On an imaginary planet, if you are in a member of a team and it is your birthday then you have to buy cake for everyone in the team.
If I am in only 1 team and I only ever eat cake when someone in my team brings birthday cake (including me) then I know I can calculate the probability of cake as:
$$ \Pr(\text{cake today}) = \Pr(\text{birthday today}) = 1 - \Pr(\overline{\text{birthday today}}) $$
If I have a team of size $n$ including myself and there are $m$ days in a year, then the probability it is not someone's birthday is:
$$ \Pr(\overline{\text{birthday today}}) = \Big({\frac{m-1}{m}}\Big)^n $$
So if the planet has 365 days a year and I am in a team of size 30, then I get
$$ \Pr(\text{cake today}) = 1 - \Big(\frac{365 - 1}{365}\Big)^{30} = 0.0790... $$
My problem is that when I try to obtain the same result by counting, I get a different result and I am not sure where I am going wrong:
$$ \Pr(\text{cake today}) = \frac{\text{# all birthday sequences given at least one birthday today}}{\text{# all possible birthday sequences}} $$
I treat # all possible birthday sequences cardinality as a bars and stars problem
$$ \text{# all possible birthday sequences} = \binom{n + m - 1}{n} $$
and then treat # all birthday sequences given at least one birthday today the same except I pre-allocate one of the team member's birthday to today:
$$ \text{# all possible birthday sequences given at least one today} = \binom{n + m - 2}{n-1} $$
$$ \Pr(\text{cake today}) = \frac{\binom{n + m - 2}{n-1}}{\binom{n + m - 1}{n}} = \frac{n}{n+m-1} $$
$$ \Pr(\text{cake today}) = \frac{30}{30 + 365 - 1} = 0.0761... \neq 0.0790... $$
Where am I going wrong?
If $n=2$ and $m=3$ then you count $\binom{2+3-1}2=6$ possible birthday sequences. I reckon that comes to the $3$ possibilities that there are $2$ birthdays on the same day plus the $3$ possibilities that there are $2$ birthdays on distinct days (repair me if I am wrong). However these possibilities are not equiprobable.
Observe that the probability that you and your team-mate both have your birthday on e.g. day $1$ is $(\frac13)^2$, but the probability that one of you has it on day $1$ and the other on day $2$ is $2\times(\frac13)^2$.