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This is the example I am looking at.

Question:
Suppose a family has $2$ children and one is a boy, and that the probability of having a child of either sex is equal and independent across children. What is the probability that they have $2$ boys?

Now going by intuition, I would say that the sample space would be this. $$\{B, B\}, \{B, G \}, \{G , B\} \text{ and }\{G , G\}.$$

So given that the first child is a boy. Then there is a $50\%$ $(1/2)$ chance for the second child to also be a boy.

But going by the Bayes' rule the problem is solved like this.
Let $\mathscr S = \left\{\{B,B\},\{B,G\},\{G,B\},\{G,B\}\right\}.$ Then

$$ \Pr(\{B, B\} \mid |B ≥ 1)=\frac{\Pr(B ≥ 1| \{B, B\})\times \Pr(\{B, B\})}{\sum\limits_{S\in\mathscr S}\Pr(B\geq 1\mid S)\Pr(S)}.$$ Therefore after reduction: $$\Pr(\{B, B\} |B ≥ 1)= 1/3.$$

The link to how this was solved is here.

I understand that when dependent events are in play we need to use the Bayes rule. When I apply the formula I get $1/3$ too. What I don't understand is what part of my fundamental understanding that is incorrect. Why is $1/2$ incorrect. When to identify and avoid such wrong intuitions.

Thanks for your time.

Em.
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maths_learner
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    Humans have terrible intuition regarding probability in general, and the only way to get better is to repeatedly work out problems like this. Even then, it's still iffy. – EuYu Jun 02 '16 at 08:44
  • Similar question here http://math.stackexchange.com/questions/1800658/probability-of-having-a-girl – Hailey Jun 02 '16 at 08:52
  • Formatting tips here. – Em. Jun 02 '16 at 08:53
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    The problem says "one is a boy" but you assume "the first child is a boy". Again, this is a matter of phrasing, but the intuitive answer matches with the Bayesian answer. – shardulc Jun 02 '16 at 09:06

2 Answers2

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$\require{cancel}$You gave the sample space $$\{\mathsf{(B, B), (B, G), (G , B),(G , G)} \}.$$

It's ordered, hence the parentheses.

Then you are told one is a boy, not that the first is a boy, and hence $$\{\mathsf{(B, B), (B, G), (G , B),\cancel{\boxed{\mathsf{(G , G)}}}} \}.$$

There is one case that meets the requirement. Assuming only two genders, equally likely, then the choice is $1/3$.

Further, the point is that you were given some information. If I told you I would give you $\$1000$ if you told me how many boys the family had, would you rather have

  1. No information, or
  2. Known that they have at least one boy?

(Rhetorical)

Em.
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For an intuitive approach:

If you know that one of the children is a boy then the possibilities are

B B
G B
B G

then the probability of both being boys is 1/3

Manoel Rebelo Abranches
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