Mom has 2 children and saids that at least one of them is a boy

timothy...@gmail.com

unread,

Sep 15, 2006, 5:38:03 PM9/15/06

to

"You meet a woman and ask her if she has any children. She replies,
"two." You ask if she has any sons and she says, "yes." So now
you know she has exactly two children and at least one of them is a
boy. What is the probability that her other child is also a boy, and
therefore that she has two sons?" - Skeptics Guide to the Universe
podcast

Any ideas on this?

Jack Tomsky

unread,

Sep 15, 2006, 6:05:54 PM9/15/06

to

Suppose that the two children are ordered. Then the problem is the conditional probability,

P(BB|BG or GB or BB) = P(BB)/P(BG or GB or BB)
=P(BB)/[1-P(GG)] = (1/4)/(3/4) = 1/3.

Jack

Reef Fish

unread,

Sep 15, 2006, 7:50:18 PM9/15/06

to

Jack Tomsky wrote:
> > "You meet a woman and ask her if she has any
> > children. She replies,
> > "two." You ask if she has any sons and she says,
> > "yes." So now
> > you know she has exactly two children and at least
> > one of them is a
> > boy. What is the probability that her other child is
> > also a boy, and
> > therefore that she has two sons?" - Skeptics Guide to
> > the Universe
> > podcast
> >
> > Any ideas on this?
> >
>
>
> Suppose that the two children are ordered.

Not sure what you mean by "ordered", as in Mom called up
Sears and Nobuck and ordered two children? ;-)

>From the answered to the question "Do you have any sons" and
she answered "yes", I don't see any ordering effect in the answer.

The only possibility that was ruled out was GG, among the
original possibility of BB, BG, GB, and GG. So, the conditional
probability that the other is a Boy, is the usual conditional
probability over the restricted space of (BB, BG, GB), and
the probability of two boys is therefore 1/3, same answer
as your gave, expressed in the language of restricted space.
That assumes the B/G ratio is 1:1 and the combinations are
independent.

If the question is asked of REAL Moms, then the probabilities
are slightly different, due to TWO empirical facts: (1) There are
more Girls than Boys and (2) The birth process is Markov rather
than Bernoulli, that if P(G|G) > 1/2 and P(B|B) is also > 1/2.
that is, boys and girls tend to have "runs" in boys or girls.

My wife's has two sisters (3 girls in the family); One of our
nieces has 3 boys, before finally having a boy; her sister has
three boys. So, if I assume the original unconditional space
to be: P(BB) = .26, P(BG) = P(GB) = .22, and P(GG) = .3,
then the conditional probability of (BB|B) would be .26/(.26+.22+.22)
or 26/70, slightly greater than 1/3. :-)

On the THIRD hand, the condition didnt state whether the boys
and girls have to be natural born by the same Mom. All the
probabilites change if adopted boys and girls are admitted as
well. :-)

Then a Portuguese Monte Carlo program will have to invoked
to prove that 1/3 < 26/70 with probability .78. :0)

-- Reef Fish Bob.

john2

unread,

Sep 17, 2006, 6:15:16 PM9/17/06

to

Is this similar to the Monty Hall problem ?

john2

Reef Fish

unread,

Sep 17, 2006, 7:58:11 PM9/17/06

to

Only if Mom is Monty Hall's wife.

If you had seen my explanation that a conditonal probability is NOTHING
but an unconditional probability on a RESTRICTED space, restricted by
the given condition, then you wouldn't have to blindly name drop.

-- Reef Fish Bob.

kennet...@sbcglobal.net

unread,

Sep 18, 2006, 10:03:13 PM9/18/06

to

Easy! 3/4.

<timothy...@gmail.com> wrote in message
news:1158356283.5...@i42g2000cwa.googlegroups.com...