Exactly $2$ pairs of identical birthdays in group of $15$ I've done a bit of searching but haven't found a solution to this question. If there is a group of 15 people what is the probability that there are exactly 2 pairs of birthdays (i.e. 2 people share one bday, another 2 share a different bday and the remaining 11 don't have a bday on either of those days nor share a bday amongst themselves)? 
When I tried to solve using combinations I came up with 0.1009609. Here is the math:  
((365 C 2) * (363 C 11)) / (379 C 15)
When I built a python script to simulate this I came up with 0.025011.
I'm trying to understand which answer is correct, or if neither is correct.
 A: There are $\binom{365}{2}$ ways to pick which two birthdays are shared among the two pairs, $\binom{15}{2}$ ways to pick the first pair of people with a shared birthday, $\binom{13}{2}$ ways to pick the second pair of people with a shared birthday, $\binom{363}{11}$ ways to pick the remaining $11$ birthdays, and $11!$ ways to permute them. This is all over the denominator, $365^{15}$, the total number of birthday outcomes.
$$P = \frac{\binom{365}{2} \cdot \binom{15}{2} \cdot \binom{13}{2} \cdot \binom{363}{11} \cdot 11!}{365^{15}} = 0.0247618$$
This result is also substantiated by this Python simulation:
from random import randint
from collections import Counter

trials = 10000000
numerator = 0

for trial in range(trials):
    birthday_freqs = Counter([randint(1, 365) for person in range(15)])
    if sorted(birthday_freqs.values())[-3:] == [1, 2, 2]:
        numerator += 1

print(numerator / float(trials)) #0.0247677

A: Assuming there are $365$ days in a year, then the result should be $$\frac{{365\choose 2}\cdot{15\choose 4}\cdot \frac{4!}{2!\cdot 2!}\cdot{363\choose 11}\cdot 11!}{365^{15}}$$
Where ${365\choose 2}\cdot{15\choose 4}\cdot \frac{4!}{2!\cdot 2!}$ is the number of possibilities where we choose two days from $365$ days and choose four people out of $15$ such that there are two share a birthday while the other two share another birthday. On the other hand, ${363\choose 11}\cdot 11!$ is the number of possibilities for us to choose $11$ days for the remaining $11$ people. 
A: As I did it, the probability of one pair sharing a birthday is $\frac{1}{365}$, and there are $\binom{15}{2}=105$ pairs. Therefore $P(\text{exactly }2\text{ pairs})=P(2\text{ pairs})*P(\text{not }103\text{ pairs})*\binom{105}{2}\text{ ways to choose the pairs}=(\frac{1}{365})^2(\frac{364}{365})^{103}\binom{105}{2}\approx3.089\%$
which is fairly close to the answer. However, I think there is something wrong here, as it is about $1\%$ off, but I'm not quite sure why it didn't work.
