Correct application of birthday problem The problem is as follows:
There are 10 shooters at a shooting range. Each shooter is given 5 bullets. They all begin shooting at 9am and end shooting at 10am, They each shoot all 5 of their bullets over that period.
The shooters each choose 5 random time points over the hour and they shoot 1 bullet per time point. The time resolution used is 1 second (3600 possible time points). So the times 09:37:53.100 and 09:37:53.707 are considered to have occurred in the same second.
What is the probability that two shooters fire a bullet at the same time point?

I thought the birthday problem could be used to solve this problem. I replaced the number of days in a year (365) with 3600, then I replaced the number of people in the room with the total number of bullets fired which is 50, resulting in a roughly 29% probability of two shots occurring at the same time.
However when I reduce the problem to just one shooter with 5 bullets, the above construction gives 0.13% probability.
This is wrong because a shooter can only shoot 1 bullet at a time, the probability should be zero, hence the above method is wrong.
I was wondering if some could suggest a way to solve this particular problem.

I think a rewording of the problem could be as follows:
There is a room with MxN people, where M < N. Each person is associated with only one of M clubs and all the people are equally divided amongst the M clubs. What is the probability that two people from different clubs share the same birthday? (day/month combination only)

UPDATE
How would one apply Poisson distribution to this problem? As it seems that the math gets pretty complicated for the exactly 2 collisions problem.
 A: Consider the probability that no shots are shot in the same second. The first shooter has $ 3600 $ shots to choose from, the next $ 3595 $, and so forth until the fifth has $ 3580 $ possibilities. Hence, the probability that none of them are in the same second is $$ \prod_{i = 0}^4 \frac{\dbinom{3600 - 5i}{5}}{\dbinom{3600}{5}} \approx 0.9327 $$ Hence, the complementary probability (the one you want) is approximately $ 0.0673 $. 
A: As with the birthday problem, let's compute the probability that there is no coincidence. 
Let the first shooter pick his five seconds.  The second shooter has a $\frac{3595}{3600}$ chance of avoiding a coincidence with his first shot.   He has a $\frac{3594}{3599}$ chance of avoiding a coincidence  with his second shot, and a $\frac{3593}{3598}$ with his third shot.  Hence the desired probability (of avoiding coincidence) is
$$1\cdot \frac{3595^\underline{5}}{3600^{\underline{5}}}\cdot \frac{3590^\underline{5}}{3600^{\underline{5}}}\cdot \frac{3585^\underline{5}}{3600^{\underline{5}}}\cdots \cdot \frac{3555^\underline{5}}{3600^{\underline{5}}}$$
This simplifies to  $$\frac{3595^{\underline{45}}}{(3600^{\underline{5}})^{9}}=\frac{3600^{\underline{50}}}{(3600^{\underline{5}})^{10}}$$
This is in contrast to the birthday problem, whose answer is $$\frac{365^{\underline{n}}}{365^n}$$
