TAOCP 6.4 Exercise 4
Let $N = 365$ and let $n$ be the number of people.
Exercise 4. [**] [M30| How many people should be invited to a party in order to make it likely that there are three with the same birthday?
Verified: no
Solve time: 1m15s
Let $N = 365$ and let $n$ be the number of people. For each fixed day, the number of unordered triples of people born on that day has expectation $\binom{n}{3}N^{-3}$. Summing over all days, the expected number of triples of distinct people sharing a birthday is
$$ \lambda = N \binom{n}{3} N^{-3} = \binom{n}{3} N^{-2}. $$
This quantity becomes of order $1$ when
$$ \binom{n}{3} \approx N^{2}, $$
that is,
$$ \frac{n(n-1)(n-2)}{6} \approx 365^{2}. $$
Replacing $n(n-1)(n-2)$ by $n^{3}$ for the scale of growth gives
$$ n^{3} \approx 6 \cdot 365^{2}. $$
Hence
$$ n \approx (6 \cdot 365^{2})^{1/3}. $$
Since $365^{2} = 133225$ and $6 \cdot 365^{2} = 799350$, one checks $92^{3} = 778688$ and $93^{3} = 804357$, so the transition occurs between these values, closer to $93$.
A group of about $93$ people makes the appearance of a triple birthday collision likely in the same sense as in the standard birthday paradox threshold. $\boxed{93}$