The cases where people share birthdays are more complicated than that. You could have three people with one birthday and two with another. All the combinations can be computed, but it is a lot of work, which is why we calculate the chance that all the birthdays are unique and subtract from $1$. The chance that there is exactly one pair who share a birthday is ${30 \choose 2}$ (the people to match)$365P29$(ways to choose the birthdays-the second of the pair has his chosen by the first)$/365^{30}$ (ways to choose the birthdays without restrictions).
In your code, each term is much greater than $1$, so cannot be a probability.
For question 2, the chance that some pair shares a birthday is $1$. Why do you ask?
For your edit, with five people you can list all the partitions into birthday groups: $11111,2111,311,41,5,221,32$ You can compute the chance of each one by the above techniques add all but $11111$. That will give the chance that there is at least one pair of people sharing a birthday. It is still more work than computing the chance of $11111$ and subtracting, but doable. The chance of $2111$ is $\frac{{5 \choose 2}365\cdot 364\cdot 363\cdot 362}{365^5}\approx 0.02695$