Denote the probability of a given person's birthday being on the $k$th day of a year of $n$ days by $p_k$. Then, the probability that two random people have the same birthday is
$$P = \sum_{k = 1}^n p_k^2.$$
In particular, when $p_1 = \cdots = p_n$, this probability is $\frac{1}{n}$.
To show that this is an underestimate, it's enough to show that $P$ is actually minimized by the probability distribution $p_1 = \cdots = p_n$ (and to know that the probabilities aren't actually evenly distributed). To set up the minimization problem, note that if we know $p_1, \ldots, p_{n - 1}$, then
$$p_n = 1 - p_1 - \cdots - p_{n - 1}.$$
So as a function of $p_1, \ldots, p_{n - 1}$, the probability that two random people have the same birthday is
$$P(p_1, \ldots, p_n) = \sum_{k = 1}^n p_k^2 = \sum_{k = 1}^{n - 1} p_k^2 + (1 - p_1 - \cdots - p_{n - 1})^2,$$ and we must minimize this quantity over the region of possible probabilities, namely the simplex $S$ defined by the inequalities $$0 \leq p_1, \ldots, p_{n - 1}, \qquad p_1 + \cdots p_{n - 1} \leq 1.$$
Now, we must solve the system
$$\frac{\partial P}{\partial p_k}, \qquad 1 \leq k \leq n - 1.$$
Differentiating the above formula for $p$ and setting to zero gives
$$0 = \frac{\partial P}{\partial p_k} = 2 p_k - 2 (1 - p_1 - \cdots - p_{n - 1}).$$ The quantity in parentheses is $p_n$, so this system gives that $0 = 2 p_k - 2 p_n$, or rather,
$$p_k = p_n, \qquad 0 \leq k \leq n - 1.$$
But this simply says that there is an critical point where $$p_1 = \cdots = p_n = \frac{1}{n},$$
as desired (in particular, this point is in the simplex $S$, i.e., it corresponds to a distribution of probabilities of birthdays). Since $P$ is quadratic in its arguments, the matrix
$$\left(\frac{\partial^2 P}{\partial p_i \partial p_j}\right)$$
of mixed partials is constant, and checking shows that it is positive definite. So, the solution
$$p_1 = \cdots = p_n = \frac{1}{n}$$
is the unique global minimum, i.e., if the distribution of birthdays is nonuniform, then $P\left(\frac{1}{n}, \ldots, \frac{1}{n}\right) = \frac{1}{n}$ is an underestimate of the probability that two random people have the same birthday.
Remark (One can avoid calculus when producing the system characterizing the critical point of the probability function $P$ simply by using that a quadratic function $q(t) = at^2 + bt + c$, $a \neq 0$ achieves its extremum at $t = -\frac{b}{2a}$.)
As it was already pointed out, the probability that at least two people out of 1000 have birthday in one day is exactly 1, since there are only 366 possible dates. Your solution
$$1-\prod^{1000-1}_{j=1} \left(1-\frac{j}{366}\right)$$
is not technically correct because even the product is $0$, the term $ \left(1-\frac{j}{366}\right)$ in not a probability for $j > 366$ any more.
If birthday and death day are independent and both uniformly distributed then joint distribution of these two dates is uniform with $366^2$ possible values. Now you can use the same approach so the probability that there are not two people having same birth and death day is:
$$1-\prod^{1000-1}_{j=1} \left(1-\frac{j}{366^2}\right)$$
Best Answer
There are $4*28+1=113$ days in four years of February.
Out of $113^2$ possible $(a,b)$, there are $28*4^2+1^2=449$ ways they can coincide. So the chance is $449/113^2=0.03516...$