Ten servers grabbing random ports

Question

Ten services each pick a random port to listen on, choosing independently and uniformly from 50 available ports. What is the probability that at least two services pick the same port?

Accepted Answer

"At least two clash" has many cases, so compute the opposite, all ten ports distinct, then subtract from 1. Assign ports one service at a time, each avoiding the ports already taken: - Service 1: probability 50/50. - Service 2: avoids 1 taken port, probability 49/50. - Service 3: avoids 2 taken ports, probability 48/50. - down to service 10, which avoids 9 taken ports, probability 41/50. Multiplying: P(	ext{all distinct}) = \frac{50}{50}\cdot\frac{49}{50}\cdots\frac{41}{50} \approx 0.382. Hence P(	ext{at least two clash}) = 1 - 0.382 \approx 0.618, about 62\%. With only 50 ports and \binom{…

Ten servers grabbing random ports

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