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December 24

Banknotes and good- enough probability

This a real problem I will face soon.

I have a pile of 20 banknotes. How many I need to check for counterfeit in order to get 95% that the entire pile is not a counterfeit?

And a side question - how to calculate this for any n banknotes with given p percent certainty?

Thanks! אילן שמעוני (talk) 13:29, 24 December 2018 (UTC)[reply]

Hi, I presume your question is how many notes do I have to check to be 95% sure there are no counterfeit notes in the pile. To answer this question, you need to know the actual (or an estimated) probability that any individual bank note is counterfeit. If it's greater then or equal to 5%, then you have to check all of them, and for very low probabilities, you don't need to check any of them (for example with n=20 a probability of 0.25% lets you check none of them with 95% certainity). Iffy★Chat -- 13:51, 24 December 2018 (UTC)[reply]

The difficult variable here is the source of the banknotes. The usual calculations are meaningless unless the twenty come from random sources. Dbfirs 14:02, 24 December 2018 (UTC)[reply]

18 gives only 90% certainty while 19 gives 95% certainty.

  odds=.[:*/!/"1&i.&(,:-)&>:
  ({.%+/){.18 odds 20
0.904762
  ({.%+/){.19 odds 20
0.952381

Bo Jacoby (talk) 23:41, 25 December 2018 (UTC).[reply]

Aren't you making the assumption that the likelihood of each one being counterfeit is 50%? Bubba73 ^{You talkin' to me?} 07:38, 26 December 2018 (UTC)[reply]

I have no prior knowledge. Before observation the numbers 0,1,2,...,20 of counterfeit notes have odds 1:1:1:...:1. Having observed that none of 19 notes are counterfeit the odds are 20:1:0:,,,:0, and the likelihood of no counterfeit notes is 20/21 = 0.952381. Bo Jacoby (talk) 05:40, 28 December 2018 (UTC).[reply]

That computation doesn't really seem to say much. To be 90% sure, you check 90% of the bills. To be 95% sure, you check 95% of the bills. Bubba73 ^{You talkin' to me?} 19:43, 29 December 2018 (UTC)[reply]

If n out of 20 notes are checked to be OK, the probability that they are all OK is (n+1)/21, rather than n/20. Bo Jacoby (talk) 22:47, 30 December 2018 (UTC).[reply]

Assuming independence of all the banknotes from each other: Let n = number checked, so number unchecked = 20 – n. Let P_all = probability that all are good (0.95 in your case), and P_one = probability that any one is good. Then P_all = P_one^20-n. Solving for n gives $n=20-{\frac {\log P_{all}}{\log P_{one}}}.$ If this gives n ≠ integer, then choose the next larger integer. Loraof (talk) 00:39, 27 December 2018 (UTC)[reply]