# 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.

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With $2$ shooters, the probability that they don't fire a bullet at the same time is $$\left(1-\frac{5}{3600}\right)\left(1-\frac{5}{3599}\right)\left(1-\frac{5}{359‌​8}\right)\left(1-\frac{5}{3597}\right)\left(1-\frac{5}{3596}\right)$$ This also happens to be equal to $$\frac{\binom{3595}{5}}{\binom{3600}{5}}$$Can you see why? Can you generalize? – Jared May 23 '13 at 0:00
@Jared is the numerator in the final expression the number of combinations that don't overlap? – Samonir May 23 '13 at 1:12

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 $0.0673$.

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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 $$\frac{365^{\underline{n}}}{365^n}$$

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Vadim thank-you for the answer, but I have one question, doesn't this result imply the probability for 2 or more collisions? – Samonir May 23 '13 at 1:09
@Samonir, this is the probability of 0 collisions; subtract from one to get the probability of $\ge 1$ collisions. – vadim123 May 23 '13 at 1:36
Thats what I meant, but the question seems to ask for exactly 2. So is the process to find the probability of >= 3 and subtract the probability of >= 1 ? – Samonir May 23 '13 at 4:16