twist is to concider the odds of aquiring that information is the first place. Assume that finding out the gender of any of the 8 children listed is equally probable, and that you find out one is a girl. That information is twice as likely to have come from the family with 2 girls, than one of the families with 1 girl. This brings the probability for real life situations, where the information is randomly aquired, back to 1/2. If you specificly request information, i.e. “Is at least one of your childer female?”, then the odds of aquiring the information are no longer relavent, and the odds are 2/3rd.

Have a computer generate random 2 child families, and give the gender of a random child. If it gives the gender as female, the other child will be female only half of the time. The reason is that there were 8 possible children it could have chosen. 4 were female, and 2 were in an all female house. So the chances of the house being all female, given that a girl has been chosen is 2/4 or 1/2.

One little addition, to compare this to the Monty hall example. This is similiar to saying “Does Monty know what is behind the door or not.” If Monty opens a RANDOM door , your odds, are 1/2 (assumeing the RANDOM door showed a goat. If it showed the prize your odds are 0). But since he knows where the prize is, and opens a door with the goat, your odds change to 2/3.

Whether the information is random, or influenced in some way, matters in both problems.

Another factor - there are more males born than females. I don’t know the present stats but 50 yrs ago more girls were alive at the end of one year.

I think there’s another point to consider. While statistically, the population at large is roughly 50-50, does it necessarily follow that any particular couple will always have a 50-50 chance?

some friends of mine were expecting their third child. their first two kids were boys. they were told by their doctor that the odds were high that the third child would be a boy, because a particular couple may have body chemistries/genetic factors that favour one gender over the other. in the population at large, these personal idiosycracies even out, but at the individual level there may be a preference - it’s not as mechanical as, say, flipping a coin.

any one with a medical/biology background who can comment? if it is true, it means you need more facts to make the prediction - rather like Cecil’s argument that you need to know what Monty’s role is in the door game.

p.s. - the third child was a boy.

If the doctor told them that, because they had had two boys, that “the odds were high that the third child would be a boy”, then the doctor needs to learn more about statistics. Even if there *are* such effects, two boys is much too likely as a merely random occurence to have any statistical significance.

John W. Kennedy

“Compact is becoming contract; man only earns and pays.”

– Charles Williams

## From Cecil’s original column:

The possible gender combinations for two children are:

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

## We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.

The interesting part (I hope I’m not repeating what others have said) is that the answer assumes an equal possibility of each of the 3 remaining occurences. If you are somehow seeing an actual female (imagine that one of those 6 kids left was selected), note that the odds are twice as great it came from choice #2 (4 females left, two of them are in choice #2). Since choice #2 is twice as likely, your overall odds of the “other one” being female are 1/2. This is one way of looking at the seeming discrepancy that small changes in language evoke.

That is, is what you first see randomly selected by the choice, or randomly selected by the child? It makes all the difference.

John W. Kennedy posted 09-05-1999 10:11 AM

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Well, it depends on how common "sex preferences" are. Suppose, as a simplicifaction, that all couples have a 75%/25% split. Those that favor males will produce two males in a row (75%)^2=56.25% of the time. Those that favor females would produce two males in a row (25%)^2=6.25% of the time. Probability of two boys, then a boy= (56.25%)(75%)+(6.25%)(25%)=43.75%. Two boys, then a girl= (56.25%)(25%)+(6.25%)(75%)=18.75%. Given two boys, probability of next child being boy= (43.75)/(43.75+18.75)=
```

=

70%

Granted, that is based on a rather high prefernce rate, and it isn’t as high as the 95% usually required for statistical significance (which actually isn’t relevant here, anyway), but it’s pretty high.