smozoma Posted June 30, 2021 Report Share Posted June 30, 2021 On 6/29/2021 at 3:37 PM, kingraph said: @Tickenest, perhaps you can help answer this question? If we assume 12 batches of 29000 copies of A-L, what are the chances/probability that we would find a duplicate serial number at some point if we kept picking one at a time? This is very similar to the Birthday Problem (you'd think it would be hard to find 2 people with the same birthday, but by checking 23 people there's a 50/50 chance of finding a duplicate) Birthdays are completely independent, but in our case we're not being exact because each time you see a code, it reduces the odds of finding the same number by 1/12 but also increases it by..1/29000? But anyway ignore that... https://betterexplained.com/articles/understanding-the-birthday-paradox/ We have about 180 codes so far, not counting the un-lettered codes. pairs: 180*179/2=16110 = A The chance of 2 codes NOT being the same (pretending it's totally independent): 28999/29000 = 0.9999655 = B Then the chances of getting this many codes without a match is: B^A = (28999/29000)^16110 = 0.5737 So at this point we have a 57% chance to have NOT seen a match. 43% chance to have seen a match. At 200 codes, we'll be pretty close to 50/50. 1 Quote Link to comment Share on other sites More sharing options...
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