THE SECRETARY PROBLEM — a dart

THE SECRETARY PROBLEM look at 37%, then leap

Applicants arrive one at a time in random order; you must accept or reject each on the spot, and you want the single best. The optimal rule is startling in its simplicity: reject the first n/e (~37%), then take the first one better than everyone seen so far. It wins with probability 1/e ≈ 0.368 — the same constant as the cutoff. The mathematics of when to stop looking and commit.

THE TECHNIQUE reject first n/e, take next best; win ~1/e

The demo computes the cutoff k = ⌊n/e⌋ and the win probability for n=10: live demo

HISTORY & CREDIT Lindley 1961 · the 1/e law

“More looking is always better.” — look too long and the best likely already passed; the optimum stops at exactly n/e, no more. cited

the look phase · reject the first n/e, remembering the best seen.
the leap · take the next candidate that beats them all — win probability → 1/e as n grows.
1961 · Lindley / Dynkin; popularized by Martin Gardner as the sultan’s-dowry problem.

The same constant sets the cutoff and the odds. optimal-stopping

RECOMMEND FOR I-13 the cutoff, on the compiler

On i-13, n=10 gives k=⌊10/e⌋=3 and a win probability ≈ 0.399:

$ i13 run gm_secretary-problem.i13 RUN OK · 158 step(s) · call depth 8 n_over_e = 3.679 k = 3 pwin = 0.3987 -- (k/n) * sum_{i=k}^{n-1} 1/i near_37pct = 1

Recommend as a NULL — an optimum (B39). The n/e cutoff and 1/e probability are pinned by the model; every correct derivation agrees. An optimal-stopping theorem, not a same-function DOF. NULL — look at 37%, then leap.