Happy Birthday to you!!

(365×364)/(365×365)

Intended solution: Let the first person pick a birthday. Now, the second person can pick any birthday EXCEPT the birthday that the first person picked; therefore the probability is simply $\dfrac{364}{365}\approx \boxed{0.99726}$

My observations: I noticed that the problem writer assumed that nobody will ever have a birthday on the leap day, February 29. I will take this extra day into account in this second solution:

Since the leap day occurs every $4$ years, it makes sense for our two people to pick from a set of dates spanning $4$ years. That is, we are picking from a set of $365+365+365+366=1096$ dates.

Case 1: the first person's birthday is not on the leap day. This has a probability of $\dfrac{1095}{1096}$ . Note that after the first person picks her birthday, the second person can pick any date except for the $4$ dates that the first person covers. This means that there is a probability of $\dfrac{1092}{1096}$ that the second person will pick a different birthday than the first person. Therefore the total probability for this case is $\dfrac{1095}{1096}\cdot \dfrac{1092}{1096}$ .

Case 2: The first person's birthday is on the leap day. This has probability $\dfrac{1}{1096}$ . Note that the second person now has a probability of $\dfrac{1095}{1096}$ of not picking the same birthday as the first person. Therefore the total probability for this case is $\dfrac{1}{1096}\cdot \dfrac{1095}{1096}$ .

Adding these two probabilities, we get the final probability to be $\dfrac{1095}{1096}\cdot \dfrac{1092}{1096}+\dfrac{1}{1096}\cdot \dfrac{1095}{1096}\approx \boxed{0.99635}$ which is indeed a different answer than the intended.

The probability that two people have the same birthday is 1/365 x 1/365. So the probability that two people don't have the same probability is 1 - (1/365 x 1/365) which evaluates to 0.9999924939...