syonidv

That leads to 1/2 unambiguously, but the OP does not support that interpretation

I believe the mere presence of "other" in the OP implies that our given information pertains specifically to one individual (not two collectively) and the question asks us to infer information about the "other" individual.

At its heart, the 1/3-ist interpretation is that the given information pertains to the joint distribution of two individuals, and the solution is an inference that can be made about the conditional joint distribution (i.e. the fact that in the FF case, we may conveniently neglect the issue of selecting an "other" and conclude "female" since both siblings are female).

But perhaps this speaks to the point of many posts in this forum, which is that the problem is poorly stated. The answer depends on what specifically is meant by "two kids; one girl"--how that information is obtained.

Thanks to all for the discussion, at any rate.

If you have 2 children, both girls, and 1 of them is a girl, then the 'other' child is also a girl. I really don't see your point.

I have 2 daughters. If someone says to me "Do you have at least 1 daughter?", I say "Yes". If they then ask what the gender of my 'other' child is, I don't say "what other child?"

Indeed, sir, but to the question "What is the gender and age of your other child?", the parents would answer, "What 'other' child?"

The problem is unique in that even if there is no specific 'other' child, we have the liberty of assuming 'female' since this is the only possibility. My point is that the problem statement specifically asks for the gender of the 'other child', indicating that the information we have pertains to one child, and not to two collectively. To wit, we are asked for the gender of the 'other child', not "What is the probability both children will be girls?"

I also hold fast to the 1/2-ist view and the hard semantics of 'other' (rather than Mr. Ben's reasonable "both can be correct" view) because applying conditionals to groups (e.g. "at least one of X or Y is") can lead to inconsistencies for even slight variations of the representative simulation.

Consider a dentist's office that serves 500 2-child families. Each child has a file. Suppose the dentist asks the receptionist to place a star on the file of any child whose family has at least one girl. Later, the dentist asks the receptionist to look through all the files with stars on them and for each one write down the gender of the patient's sibling.

In this case, the list of siblings would be roughly 2/3 female and 1/3 male.

I see where you're coming from, Ben, but let me play the contrarian.

The specific "condition" statement offered by the problem is: "They have two kids, one of them is a girl."

The question to be answered (Q2BA) is: "What is the probability that the other kid is also a girl?"

I agree with your assessment on the difference between the 1/3-ist question and the 1/2-ist question, but the Q2BA makes it clear that the 1/3-ist question is not well-defined in all contexts.

Suppose I ask the 1/3-ist question: "Do you have at least one girl?" and the parents reply "Yes." We'll call this a "yes" family.

Then I ask: "What is the gender of the other child?"

This question is well-defined if the parents have one boy and one girl, but for the approx. 1/3 of the time that the "Yes" families have two girls, they'll ask what 'other' child you're talking about. You didn't specify a 'first' child such that there can be an 'other' child in this case.

Or to make the point clearer, suppose the second question is: "What is the gender and age of the other child?"

If you asked this question, the parents' response would appropriately be "Uh... actually.. we have two daughters. Which one do you mean?", which is ostensibly them telling you "there is no 'other' child; you didn't specify one to begin with".

The 1/2-ist perspective doesn't suffer this drawback. For the 1/2-ist question: "Think of which of your kids did <totally random action>. Is this kid a girl?" (e.g. "Think of which of your kids you saw chewing bubblegum last. It is a girl?", or "Which kid do you have with you? Is it a girl?", etc.) and they answer 'yes'.

Now you have a well-defined 'other' in all contexts, as well as the condition statement being satisfied.

And since families with two girls will be "Yes" families twice as often than 1-boy 1-girl families (per Bayes' rule) under this scheme, our 1/2-ist philosophy comes by its name rightfully.

My $0.02, at any rate.

Wow, I was a doubter, but I have to say I'm impressed.  The question I will ask all of you, before all of your bickering and mathematics, is have you tried it out?  I wrote a simple program to chuck out the numbers, and the results are pretty astounding.

Your simulation is flawed, specifically at the point where you query

if( family.Child1 || family.Child2 )

The 'or' condition encapsulates the condition that the family has at least one girl, but it does not encapsulate the restriction that once a girl has been observed, we need to definitively assign this girl as "sibling #1" or "sibling #2".

Consider the corrected code, which honours the constraint that the observed child must be either child 1 or child 2 (I wrote it in javascript, and you can run it by simply pasting it into a site such as this one:

(function() {
    var BOY = 0, GIRL = 1;
    var FIRST = 0, SECOND = 1;

    var second_girls = 0, second_boys = 0;

    for( var i = 0; i < 100000; i++ ) {
        var family = {
            child1: Math.random() < 0.5 ? BOY : GIRL,
            child2: Math.random() < 0.5 ? BOY : GIRL
        };

        var child_we_happen_to_see = Math.random() < 0.5 ? FIRST : SECOND;
        var gender_of_child_we_see = (child_we_happen_to_see == FIRST) ? family.child1 : family.child2;

        if( gender_of_child_we_see == BOY )
            continue; /* this is our conditioning; omit any trial where a boy is observed */

        var gender_of_other_child = (child_we_happen_to_see == FIRST) ? family.child2 : family.child1;
        if( gender_of_other_child == BOY )
            second_boys++;
        else second_girls++;
    }

    alert( "Number of boys is " + second_boys + ".\n" +
        "Number of girls is " + second_girls + ". This is " +
        (new Number( 100*second_girls/(second_girls + second_boys))).toFixed( 2 ) + '%.' );
})();

In my run, I get "Number of boys is 25020. Number of girls is 24799. This is 49.78%."

This makes intuitive sense. The number of included trials is approx. 50,000--all those cases where we happened upon a family and happened to observe a girl child with them. And of those families, approximately half have a boy as the second child.

I believe somebody mentioned this earlier, but an excellent analog that helps make sense of "why you need to constrain the observed child to either child 1 or child 2" is called the "Gambler's Ruin" problem.

Consider I have two boxes, A and B. I tell you that there is some non-zero amount of money in the boxes, and that one box contains exactly twice the amount of money as the other one. But I don't tell you which one is which.

Suppose you pick box A. It has some unknown amount of money, X, in it. But then you think: wait a minute, there's a 50% chance that box B has twice as much money, or 2X, and a 50% chance that it has half as much money, or X/2. So if we calculate the expected value for the amount of money in box B we get

E(B) = 0.5*(2X) + 0.5*(X/2) = 1.25X

Or, roughly 25% more money than in box A. So we decide to swap. But now we let our (slightly larger) amount of money be X and the same logic applies to swapping back to box A.

Hence, by simply swapping A with B again and again, we can make our amount of money grow to infinity! We're rich!

Of course, we can disprove this fallacy by honouring the "our box must be only one of either A or B" constraint. That is, the calculation

E(B) = 0.5*(2X) + 0.5*(X/2)

assumes that the box we're holding is in a fuzzy state of simultaneously being both the greater AND the lesser money box (X/2 is only valid if the former is true, and 2X is only valid if the latter is true).

In the same way, "if( family.Child1 || family.Child2 )" assumes the child we've observed is in a fuzzy state of simultaneously being both the first AND second child. But it has to be one or the other.

Let mine be one more vote for itsclueless's correct (also, rational ) answer. The probability of the "other child" being a girl is 1/2.

The answer offered by brainden.com is incorrect, and even the cusory consideration that observing one sibling changes the probability of the other's gender should set off red flags in people's minds.

The proof that I give to people is as follows:

The two facts we know are:

two children have been born (fact A)
at least child is female (fact B)

Fact A gives us the four possible populations (male/female configurations):

M-M
M-F
F-M
F-F

It then also seems logical to eliminate any combination where at least one female is not present (i.e. M-M), leaving,

M-F
F-M
F-F

and then concluding a 67% probability of 2M and 1F. The problem is that these three remaining combinations are not equally likely. That is, we've botched the application of Bayes' rule.

The correct approach is to start with as our four equally-likely populations:

M-M
M-F
F-M
F-F

We now sample randomly from one of these four possible populations. We sample a 'F'. (This sampling process is completely equivalent to the statement "there is at least one female in the population". No more and no less information is gained.)

Since we know the a priori probabilities of each of the four configurations (populations) and the probability of sampling an F given each configuration, Bayes' rule tells us what the (conditional) probability of each configuration is after we've observed an 'F'.

I turns out that prob( M-M ) = 0, as expected (i.e. you can't sample a male from female-only population), but for the remaining three,

prob( M-F ) = 1/4
prob( F-M ) = 1/4
prob( F-F ) = 1/2

Or stated more intuitively: the fact that we sampled a female member gives us a higher probability that we sampled it from a female-rich population. In this case, the probability of the F-F config is exactly the same as the combined probabilities of the F-M and M-F configs, giving 50%/50% as expected.



Hence, to summarize, we started out with four populations of equal a priori probability:

M-M : probability 1/4
M-F : probability 1/4
F-M : probability 1/4
F-F : probability 1/4

We observed that at least one of the two children is female, which, through a straightforward application of Bayes' rule, yields our post-observation probabilities:

M-M : probability 0
M-F : probability 1/4
F-M : probability 1/4
F-F : probability 1/2

In configurations 2 and 3, the "other child" is always male. In configuration 4 (which is twice as likely given our observation), the "other child" is always female. Hence:

probability( second child is male ) = 1/4 + 1/4 = 1/2
probability( second child is female ) = 1/2 = 1/2

Q.E.D.