Problem $71$ of Chapter 4 from Introduction to Probability by J. Blitzstein and J. Hwang.
In a group of $90$ kids, what is the approximate probability of there being at least one pair of kids born on the same day, whose mothers are also born on the same day?
Could someone validate my attempt below? Thanks!
Attempt
Let $I_{j}$ be the indicator random variable for pair $j$ having the aforementioned property. $P(I_{j}=1) = \frac{1}{365^{2}}$,
under the assumption that the probability of being born on a particular day is $\frac{1}{365}$. Note that since we don't know anything about the age of the kids, we are assuming their mothers are also equally likely to be born on any of the $365$ days.
Then, the expected number of pairs with the aforementioned property is $\text{E}(X) = \binom{90}{2}\frac{1}{365^{2}} \approx 0.03$.
Let $Z \sim \text{Poiss}(0.03)$ model the distribution of pairs with the desired property. Then, probability that there is at least one such pair is $1 – P(Z=0) = 1 – e^{-0.03} \approx 1 – (1 – 0.03) = 0.03$.
Best Answer
Looks okay to me. Let's look at the precise answer and see how it agrees with your approximation.
Assuming no leap years, this is equivalent to the vanilla birthday problem with $N:=365^2$ days, accounting for all (birthday, mom's birthday) possibilities. So the exact probability that at least two kids among the $k:=90$ kids share the same (birthday, mom's birthday) tuple by the standard birthday problem is
$$1-\frac{N}{N}\times \frac{(N-1)}{N}\times...\times \frac{(N-k+1)}{N}\\ =1-1\times \left(1-\frac{1}{N}\right)\times...\times \left(1-\frac{k-1}{N}\right) \approx 0.029621,$$
which is approximated as
$$\approx 1-\exp{(0)}\times \exp{\left(-\frac{1}{N}\right)}\times \exp{\left(-\frac{k-1}{N}\right)}\\ =1-\exp\left(-\frac{k(k-1)}{2N}\right) \approx 0.029615.$$
As you have noted, this is just the Poisson approximation for a binomial model, where the number of matching pairs of kids is modeled as binomially distributed with $k\choose 2$ "trials" and a $1/N$ probability of "success."