Total ways to select birthdays for all persons: $$365^{n}$$ If $n\ge365$, there's $100\%$The probability by piegeon hole principle. If $n<365\iff n\le364$ ways to have atleast one person with same birthday as you is the same as total ways minus the no. of ways in which none of them have same birthday as you: $$365^{n}-\binom{364}{n}(n)!$$ The probability is thus
[Explanantion: $$P(E)=\begin{cases}\begin{align}&1-\frac{\binom{364}{n}(n)!}{365^n}\quad &n\le364\\&1\quad &n\ge365\end{align}\end{cases}$$Spoiler]
Choose n birthdays from 364 days(except the one which you have), all different dates then select one person whose birthday you want to change to yours and arrange the rest.Total ways of selecting birthdays will be $365^n$. In case of 366 persons select one person to have same birthday as you and arrange the rest 364 persons with the remaining 364 dates.
$$\frac{\binom{364}{n}\binom{n}{1}(n-1)!}{365^n}:\quad n\le 364\\\frac{\binom{365}1364!}{365^n}:n=365\\0:n\ge366$$ Sorry I know combinatorics and haven't yet covered probability or expected value.