Sex

[factor-fiction]

Following on from the entertaining Dice and Socks threads earlier this week I've now found the problem about the sex of children.

One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"

Thoughts please?

[-Talbot-]

One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"


I can't see how the answer to FF's question is anything other than 50/50.

You are stood with Mr Smith and his son waiting for Mrs Smith to come along with his other child. She will turn up with either a boy or a girl...50/50.

[factor-fiction]

Jim- thanks for your analysis. The consensus seems to be for 1/3 with a few preferring 1/2.

Jim- you went for 1/3. I still feel it depends on how we know that 'at least one is a boy'

Suppose all I know initially is that he has two children. I then walk past his house and see a young boy standing at the window in his pyjamas. So now I know at least one of his children is a boy. But that tells me nothing about the sex of the other child- and isn't there pretty much a 50-50 chance of a child being a boy or a girl?

[sunny-dave]

Correct ludwig - but the "1 of each" is Twice as Likely ...

[jim360]

It's a labelling issue again.

In the manner of the socks solution:

P(B then B) = 1/4
P(B then G) = 1/4
P(G then B) = 1/4
P (G then G) = 1/4

One is a boy, so GG excluded. Order does not matter, so label options as BB = S(ame) and BG/GB = D(ifferent).
Probabilities change* to:

P(B then B) = 1/3
P(B then G) = 1/3
P(G then B) = 1/3

P(D) = P(B then G) + P(G then B) = 1/3 + 1/3 = 2/3.
P(S) = P(B then B) = 1/3.

So it's 1/3 not 2/3

*(The probabilities aren't actually changing but are just rescaled using conditional probabilities.)

[ToraToraTora]

ludwig, for your question, using old style prob theory from school we are looking at the chance of a boy 1/2 AND a boy 1/2 when the word is AND we multiply, hence 1/2 X 1/2. your options are BB/GB/BG/GG. you are looking at it after the event rather than before.

[sunny-dave]

I refer you to my answer at 12:39, factor.

You've selected the family at random, excluded the GG option - so what are you left with?

[Old_Geezer]

The other one is also one of each ludwig, except they came in a different order. An end result of something is not the same as the number of ways to get it.

[ToraToraTora]

I too started off like that talbot, now I see, it is 1/3 for For Funks Sake question

[jim360]

It does depend on how we know to an extent, but the question implies that we only know that "at least one is a boy". Absent any further information we have to calculate on the basis that it's either Girl/ Boy or Boy/ girl.

Incidentally, the question has excluded the possibility of twins, which probably screws things up yet again as it's more likely to have identical twins than fraternal twins of opposite gender. But, again, we calculate based on the assumption that there were no twins involved, and as long as we state this then the probability is 1/3, not 1/2.

[pixie373]

Ummmm, thanks Dave. Logically I'm still with Factor and Talbot - but it did take Newjudge about three months to explain that goat/car one to me.

[ToraToraTora]

sorry I put for factorfiction's question but I used the initials and that got translated by AB, apologies!

[pixie373]

No, fraternal twins are more likely, Jim :-)

[ludwig]

// Correct ludwig - but the "1 of each" is Twice as Likely ... //

Oh bu99er.

Stupid question.

[woofgang]

Ludwig two girls is ruled out by the OP

[woofgang]

Yes!!! this
OG "The other one is also one of each ludwig, except they came in a different order. An end result of something is not the same as the number of ways to get it."

[pixie373]

So the options left are two boys- or one of each.

[jim360]

Is it? Well, there you go. I should perhaps have looked it up -- but anyway the point is that including twins would change the question.

The answer is, anyway, 1/3 IF we assume no twins, that we only know that one is a boy, and that boys and girls are equally likely. All of these are idealised, but in this ideal world the answer is correct.

[ToraToraTora]

ok I know a guy who has 5 daughters! 1/32! from the outset!

[sunny-dave]

OK Woofie - there are two possible results :

I win the lottery

I don't win the lottery

By your analysis (as the only two possible outcomes) they are equally likely?

[sunny-dave]

There are also only two outcomes of this problem - but (unless we know whether the boy we have seen was born first or second) they are not equally likely.