Secretary problem

You are interviewing n candidates for a position, arriving in uniformly random order. After each interview you must hire or reject immediately — no going back. You want to maximise your probability of hiring the single best candidate.

The optimal strategy has a clean structure: observe and reject the first r candidates (the “scouting” phase), then hire the first candidate strictly better than all of them.

The success probability for cutoff r is:

P (hire best ∣ r) = \frac{r}{n} j = r \sum n - 1 \frac{1}{j}

Optimal cutoff r*

11 of 30

≈ n/e = 11.0

P(best) theory

37.9%

simulated rate

40.3%

n (candidates) 30 simulations 2000

Observe the first r* ≈ n/e candidates without hiring. Then hire the next one better than all of them. Achieves P(best) → 1/e ≈ 37% as n → ∞.

The 37% rule

The optimal cutoff is:

r^{*} \approx \frac{n}{e}

and the resulting success probability converges to:

P^{*} \approx \frac{1}{e} \approx 37%

The red dot on the curve marks the optimal r for your chosen n. Notice it always sits at roughly 37% of the way through the candidate list, and achieves roughly 37% success probability — a coincidence of $1/ e$ appearing twice.

What to notice

Too early and you hire the wrong person. With a tiny cutoff, you almost always settle for a mediocre candidate seen early.
Too late and you miss the best. With a large cutoff, the best candidate often appears in the scouting phase and you never hire them.
The curve is asymmetric. The right tail falls sharply — cutting off too late is worse than cutting off a bit too early.
The rule is n-independent in the limit. Whether you interview 20 or 100 candidates, always observe ~37% first.

Real applications

The secretary problem formalises optimal stopping, which appears in:

Choosing when to sell a house (after seeing enough offers to set a benchmark)
Deciding when to commit in a relationship
Algorithm design (online algorithms that must make irrevocable decisions with incomplete information)