- 30 Sep '11 22:58A survey covered all the two-child families in the country. It can be assumed that the birth rates of boys and girls are the same, and there is no correlation of birth genders of elder or younger sibling.

In one of the families was a girl child called Mary. How likely is it that Mary has a brother? - 01 Oct '11 06:16

wrong. it's 50:50, 1/2.*Originally posted by Thomaster***25% of the families has boy - boy**

25% of the families has boy - girl

25% of the families has girl - boy

25% of the families has girl - girl

If Mary's a girl, she's either in a boy/girl or girl/boy family, or in a girl-girl family. That makes 25+25/(25+25+25) = 2/3

those odds are answer the question: how like is it that Mary's family has two daughters?

Mary's sibling, a brother or sister? that just leaves the odds at brother or sister.

There could be some more specific odds calculation, like .49999999999999999999, if it was stated that there were precisely an equal number of males and female children, but without that, it's always 50:50, because the births are random. - 01 Oct '11 09:22 / 2 editsThis problem is complicated on many levels. For background reference on how specification of the name (or more generally the subject of 'identification' can alter the problem, I would recommend a paper like "On the so called Boy or Girl Paradox" by D'Agostini.

Anyway, based on analysis like D'Agostini's, my answer would be as follows.

Let m be the fraction of girls possessing the name Mary within the relevant population. If we do not allow that two children within the same family can have identical names, then m eventually drops out of the analysis and the answer is 1/2. If we do allow that two children within the same family can have identical names (which is rather bizarre in practice), then the answer is 1/(2-m/2). Note that in the case of allowing identical names within a family and in the extreme where m = 1 and the specification of Mary loses whatever power it had of unique identification, the answer = 2/3. - 01 Oct '11 15:09How was Mary selected?

Was it:

Pick a random family, find out the name and sex of the first child, and make some statement like "One of the families has a child called Tom".

If that is how it was done then the first child is not correlated with the second child : 50/50

Or was it:

Pick a random family, if it contains 2 boys then pick another, if not give the name of the oldest girl.

If that is how it was done, then the 2 boy families are gone, giving only 3 viable combinations (BG; GB; GG), so there is a 2/3 chance Mary has a brother.

Or did we have a list of all children, find one called Mary, and find the family she is in. In this case we are back to a 50/50 chance. - 01 Oct '11 16:31yeah, and the problem also doesn't state whether or now Mary has a little lamb and whether the lamb has been sheared

the answer is 50:50 with some X-1 factor, as Mary, presumably, is female, and also, presuming that there really is no significance to the name of some female being Mary - 01 Oct '11 17:20 / 2 editsSomething to specify then for next time.

* all the girls in the country are called Mary. Half the boys are Peter, the other half Paul.

* Mary was a child selected by random of all the children of two-child families in the country to present a bouquet of flowers to the great leader.

* when the great leader received the bouquet, he wanted to ask about Mary's sibling. But he had forgotten the gender of the sibling from his briefing data, and wanted to look good.. so he chose to ask randomly. But it was nice to optimise odds so he asked, "How is your brother, little Mary?"

A variant of the same is a magician has three playing cards. One is white on both sides, one is white on one side and black on the other, one of black on both sides. he picks a card at random and sees that the visible side is white. The odds of the other also being white is..

a. 1/2 as there are two cards, white-black and white-white; or

b. 2/3 as there are three sides of cards, WHITE-white, white-WHITE, and white-black.

Yet another variant is the famous two-goats-and-a-car gameshow.

There are three curtains, A, B, and C. One has a car behind it, the other two have goats (or gift certificates to get a car or goat so the size of the curtain and animal smell and sounds are no hint). Contestant can pick what is behind a curtain of his choice. He chooses A, at which point the host removes a curtain.. say, curtain C.. and show that that one was a goat. The contestant can change his mind at this point and pick B instead of A. Is it beneficial for him to do so? - 01 Oct '11 23:22

your final version gets to the point.*Originally posted by talzamir***Something to specify then for next time.**

* all the girls in the country are called Mary. Half the boys are Peter, the other half Paul.

* Mary was a child selected by random of all the children of two-child families in the country to present a bouquet of flowers to the great leader.

* when the great leader received the bouquet, he wanted to ask about Mary ...[text shortened]... can change his mind at this point and pick B instead of A. Is it beneficial for him to do so?

in the gameshow, the host removes a choice. the original chance of winning was 1 in 3. after removal, the player should "switch" as the other side has a 2 in 3 chances of winning. the 1 in 3 chances doesn't change. a good book and a lot of articles and blogs and link sites have discussed this over the last 10 years or so.

your first puzzle is completely different. it's 50:50, or maybe, as said before, something like 49.9999:50.00001, depending on the population and an even birth count. - 02 Oct '11 02:45

WRONG*Originally posted by Thomaster***2/3**

If that were true it would mean that every 2 child family with a daughter had a 2/3 chance of having a son and daughter.

Switch genders and we can then say every 2 child family with a son has a 2/3 chance of having a son and a daughter.

So therefore EVERY 2 child faamhaving a son and daughter - which is clearly false. - 02 Oct '11 08:22 / 1 editFor simplicity let's say that there are only four families.

* Mary One and Mary Two

* Paul One and Mary Three

* Mary Four and Paul Two

* Peter One and Peter Two

one of the eight children is selected randomly. And turns out to be one of the girls.

Mary One and Mary Two have a sister.

Mary Three and Mary Four have a brother.

Makes a compelling case in favor of 50 - 50.

Two more variants.

* I toss two coins on the table, and show the first to you. It's heads. Likelihood of the second being tails? One in two.

* I toss two coins. You ask whether I tossed heads. I show you one of them and indeed, it's heads. Likelihood of second being tails? Two in three. - 02 Oct '11 16:23

maybe you don't fully grasp randomness?*Originally posted by talzamir***For simplicity let's say that there are only four families.**

* Mary One and Mary Two

* Paul One and Mary Three

* Mary Four and Paul Two

* Peter One and Peter Two

one of the eight children is selected randomly. And turns out to be one of the girls.

Mary One and Mary Two have a sister.

Mary Three and Mary Four have a brother.

Makes a compellin ...[text shortened]... show you one of them and indeed, it's heads. Likelihood of second being tails? Two in three. - 02 Oct '11 22:33

Returning to the subject of 2 children families, do you think the following questions yield the same answer?*Originally posted by coquette***maybe you don't fully grasp randomness?**

Question 1: What is the probability of brother & sister, given that one of the children is called Mary?

Question 2: What is the probability of brother & sister, given that at least one child is a girl?