The Secretary Problem and the 37% Rule
After reading you will know why skipping the first 37% of candidates and then grabbing the next record-breaker gives you the best shot at the single best pick, how to derive that number, and where the rule quietly fails.
What the problem asks
You interview candidates one at a time in random order. After each interview you must decide on the spot: hire this person and stop, or reject them forever and move on. You cannot recall a rejected candidate. You cannot see anyone you have not yet interviewed. You only learn relative ranks: after seeing five people you know who was best of those five, but not how they compare to the ones still waiting.
Your goal is narrow and strict: land the single best candidate of the whole pool. Second best counts as a total loss. With 100 candidates and blind guessing you would win 1 time in 100, a success rate of 0.01. The surprising result is that a simple rule pushes that up to about 0.37, no matter how large the pool.
The rule has two phases. First, look: interview and reject a fixed fraction of the candidates while remembering the best score you have seen. Then leap: hire the first later candidate who beats every candidate from the looking phase. If nobody beats them, you are stuck with the last person.
When the rule applies, and when it does not
The 37% rule is exact only under a tight set of assumptions. Break any one and the number moves.
- Fixed, known count
- You must know n, the total number of candidates, in advance. If candidates keep arriving with no known end, the cutoff changes.
- Random order
- Every arrival order must be equally likely. If the best candidate is more likely to come early or late, the optimal cutoff shifts.
- No recall
- A rejected candidate is gone. If you can call people back, the whole problem changes and you should never commit early.
- Rank-only, all-or-nothing
- You see only relative order, and only the very best pick counts. If you have real scores, or you would happily settle for top-10%, use a different strategy.
Real hiring, dating and apartment hunting rarely satisfy all four. Treat the model as a clean statement about optimal stopping under uncertainty, not as career advice. The footnote on the tool page makes the sharpest limit clear: the rule maximises the chance of the best pick, not the average rank of your hire.
If landing second best would still make you happy, this rule is wrong for you. It treats ranks 2 through n as identical failures, so it sometimes rejects an excellent candidate while holding out for the one true best.
The formula and where it comes from
Let n be the number of candidates and let you reject the first k of them, then hire the next record-breaker. Success means you hire the overall best candidate. That happens exactly when the best candidate sits at some position i \gt k, and the best of the first i-1 candidates fell inside the looking phase (positions 1 through k). If instead the best of the first i-1 sat in the leaping zone, you would have already hired that person by mistake.
The probability the best candidate is at position i is 1/n. Given that, the chance the best of the earlier i-1 is among the first k is k/(i-1). Summing over all valid positions gives the success probability:
Here k is the number rejected, n the total, and the sum runs over every position i where the true best could sit and still be caught. Each term 1/(i-1) is the chance the leading candidate so far landed in the looking phase.
For large n, write the cutoff as a fraction x = k/n. The sum approaches an integral, and the whole expression becomes:
To maximise, differentiate and set to zero: -\ln x - 1 = 0, so \ln x = -1 and x = 1/e \approx 0.3679. Plug that back in: P(1/e) = -(1/e)\ln(1/e) = 1/e \approx 0.3679. Both the best cutoff and the best success rate equal 1/e. That coincidence is why the number shows up twice on the tool page.
A worked example with 100 candidates
Reproducing the demo with n = 100
Take the default pool of 100 candidates. The rule says reject the first k = 37 (since 0.368 \times 100 \approx 37), then hire the next record-breaker. Evaluate the exact formula.
- Set n = 100, k = 37.
- Compute the harmonic tail \sum_{i=38}^{100} 1/(i-1) = \sum_{j=37}^{99} 1/j. This equals H_{99} - H_{36} \approx 5.1774 - 4.1996 = 0.9778.
- Multiply by k/n = 0.37: 0.37 \times 0.9778 \approx 0.3618.
So hiring after a 37-candidate look wins about 36.18% of the time. Compare a few nearby cutoffs computed the same way:
| Reject first k | Fraction k/n | Success P(k) |
|---|---|---|
| 10 | 0.10 | 0.2321 |
| 25 | 0.25 | 0.3452 |
| 37 | 0.37 | 0.3618 |
| 50 | 0.50 | 0.3499 |
| 75 | 0.75 | 0.2264 |
The curve is flat near the top. Cutting at 25 or 50 instead of 37 costs only a couple of percentage points, so the exact 1/e is not fragile. When the simulator runs thousands of trials per cutoff it will scatter around these exact values, with the peak sitting near k = 37.
Explore the cutoff yourself
Reading the simulator output
The simulator runs many trials at each candidate cutoff and plots the empirical success rate. Three things are worth checking against the math.
First, the peak location. With enough trials the maximum should land near a cutoff fraction of 0.37. For n = 100 that is 37 candidates. For n = 20 the exact optimum is k = 7 (fraction 0.35), so do not expect the fraction to hit 0.368 precisely at small n; the continuous approximation only kicks in as n grows.
Second, the peak height. It should hover near 0.36 to 0.37. If you run only 200 trials per cutoff, the sampling noise on a rate near 0.36 has standard error about \sqrt{0.36 \times 0.64 / 200} \approx 0.034, so bars can jitter by three or four points. Raise the trial count to shrink that.
Third, the flat top. The bars around the peak should be nearly level, matching the table above. A sharp single spike would be a sign of too few trials, not a real feature.
To confirm your reading is stable, run the same cutoff twice with different trial counts. The rate at k = 37 should stay near 0.36 both times. If it swings by more than your estimated standard error, you need more trials.
Common mistakes
Confusing "best so far" with "good enough". The leap rule fires on the first candidate who beats everyone in the looking phase, not on the first candidate who seems strong. If the looking phase happened to contain the overall best, no later candidate ever beats them, and you are forced to take the final person. That failure mode is baked into the 0.37 loss rate.
Expecting a high win rate. Winning 37% of the time still means losing 63% of the time. The rule is optimal, not reliable. Blind guessing on n = 100 wins 1%, so 37% is a huge improvement, but it is not close to certainty.
Applying it when you have real scores. If you can measure quality on a numeric scale rather than mere rank, you can do far better with a threshold rule, because you know how good "good" is. The secretary problem assumes you are blind to absolute quality.
Rounding the cutoff badly at small n. For n = 3 the optimal k is 1, giving success rate 0.5, not 1/e. Small pools deserve the exact formula, not the 37% shortcut.
Related tools
Optimal stopping is one corner of a larger family of probability toys on this site. To watch a fixed win rate emerge from repeated play, try the Monty Hall Simulator, where switching settles on 2/3. To see averages converge as trials pile up, the Law of Large Numbers shows the running mean lock onto its expected value. For estimating a fixed number by throwing random samples at it, the Monte Carlo Playground is the natural next step, and it uses the same "many trials, read the rate" loop this simulator does. If you enjoy counterintuitive odds, the St. Petersburg Paradox pairs an infinite fair price with a payout that is almost always tiny.
Frequently asked questions
Why is the answer 1/e and not something rounder?
Because maximising -x \ln x gives \ln x = -1, which is x = e^{-1} \approx 0.3679. The number e enters through the harmonic sum becoming a logarithm as the pool grows.
Does the pool size change the best fraction?
Barely, once n is above about 20. For n = 100 the optimum is k = 37 (0.37). For n = 1000 it is k = 368 (0.368). Small pools like n = 3 or n = 5 need the exact formula because the fraction has not settled yet.
What if I would accept the second best?
Then this rule is not optimal for you. A version that rewards good ranks rather than only the top rank uses a smaller looking phase and a different objective. The classic 37% rule strictly maximises the chance of the single best.
Why does the simulator give slightly different numbers each run?
It estimates each rate from a finite number of random trials. A rate near 0.36 from 500 trials has a standard error of about \sqrt{0.36 \times 0.64/500} \approx 0.021, so two points of wobble is normal. More trials tighten it.
Is 37% the chance I get a good hire, or the best hire?
The best hire, exactly. The rule delivers the single top candidate about 37% of the time and delivers someone else (or the forced last pick) the other 63%.
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