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### Probability of Two Male Children

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Date: 7/5/96 at 17:56:22
From: Bret McClory
Subject: Changing Probability: Two Boys

Hi! My name is Bret McClory, I'm 16, and I have a math problem that
has been bugging me.  It's more of a logic question.

Several weeks ago, in one of the weekly periodicals I read, a person
presented this logical statement: If a family has two children, and
the older child is a boy, there is a 50 percent chance the family will
have two boys.  However, in a family with two children, if all we know
is that one child is a boy (no age specified) there is a 1/3 chance of
that family having two male children.

This boggles me, as I can plainly see the first scenario - it's quite
simple: the younger child may be a boy or girl (1/2).  It is a mystery
to me how by not knowing whether the known boy is older or younger
could change the probability of the family as a whole.

However, I have discussed this problem with my grandfather and he
explains it like so: any two-child family has four possible outcomes:
b:b, g:g, b:g, g:b. He states that, in the first case where the known
boy is also known to be the older child, we thereby eliminate the g:g
and g:b combinations, leaving two left. And obviously one of those
two is b:b so our odds of having two male children is 1/2.

By his same logic, in the 2nd instance all that is known is that one
child is a boy, therefore of course the g:g combination cannot exist,
leaving b:b, b:g, g:b.  Hence, there is a 1/3 chance of two boys.

Although his explanation is very clear, I fail to understand how the
odds of the entire family can be swayed by whether or not a boy is
older or not.  Example:  I see a common ordinary boy.  I am told that
he has a sibling.  Am I to believe now that his sibling has a 2/3
chance of being a girl??  Now I am told he is the older child.
Magically the odds change and his younger sibling has a 1/2 chance of
being a girl now?

It baffles me.

Bret McClory
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Date: 7/5/96 at 20:8:17
From: Doctor Anthony
Subject: Re: Changing Probability: Two Boys

This is an example of conditional (Baysian) probability.  You are told
additional information (a condition e.g oldest child is a boy), and
this restricts the sample space, changing the probabilities.  A simple
example will illustrate the point.

Suppose I have 5 coins in a bag, one of which is a double-headed
penny.  I choose a coin at random and toss it four times. Each time I
get a head. What is the probability that I have the double-headed
penny?

The sample space is now reduced to getting 4 heads, and we have the
two possibilities to consider:

I have the double-headed coin and throw 4 heads = 1/5 * 1 = 1/5

I have a normal coin and throw 4 heads = 4/5 * (1/2)^4 = 1/20

Since one or other of these MUST happen we say:-

The sample space is now 1/5 + 1/20 = 1/4  and this is the denominator
of the probability calculation

Probability that I have double-headed coin = (1/5)/(1/4)

= 4/5

The more information you provide, the more you change the sample space
(the denominator) of the probability calculation.  In the case of sons
or daughters in a family of four, if you exclude gg or gb, then as you
said you can only have bb or bg, so 1/2 probability that second child
is a girl, whereas if one child (unspecified) is a boy, the
probability space is now bb, bg, gb and the chance of two boys is now
only 1/3.

The thing to remember is that CONDITIONAL probability can dramatically
change the commonsense idea of what a particular probability should
be.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Logic
High School Probability

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