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From: riderofgiraffes on 26 May 2010 11:28
A relative of the Monty Hall problem, and currently
causing some division and argument. Originally posed
by Gary Foshee.

Suppose I have two children, and one of them is a boy
born on a Tuesday. What is the probability that both
my children are boys?

Don't be too quick to react. Just as with Monty Hall
some well-known and well-respected mathematicians are
quite definite, it's just that they don't all get the
same answer.
From: dannas on 26 May 2010 16:27

"riderofgiraffes" <mathforum.org_am(a)solipsys.co.uk> wrote in message
news:560624655.228654.1274902118922.JavaMail.root(a)gallium.mathforum.org...
>A relative of the Monty Hall problem, and currently
> causing some division and argument. Originally posed
> by Gary Foshee.
>
> Suppose I have two children, and one of them is a boy
> born on a Tuesday. What is the probability that both
> my children are boys?

tuesday is immaterial.
being born is immaterial.

the paragraph below is immaterial.

>
> Don't be too quick to react. Just as with Monty Hall
> some well-known and well-respected mathematicians are
> quite definite, it's just that they don't all get the
> same answer.


From: danheyman on 26 May 2010 16:36
On May 26, 3:28 pm, riderofgiraffes <mathforum.org...(a)solipsys.co.uk>
wrote:
> A relative of the Monty Hall problem, and currently
> causing some division and argument.  Originally posed
> by Gary Foshee.
>
> Suppose I have two children, and one of them is a boy
> born on a Tuesday.  What is the probability that both
> my children are boys?
>
> Don't be too quick to react.  Just as with Monty Hall
> some well-known and well-respected mathematicians are
> quite definite, it's just that they don't all get the
> same answer.

I claim the correct answer is 1/3 (with the usual assumptions of
independence and P{boy}=1/2).
The formal way to get the answer is with Bayes' theorem. The easy way
is to start with the four a priori equally likely events (in birth
order)

BB BG GB GG

The last one is ruled out and the other 3 are still equally likely.

Note that if we're told he is the older child, the answer is 1/2
because the second event is also ruled out.
From: W^3 on 26 May 2010 17:30
In article
<560624655.228654.1274902118922.JavaMail.root(a)gallium.mathforum.org>,
riderofgiraffes <mathforum.org_am(a)solipsys.co.uk> wrote:

> A relative of the Monty Hall problem, and currently
> causing some division and argument. Originally posed
> by Gary Foshee.
>
> Suppose I have two children, and one of them is a boy
> born on a Tuesday. What is the probability that both
> my children are boys?
>
> Don't be too quick to react. Just as with Monty Hall
> some well-known and well-respected mathematicians are
> quite definite, it's just that they don't all get the
> same answer.

My interpretation is to consider the collection of 4-tuples (s1 d1 s2
d2), where s1, s2 lie in {b, g} (the sexes of the children), and d1,
d2 lie in {1, 2, ..., 7} (the days of the week these children are
born). All of these variables should be independent, so each 4-tuple
has probability 1/2*1/7*1/2*1/7.

Let E be the event {s1 = b, d1 = 3} U {s2 = b, d2 = 3}. The
probability we seek is P({s1 = b, s2 = b} intersect E)/P(E). I get
13/27 for this.
From: Rob Johnson on 26 May 2010 18:04
In article <560624655.228654.1274902118922.JavaMail.root(a)gallium.mathforum.org>,
riderofgiraffes <mathforum.org_am(a)solipsys.co.uk> wrote:
>A relative of the Monty Hall problem, and currently
>causing some division and argument. Originally posed
>by Gary Foshee.
>
>Suppose I have two children, and one of them is a boy
>born on a Tuesday. What is the probability that both
>my children are boys?
>
>Don't be too quick to react. Just as with Monty Hall
>some well-known and well-respected mathematicians are
>quite definite, it's just that they don't all get the
>same answer.

I believe that implicit in the question (whether actually true or
not) is that a boy is as likely as a girl and that no day of the
week is any more likely for a boy to be born than any other.

To make sure we have disjoint cases, let's divide the possibilities
of having two children, one of them a Tuesday Boy, as follows:

Girl & Tuesday Boy = 7/14 * 1/14 = 7/196
Tuesday Boy & Girl = 1/14 * 7/14 = 7/196
Non-Tuesday Boy & Tuesday Boy = 6/14 * 1/14 = 6/196
Tuesday Boy & Non-Tuesday Boy = 1/14 * 6/14 = 6/196
Tuesday Boy & Tuesday Boy = 1/14 * 1/14 = 1/196

Thus the probability of a Tuesday Boy is 27/196.
The probability of a Tuesday Boy and two boys is 13/196.

Thus, given two children, one a Tuesday Boy, the probability of two
boys is 13/27.

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
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