Another interesting puzzle

My own reason and justification... (I quoted the above because i wanted to make sure that you are on my side when arguing ;)

Flaw with celestine's Solution:

you have taken P(b) remember that B is not the event that your friend has a boy .. But it is the probability that u meet a boy

P(bg/b) = P(bgΛb)/P(b) = 2/3

P(bg/b) is not the probability of having a boy and girl given that he has a boy... but the probability of having a boy and girl given that you have met a boy!!

Now I think u can see the subtle difference? and the mistake in ur proof?

The sibling mystery

We discussed several of the paradoxes in probability, beginning with the Sibling Mystery:

A boy you meet on the street tells you he comes from a family of two children. What is the probability he has a sister? (What assumptions are needed for this question even to make sense?)

We agreed that in two-child families, the frequencies of boys among the first children are about 0.50 and the frequencies of boys among the second children are also about 0.50. These values cannot be derived theoretically, but are statements of fact about the world that have to be learned by observation.

We also assumed that the gender of the second child in a family is independent of the gender of the first. This assumption, too, is subject to empirical testing, but our experience indicates this is at least approximately true.

These frequencies and this independence assumption let us determine the frequencies of the four kinds of two-child family. Writing the gender of the eldest sibling first, these four kinds are boy-boy, boy-girl, girl-boy, and girl-girl, each occurring with a frequency of 0.50 * 0.50 = 0.25. Therefore, the frequency of two-boy families is 0.25, of two-girl families is 0.25, and of boy-girl families (in any order) is 0.25 + 0.25 = 0.50.

We modeled this problem using tickets in a box. The box has one ticket for every two-child family. On the ticket is written the family composition. We have just deduced that about 25 percent of those tickets say "two boys." about 25 percent say "two girls," and the remaining 50 percent say "boy and girl."

Warning! The following analysis is incorrect. Keep reading to find out why.

The problem situation can now be rephrased entirely in terms of drawing a ticket out of the box. The ticket has the word "boy" on it. What are the chances that it is one of the "boy-girl" tickets? (The extensive debate over the "correct" solution to this problem revolves ultimately around whether this model is appropriate. It is possible to construct alternative scenarios that require a different probability model.)

If we were to repeat the drawing many, many, times (each time replacing the previous ticket, to leave the box contents unchanged), then evidently we would observe about twice as many boy-girl tickets as boy-boy tickets, because there are twice as many boy-girl tickets as boy-boy tickets in the box. Thus the chance that the boy has a sister is 2/3 (about 67 percent). In terms of the proportions in the box, this number is 0.25 / (0.25 + 0.50).

source of the above discussion!!

http://www.quantdec.com/envstats/notes/class_04/prob_sim.htm

Yes I agree that online sources are not reliable.. But I was having a tough time to figure out what was wrong!!

Nick Hobson (to whom I am most grateful for his time and attention) has been kind enough to point out the flaw in the previous argument. The ticket-in-a-box approach was not incorrect; it was just incorrectly executed! The process of encountering a boy (which is known as a "convenience sample" in the statistical literature) is not the same as noticing the ticket has the word "boy" on it. The reason is that when we encounter the boy on the street, we see his gender, but not his sibling's gender. In terms of tickets, we chance to notice only half the information on the ticket.

Because this can be confusing, let's use a ticket-in-a-box model that more faithfully represents what is going on. Previously, we let tickets represent families. In order to model our sampling correctly, the tickets need to add one more piece of information: namely, which sibling we encounter on the street.

Let's do this carefully. The purpose is to model the encounter in the street, without yet taking into account it's a boy we meet. So, in a quarter of the cases (corresponding again to a quarter of all two-child families), the ticket will say "two boys". We replace each of those tickets with two tickets. Each says "two boys" on the back, but on the front one of them, corresponding to meeting the older child, says "you meet a boy" and the other, corresponding to meeting the younger child, also says "you meet a boy".

In another quarter of the cases, the ticket will say "boy and girl." We replace each of those again with two tickets. On the front one of them says "you meet a boy" but the other one says "you meet a girl." We continue like this with the other two kinds of tickets, "girl and boy" and "two girls."

In this fashion the box becomes populated as follows:
Proportion Back Front
1/8 Two boys "You meet a boy."
1/8 Two boys "You meet a boy."
1/8 Boy and girl "You meet a boy."
1/8 Boy and girl "You meet a girl."
1/8 Girl and boy "You meet a girl."
1/8 Girl and boy "You meet a boy."
1/8 Two girls "You meet a girl."
1/8 Two girls "You meet a girl."

Meeting a boy on the street is tantamount to removing all the tickets that say "you meet a girl." The contents of the box are now
Proportion Back Front
1/4 Two boys "You meet a boy."
1/4 Two boys "You meet a boy."
1/4 Boy and girl "You meet a boy."
1/4 Girl and boy "You meet a boy."

That leaves half of the tickets saying either "boy and girl" or "girl and boy" on the back. We conclude that the probability the boy has a sister is 1/4 + 1/4 = 1/2, not 1/3.

I was able to reconcile the first (mistaken) analysis with this (correct) analysis by realizing that a two-boy family is twice as likely to have a boy on the street as a boy-girl family. This doubles the probability of encountering a boy from a two-boy family, thereby raising the chance he has a brother from 1/3 = 0.25 / (0.25 + 0.50) to [2*0.25] / ([2*0.25] + 0.50) = 1/2, implying the chance he has a sister is 1 - 1/2 = 1/2.

By the way, Hobson's analysis was much simpler. He reasoned by analogy with flipping coins, arguing that if "A friend grabs one (without looking at the other) and announces that it shows heads... [then] the probability that [the other coin shows tails] is 1/2." That's clear, because the coin tosses were independent. However, the whole point of this page is to show how we can use ticket-in-a-box models to solve problems in probability so that when we encounter much trickier situations, where coin-flipping analogies and the like become suspect (or harder to prove correct), we have a hope of deriving a correct solution.

The lesson I learned in making this mistake is that one must be careful to ensure that the process of drawing the tickets from the box perfectly emulates the process by which information is actually obtained; it does not suffice just to populate the box with the correct proportions of tickets.

(Updated 26 August 2004.)

Yeah all that is written there takes a lot of patience to read and understand.. but read my last comment on the previous page. That will explain what is wrong with the solution 2/3

Initially Even I had (for about 1 min) got the answer as 2/3.. Then I thought it and put the answer as 1/2 :)

This time around some of you picked this question from the old data and then it took me time to realise what was going on.. cos for aobut 10 minutes i was not able to realise what was wrong with celestine's proof :)

But it was really good we discussed... This was a wonderful question :)

@gaurav... this is a classical case of misunderstanding the question....

The statement of the question here is different from the statement of the wrong solution...

It is 1/2