The fifth user is first to repeat a code

Question

Users are given codes one at a time, each a uniform random $2$ -digit code from $00$ to $99$ (one of $100$ codes).

What is the probability that the fifth user is the first to receive a code that a previous user already has? (The first four are all distinct, and the fifth matches one of them.)

Accepted Answer

Two things must happen: the first four codes are all distinct, then the fifth collides with one of them. Part 1, the first four all distinct: P(	ext{first 4 distinct}) = \frac{100}{100}\cdot\frac{99}{100}\cdot\frac{98}{100}\cdot\frac{97}{100} \approx 0.9411. Part 2, given four distinct codes are out, the fifth matches one of those four: P(	ext{5th matches} \mid 	ext{4 distinct}) = \frac{4}{100} = 0.04. Multiply: P(	ext{first collision at user 5}) \approx 0.9411 	imes 0.04 \approx 0.0376, about 3.8\%. Takeaway. "First repeat exactly at step k" equals (chance the first k-1 are distinct) ti…

The fifth user is first to repeat a code

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