Boy Or Girl Paradox

Probability Level 3

Mr. Jones has two children.
The probability that he has a girl knowing that he has a boy is $\dfrac{2}{3}$ .
What is the probability that he has a girl knowing that he has a boy born on Thursday?

Assume that it is equally likely that a child of any sex is born on any day.

\frac{14}{27}

\frac{2}{3}

\frac{12}{25}

\frac{1}{2}

\frac{3}{4}

1 solution

Romain Milon
Jul 9, 2016

Relevant wiki: Conditional Probability - Problem Solving

MR. Jones can have either a two boys, a boy and a girl, a girl and a boy or two girls. We will denominate those pairs this way :

BB

BG

GB

GG

We know that MR. Jones has a boy, so we can eliminate the two girls possibility ; and our sample space is now:

BB

BG

GB

We can easily find the probability that MR. Jones has a girl by isolating favorable events (which are $BG$ and $GB$ ). Thus, the probability is $\frac{2}{3}$ .

But knowing that the boy is born on Thursday changes everything. We now need to break down each pair of children possible and indicate the day each child is born on. To do this, we will order the day of the week (Sunday to Saturday) and denote each child this way: $B_k$ for a boy born on $k^{th}$ day of week and $G_k$ for a girl born on $k^{th}$ day of week ( $1 = Sunday$ to $7 = Saturday$ ). We now have a sample space that looks like:

$B_5B_1$	$B_1B_5$	$B_5G_1$	$G_1B_5$
$B_5B_2$	$B_2B_5$	$B_5G_2$	$G_2B_5$
$B_5B_3$	$B_3B_5$	$B_5G_3$	$G_3B_5$
$B_5B_4$	$B_4B_5$	$B_5G_4$	$G_4B_5$
$B_5B_5$	$B_5B_5$	$B_5G_5$	$G_5B_5$
$B_5B_6$	$B_6B_5$	$B_5G_6$	$G_6B_5$
$B_5B_7$	$B_7B_5$	$B_5G_7$	$G_7B_5$

Because the first as well as the second boy can be the one born on Thursday, we have two columns for the possibility (pair) $BB$ . We note that we get two identical pairs ( $B_5B_5$ ). We can get rid of one ; if both children (being boys) are born on thursday, it does not matter if we know of the first one or the second one is born on thursday. Thus, our final sample space is:

$B_5B_1$	$B_1B_5$	$B_5G_1$	$G_1B_5$
$B_5B_2$	$B_2B_5$	$B_5G_2$	$G_2B_5$
$B_5B_3$	$B_3B_5$	$B_5G_3$	$G_3B_5$
$B_5B_4$	$B_4B_5$	$B_5G_4$	$G_4B_5$
$B_5B_5$		$B_5G_5$	$G_5B_5$
$B_5B_6$	$B_6B_5$	$B_5G_6$	$G_6B_5$
$B_5B_7$	$B_7B_5$	$B_5G_7$	$G_7B_5$

We see that 14 pairs out of the 27 possible combinations of children contains a girl. Therefore, our probabilty is $\frac{14}{27}$ .

Logically this does not make sense although mathmatically you are absolutely accurate. Knowing he has a boy means that there is an 18/27 chance that his second child is a girl but what you are postulating is that if you know what day the boy was born the chance his second child is a girl drops to 14/27. Logically that would mean it would have to always be 14/27 as you will always know the boy was born on a day. What if you knew the boy was born on the 256th day of the year then the probability drops to 730/1459 which again means it would always have to be 730/1459 as the boy will always be born on a day in the year. The problem with this puzzle is that it assumes that the day of birth of one child has some bearing on the probability of gender of the second child, it does not, therefore children are a bad context for this puzzle.

Darren Sanderson - 4 years, 11 months ago

The logic of this problem is really subtle, i didn't really understand myself why it works like this at first, but try to think about it this way: having two boys makes it more likely to have at least 1 of them born in a given day, so that affects the chances in favour of that possibility. Actually, it's the exact same line of thinking wich gave us that 2/3 at the beginning

Matteo Verni - 4 years, 11 months ago

In an event of an equal chance sample space, it is. In a sample space, there are the same amount of boys in a family with 2 boys and in families that have 1 boy. So, the more you narrow the traits of the baby down, you screen out families with only 1 boy much faster than family with 2 boys. In other word, if you have 14x14 families with unique pair of child (gender and week day), the odds is as given. But however, if 1 of those families were to have a new child, the date of birth and gender would be 1/2 and 1/7 respectively.

Lam Nguyen - 4 years, 10 months ago

Nicely illustrated solution.

Agnishom Chattopadhyay - 4 years, 11 months ago

Why is it assumed they were born in the same week? Why wouldnt it just be 2/3 because the question is literally just what was given to you (2/3 chance girl knowing he has a boy), verbatim, plus some extra info that should be useless? I wish it was easier to decipher the questions sometimes, because once i found out what you interpreted the question as, it became pretty simple.

Henry Lembeck - 4 years, 11 months ago

The extra info seems useless, but it actually provides valuable data.

Siva Budaraju - 4 years, 1 month ago

If a boy was born on thursday, having another boy would double the chances of that happening.

Siva Budaraju - 4 years, 1 month ago

Boy Or Girl Paradox

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