I looked online, and found more than one and * inconsistent* answers to the Birthday Paradox when we throw the leap year into the mix. None of the answers I saw match with my own. I am posting my solution to see if it is correct, or if I am missing something.

**Question:**

Assume that the leap year occurs every four years. (i.e. ignore the 100 and 400 year rule). Also assume that the number of people born each day is the same. What is the probability that in a group of $n$ people (each one selected randomly), * no two people share the same birthday*?

**My Solution:**

Let $\mathcal{D}$ be the set of all *possible dates* in an year. (Thus $\mathcal{D}$ contains 366 elements. Note that these possibilities are not equally likely, since a person is four times as likely to be born on (say) Jan 1 than on Feb 29. This is true for any given day other than Feb 29, and it is encoded in the probability assignments given below.)

Now probability that a randomly selected person is born on Feb 29 is $\frac{1}{1 + 4*365} = \frac{0.25}{365.25}$.

Also, the probability that a randomly selected person is born on a given day other than Feb 29 is $\frac{1}{365.25}$.

Now, for a group of $n$ randomly selected people, the * Sample Space* of birthdays is $\mathcal{D}^n$. Let $\mathcal{A} \subset \mathcal{D}^n$ be the subset such that no two people share the same birthday.

Divide $\mathcal{A}$ into two disjoint sets $\mathcal{A}_1$ and $\mathcal{A}_2$ such that

$\mathcal{A}_1 = \{\xi: \xi \in \mathcal{D}^n \text{ and no two people have same birthday, and none is born on Feb 29} \}$, and

$\mathcal{A}_2 = \{\eta: \eta \in \mathcal{D}^n \text{ and no two people have same birthday, and exactly one is born on Feb 29} \}$

Now, $\mathbb{P}(\xi) = \frac{1}{(365.25)^n}$ for each $\xi \in \mathcal{A}_1$, and $\mathbb{P}(\eta) = \frac{0.25}{(365.25)^n}$ for each $\eta \in \mathcal{A}_2$.

Also, $|\mathcal{A}_1| = \; ^{365}P_{n}$, and $|\mathcal{A}_2| = \; n \; \cdot \; ^{365}P_{n-1}$

Finally, $\mathcal{A}_1$ and $\mathcal{A}_2$ being disjoint, it follows that

$$\mathbb{P}(\mathcal{A}) = \mathbb{P}(\mathcal{A}_1) + \mathbb{P}(\mathcal{A}_2) = \frac{^{365}P_{n}}{(365.25)^n} + \frac{0.25 \; \cdot \; n \; \cdot \; ^{365}P_{n-1}}{(365.25)^n}$$

**PS:**

As usual, if you want to find out the probability that at least two people share the same birthday then you would calculate $1 - \mathbb{P}(\mathcal{A})$. Interestingly, the number of people required so that this probability is more than 0.5 is still $n = 23$, same as the birthday paradox without leap year.

**Please let me know if the solution above looks accurate.**