I agree with your intuition. I think the author must either be looking at statistical noise or have made a mistake. At least for the case of two people, if we

I agree with your intuition. I think the author must either be looking at statistical noise or have made a mistake. At least for the case of two people, if we say that a_n is the probability of a person's birthday being chosen (so that ∑(a_n) = 1), then the chance of a match is ∑(a_n^2). To construct a rigorous argument, one could bring in the power mean theorem[1], which states that, if x > y and the sequence {a_n} is all nonnegative, then (avg(a_n^x))^1/x ≥ (avg(a_n^y))^1/y, with equality if and only if all elements of {a_n} are the same. This can be used to show that, if ∑(a_n) is fixed, then ∑(a_n^2) (i.e. the probability of a match) is minimized only when {a_n} are all equal. It's less obvious how to make a rigorous argument for the cases of 3 and more people, but https://en.wikipedia.org/wiki/Muirhead%27s_inequality https://en.wikipedia.org/wiki/Muirhead%27s_inequality might help (if you trust the proof of it). [1] https://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality https://en.wikipedia.org/wiki/Generalized_mean#Generalized_m...