Minimising the probability of two people having the same birthday? I was asked a particularly weird question at an interview today:

Suppose Alice and Bob live on a universe in which a year has $k$ days. What probability distribution of birthdays would you choose to a) maximise; and b) minimise the probability of the two of them to have the same birthday?

For a), I said Dirac delta, which was correct. For b), my interviewer said it should be a uniform probability distribution. I can’t see why at all. Is there a mathematical proof of this? Not too sure how to even begin. Thank you!
 A: This is how you do that formally:
Let $P_1,....,P_k$ be the probabilities of being born in day $1,...,k$.
Then the probability that Alice and Bob were born the same day is
$$\sum_{i=1}^k P_i^2$$ where $P_i^2$ is the probability that Alice and Bob were born in day $i$.
Hence you want to minimize $\sum_{i=1}^k P_i^2$ given that $P_1+...+P_k=1$. Using Lagrange multipliers you can show that the minimum is achieved where $P_i=\frac{1}{k}$ (If you want to be completely rigorous then see my comment to my answer below).
Indeed let $f(P_1,...,P_k):=\sum_{i=1}^k P_i^2$ and $g(P_1,...,P_k)=\sum_{i=1}^k P_i -1$ Let $L(P_1,...,P_k,\lambda) = f(P_1,....,P_k)-\lambda g(P_1,...,P_k)$ We compute the partial derivatives
$$\partial_{P_i} L = 2P_i -\lambda P_i$$ and $$\partial_{\lambda} L = g(P_1,...,P_k)$$
Hence we have $2P_i-\lambda = 0$ and so $\lambda = 2P_i$ this implies that all $P_i$ are equal. Since $g(P_1,...,P_k)=0$ we have that $P_i=\frac{1}{k}$ for all $i$.
The intuition behind this is that if the probability to be born in one day is higher than the others it is more likely that Alice and Bob will be born in this day (the extreme case, as you mentioned is the dirac measure which means that the probability to be born at the $i$'th day is $1$, hence Alice and Bob have to be born at the same day).
