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If I have a device that displays a random number from 1 to 10 billion and I use that device X times, what is the probability that not all of the numbers displayed are unique?

Andrew Downes
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    https://en.wikipedia.org/wiki/Birthday_problem – Alex R. Feb 06 '19 at 19:34
  • It seems like it is explained here: https://en.wikipedia.org/wiki/Birthday_problem#Cast_as_a_collision_problem

    But I don't really understand those formulas.

     – Andrew Downes Feb 08 '19 at 10:25
  • Thanks for sharing that. The article says "By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are only 366 possible birthdays, including February 29). However, 99.9% probability is reached with just 70 people, and 50% probability with 23 people." Can I assume that those proportions also apply if we increase from 366 options to 10 million? I.e. when we have used 23/366 * 10,000,000 = 630,000,000 ish random numbers, the probability of collision is 50%? – Andrew Downes Feb 08 '19 at 11:23
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    If you want to find out of $10^9$ how many you need to pick to make the probability of collision 50%, you can use the formula (given on the Wikipedia page) $n(p; d)\approx\sqrt{2d\cdot\ln\left(\frac{1}{1 - p}\right)}$. Since $n(0.5; 10^9) \approx 37233$, there will be a 50% probability of collision when you have around 37233 items. – leekaiinthesky Feb 08 '19 at 20:17

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