Thought Toys · Strategy & computation · Exhibit 47

Reject the first 37%

Applicants arrive one at a time in random order. You must hire or pass on each the moment you meet them — no callbacks. It sounds hopeless, yet one simple rule catches the single best applicant more than a third of the time, however long the line.

The line-up — bar height is quality, revealed left to right lookhiredthe best

Chance of hiring the very best vs. how long you look exactyour runs

your turn — slide the look phase and watch the peak sit at 37%

What you're seeing

Every applicant has a hidden quality; the tallest bar is the best of the bunch. You interview them left to right and can only ever hire the person in front of you. The rule on trial: look at a fixed fraction and hire no one, just remember the best you've seen — then leap, hiring the first later applicant who beats that running best. Hit "Deal one" and watch it play out: the grey look-phase goes by, then the first applicant taller than everything before them is hired in amber. Green marks who was actually best. You win only if amber lands on green.

Any single deal is a coin toss of luck — but slide the look phase and hit "Run 500," and the pattern is iron. Look too little and you leap at someone merely good early on; look too long and the best has usually already walked past during the look phase. The sweet spot is 1/e ≈ 37%, and there the odds of catching the single best applicant are also about 37% — not just "pretty good," but provably optimal — no strategy of any kind does better. The same look-then-leap arithmetic is why "reject early options to calibrate, then commit" is such durable advice for hiring, apartments, and more.

The rule, exactly. Reject the first m applicants (the look phase), then hire the first one who beats all m. The chance of hiring the overall best is P(win) = (m/n) · Σj=mn−1 1/j which, as n grows with m/n = r, tends to r·ln(1/r) — maximized at r = 1/e ≈ 0.368, giving P → 1/e too. Verified in node (improve/verify/47-secretary.js): at n=100 the best cutoff is m=37 with P≈37.1%; a 200,000-deal Monte Carlo of the actual game matches the formula; as n→∞ both the cutoff fraction and the win probability converge to 1/e; and an independent backward-induction dynamic program — free to accept or reject at every step reaches the identical optimum, so the 1/e rule is genuinely best, not merely good. Negative controls: "take the first," "take the last," and "take a random one" each win exactly 1/n of the time — at n=100 the 1/e rule beats them by more than 30×; and any cutoff far from 1/e (too greedy or too patient) is measurably worse.

Also in Strategy & computation: Freeze too fast, stay stuck →

All 23 in Strategy & computation
  1. 10The evolution of trust
  2. 23Sorting algorithms
  3. 24PageRank & the random surfer
  4. 25Huffman coding
  5. 26Dijkstra's shortest path
  6. 27Nash equilibria
  7. 33The learning-rate cliff
  8. 36A* pathfinding
  9. 37Braess's paradox
  10. 44Diffie–Hellman key exchange
  11. 45Preferential attachment
  12. 46Aliasing & the Nyquist limit
  13. 47The secretary problem — you are here
  14. 49Freeze too fast, stay stuck
  15. 51Cross one line, and its territory closes
  16. 56Catch one error, miss the next
  17. 57Why more processors stop helping
  18. 58Why a busy line explodes
  19. 59The set that's only sure when it says no
  20. 60The fit that memorizes instead of learns
  21. 61When the wire breaks, pick one
  22. 67Better at both, and still better off trading
  23. 69Everyone was consistent. The vote wasn't.

← the cabinet · Thought Toys — a cabinet of explorable explanations. Exhibit 47.