The twentieth PIN is the first duplicate

Question

PINs are issued one at a time, each a uniform random $3$ -digit code from $000$ to $999$ (one of $1000$ codes).

What is the probability that the twentieth PIN is the first to duplicate an earlier one? (The first nineteen are all distinct, and the twentieth matches one of them.)

Accepted Answer

Two parts must both occur: the first nineteen PINs are all distinct, then the twentieth collides with one of them. Part 1, the first nineteen all distinct: P(	ext{first 19 distinct}) = \frac{1000}{1000}\cdot\frac{999}{1000}\cdots\frac{982}{1000} \approx 0.8419. Part 2, given nineteen distinct codes are out, the twentieth matches one of those nineteen: P(	ext{20th matches} \mid 	ext{19 distinct}) = \frac{19}{1000} = 0.019. Multiply: P(	ext{first duplicate at PIN 20}) \approx 0.8419 	imes 0.019 \approx 0.0160, about 1.6\%. Takeaway. First repeat exactly at step k equals (chance first k-1 d…

The twentieth PIN is the first duplicate

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