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[A little probability puzzle; I came up with it, but I cannot solve it.]

A class has $N=100$ students. To complete the course, the students have to pass a test, which is hard: at any attempt, any student has a probability $p=1/6$ of passing the test. On the positive side, students are allowed to take the test as many times as they like, until they pass. The students don't learn anything from previous attempts (so $p$ does not change), but they never give up: each student attempts the test until he/she passes once.

A simple probability problem is to find the expected value of the number $m$ of attempts the average student will attend the test before passing. This, however, is merely a side exercise. My question here is instead: what is the expected value of $m$ for the worst student in the class of 100, that is, the student which attempts the test for the highest number of times?

GioMott
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  • An obvious generalization is to find the expected value of m for generic N and p. Clearly, m increases as p decreases (the test becomes harder) and also as N increases (in a larger class, the worst student will reach more extreme values for m); if N tends to infinity, so does the expected value of m. However, I do not know how to analyze this problem. – GioMott May 01 '23 at 17:50
  • The expectation of $m$ in general is equal to $6$, this is a standard result obtained by direct calculation involving the geometric distribution – FShrike May 01 '23 at 17:53
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    It's relatively easy to compute the probability that they are all passed within $i$ trials. Subtracting consecutive terms in that sequence gives you the probability that the worst student (possibly not unique) takes exactly $i$ trials. – lulu May 01 '23 at 17:55
  • @FShrike you are right, that is the "simple" part of my problem. – GioMott May 01 '23 at 18:03
  • @lulu: could you kindly expand your comment in a full answer? I would be interested in a derivation of the solution. – GioMott May 01 '23 at 18:03
  • I should add that I have performed some quick computer simulations, so I have a numerical estimate. The result for the case of p=1/6 and N=100 turned out to be much higher than I anticipated. – GioMott May 01 '23 at 18:05
  • It's a numerical computation, the sum I get looks hard to evaluate in closed form. Maybe it can be rewritten to give a geometric series, but I doubt it. I get about $28.95177374$ – lulu May 01 '23 at 18:06
  • The solutions written out in the duplicate follow the path I sketched. – lulu May 01 '23 at 18:07
  • @lulu yes, I was interested in a closed-form solution, if it exists. I confirm that your numerical result is close to what I get. – GioMott May 01 '23 at 18:08

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