Assume \(d\) equally likely days.1 Seat people one at a time; the \(i\)-th person avoids everyone before them with probability \(1 - i/d\). So
\[\Pr[\text{no collision}] \;=\; \prod_{i=0}^{n-1} \left(1 - \frac{i}{d}\right). \label{eq:exact}\]Using the inequality \(1 - x \le e^{-x}\), a consequence of convexity,2 on each factor of \eqref{eq:exact} and summing the exponents gives a clean upper bound:
\[\Pr[\text{no collision}] \;\le\; \exp\!\left(-\frac{n(n-1)}{2d}\right). \label{eq:bound}\]With \(d\) equally likely values and \(n(n-1) \ge 2d \ln 2\), two of the \(n\) values coincide with probability at least \(1/2\). In particular \(n \approx \sqrt{2d \ln 2} \approx 1.18\sqrt{d}\) suffices.
By \eqref{eq:bound}, \(\Pr[\text{no collision}] \le \exp(-n(n-1)/2d) \le \exp(-\ln 2) = 1/2\).
For \(d = 365\) the condition reads \(n(n-1) \ge 505.99\ldots\), and \(n = 23\) is the first value that meets it, since \(23 \cdot 22 = 506\).
Drag the slider to 23 and notice that the curve crosses one half there, then drag it to 57: the probability is already above 99%.
Replace “birthdays” with “hash values”. A \(b\)-bit hash has \(d = 2^b\) outputs, so by Theorem \ref{thm:sqrt} a collision is more likely than not after about \(1.18 \cdot 2^{b/2}\) items. A 64-bit hash, which sounds enormous, gets there after roughly \(1.18 \cdot 2^{32} \approx 5\) billion items. That is a normal day for a large database.
import math
def items_until_likely_collision(bits: int) -> float:
"""Smallest n (as a real number) with n^2 >= 2 * 2**bits * ln 2."""
return math.sqrt(2 * 2**bits * math.log(2))
print(f"{items_until_likely_collision(64):.3e}") # 5.057e+09
The same square root explains why birthday attacks halve the security of a hash function, and why 128-bit identifiers are the usual choice when collisions must essentially never happen.
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