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A person's Facebook account has 435 connections.

Of those connections 3 have given birth to daughters in Feburary.

Of those three births two have occurred on the mother's birthday.

Question: What is probability of two women giving birth to girls on their birthday?

Here's my understanding: Assumes probability of boy/girl each at 0.5

You have two independent events.

Event #1:Mother #1 giving birth to female baby on Mother #1 birthday.

Event #2:Mother #2 giving birth to female baby on Mother #2 birthday.

So what is the probability of event #1?

It is the probability of choosing two people Mother #1 and Baby Female #1 at random on same birthday.

The probability of Mother #1 birthday any specific day of the year is 1/365
The probability of Female Baby #1 birthday any one specific day of the year is 1/365 * 1/2

Event #1 = Probability of Both A1 and B1 = P(A1) * P(B1) = 1/365 * 1/365 * 1/2 = 1/266,450

So what is the probability of event #2?

The same as event #1.

What is the probability of Event #1 AND Event #2

Probability = P(event #1) * P(Event #2) = 1/266,450 * 1/266,450= 1/70,995,602,500

Approximately 1 in 71 Billion

It this an accurate explanation?

Would the probability change if:

  1. Only 3 of the 435 connections were female?

  2. All 435 connections were female?

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    The specific birthday of the mother doesn't matter as the mother birthday could be any date. – 2017-02-23
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    Assuming every woman has exactly one child (why are you assuming that?) Then the probability of a woman have a daughter with the same birthday is 1/365*1/2 (1/2 for a girl 1/365 child has same birthday). Probability that is not the case is 729/730. Probability of 0 out of 200 women don't share is (729/730)^200 and the probability of at least one is 1-(729/730)^200 about 23%. prob exactly 1 is 200*(729/730)^199 1/730. At least 2 is 1-(729/739)^200-200 (729/730)1/730 approx 3%. It's unusual but not unheard of. – 2017-02-23
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    If all 435 are women then probity of at least 1 is about 45 percent. Of at least 2 is about 12 percent. – 2017-02-23
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    I'm not making any assumption about the number of children a woman is having. I'm just asking about two women in a set of 435 who are giving birth to children on either the same day or on separate days where both births are on the mother's birthday. The question is about the probability of two specific births, regardless of number of previous births. – 2017-02-23
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    Then the number of women are irrelevant as you are not looking at *women* you are looking at births. And the number of births is also irrelevent as you are not looking at general births as you are looking at *specific* births. And each of your probabilities are 365 times too small as the specific date doesnt matter. – 2017-02-23

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There are two main sources of fuzziness in your question. The first is how many of the 435 contacts are females; some people have a very significant skew in the gender of their contacts. The second is how many children (and in particular daughters) those females have given birth to; the answer if obviously quite different if we are looking at the contacts of a $15-$year old, or at the contacts of a $51-$year old.

In general, a mother of $d$ daughters has probability $\approx$ $(\frac{364}{365})^d$ of having none of them born on her birthday. The $\approx$ is due to some years having $366$ days (which increases the probability) and to the fact that children are not born evenly throughout the year (though close enough). For smallish $d$, we can then approximate the probability of a mother of $d$ daughters having at least one born on her own birthday as $1-(\frac{364}{365})^d\approx \frac{d}{365}$. If we have $m$ such mothers, the probability that exactly $s$ have the same birthday as a daughter is then $\approx{m\choose s} (\frac{d}{365})^s (1-\frac{d}{365})^{m-s}$. For smallish $s$ and $d$, and for $md$ significantly smaller than $365$, we can approximate this further as $\approx\frac{1}{s!}(\frac{md}{365})^s$.

So, if all the contacts are mothers (i.e. $m=435$), and each has $1$ daughter ($d=1$), for $s=2$ we have a probability $\approx{435\choose 2} (\frac{1}{365})^2 (1-\frac{1}{365})^{435-2}$ or a little over $20\%$ that exactly $2$ of them will have a daughter with the same birthday. Not that unlikely! Note that $md > 365$ in this case, so we can't quite use the most simplified formula. On the other hand, if we only have $3$ mothers ($m=3$), we can use the most simplified formula and compute the probability as $\approx \frac{1}{2!}(\frac{3\cdot 1}{365})^2$ or about $1$ in $30000$ - far less likely, but not quite impossible.