Virtual laboratory/Mathematics/Probability/The Secretary Problem — The 37% Rule

The Secretary Problem — The 37% Rule

An extension of the topic of probability and the number e. To maximize the chance of choosing the best candidate seen one at a time (without recall), you should reject the first ~37% (1/e), and then pick the first candidate better than all those seen so far. The optimal threshold tends to 1/e, and the chance of success is also ~1/e. The problem was popularized by Martin Gardner in 1960.

The Secretary Problem — The 37% Rule
Mathematics · Probability · The secretary problem — the 37% rule (1960)
Number of candidates nn12 cand.
The optimal strategy is to reject at the start (merely observing) about…
candidates n
12
reject the first ⌊n/e⌋
4
optimal threshold
33%
chance of success
40%
🎯 Challenge: hit the observation phase
streak: 0

I’ll give you a number of candidates to reject — set the slider n so that the optimal threshold round(n/e) gives exactly that.

What just happened?

You review nn candidates one at a time, in random order, and after each you must immediately decide “take” or “reject forever” — with no going back. You want to pick the best. A surprisingly simple and elegant optimal strategy: reject the first ~37% of candidates, merely observing them and remembering the best one as a “benchmark”, and then pick the first one who surpasses all seen so far. The scene shows two phases — observation (rejection) and selection — split by a threshold, plus the success-probability curve with a clear maximum at 37%.

The optimal cutoff threshold:

r∗≈ne,r∗n→1e≈0.368r^{*} \approx \dfrac{n}{e}, \qquad \dfrac{r^{*}}{n} \to \dfrac{1}{e} \approx 0.368

— you reject the first n/en/e candidates (where e≈2.718e \approx 2.718 is Euler’s number), which for large nn gives a threshold converging to 37%.

The chance of choosing the best:

P(r∗)=r∗n∑i=r∗+1n1i−1  ⟶  1e≈0.368P(r^{*}) = \dfrac{r^{*}}{n}\sum_{i=r^{*}+1}^{n}\dfrac{1}{i-1} \;\longrightarrow\; \dfrac{1}{e} \approx 0.368

This strategy gives a ~37% chance (exactly 1/e1/e) of landing on the absolute best candidate — regardless of whether there are 10, 100, or a million of them! This is remarkable, because naively one would think that, seeing each only once, the chance should be negligible (≈1/n\approx 1/n).

Why exactly 1/e? (the invariant)

The magic number 1/e1/e (the reciprocal of Euler’s number) appears here naturally from probability: maximizing P(r)P(r) over the threshold rr, we get a condition whose solution tends to r/n=1/er/n = 1/e. At that point the chance of success itself is also 1/e1/e. The panel on the right shows both bars glued to the 1/e1/e line — and that regardless of nn.

History and applications

The secretary problem (also known as the marriage problem, the sultan’s dowry problem, or the “best choice problem”) is a classic optimal-stopping problem, popularized by Martin Gardner in 1960. It has real applications in decision theory, economics, and even in life choices (hence the playful “37% rule” for dating — after seeing 37% of the options, pick the next better one). It is a beautiful example of how mathematics gives a concrete, optimal strategy in a seemingly hopeless situation.

Why this matters for your exam:

This is an extension of the topic of probability and the number ee. It shows that even with irreversible “live” decisions a simple threshold rule beats random choice — and that the constant ee pops up in places where you would never expect it.

Remember: observe and reject the first n/e≈37%n/e \approx 37\%, then take the first one who is better. Chance of the best: 1/e≈37%1/e \approx 37\%.

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Theory & formulas

The 1/e rule

Candidates are seen one at a time, without recall. Optimal strategy: reject the first ≈ne\approx\dfrac{n}{e} (about 37%37\%), remembering the best among them, and then pick the first candidate better than all those seen so far.

Chance of success

For large nn the threshold and the probability of choosing the best both tend to: 1e≈0.368\frac{1}{e}\approx 0.368

Where e comes from

Maximizing the chance leads to an equation whose solution is the threshold n/en/e — hence the constant ee in this optimal-stopping problem.

Notation: nn number of candidates, e≈2.718e\approx 2.718.

Frequently asked questions

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How many candidates should you reject at the start?

About 37% (exactly n/e), merely observing them, and then pick the first candidate better than all seen so far.

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What is the chance of choosing the best candidate?

About 1/e ≈ 37% — independent of the number of candidates (for large n).

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Where does the number 1/e come from in this problem?

From the probability theory of optimal stopping — the optimal threshold and the chance of success both tend to 1/e.

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