Expected number of record highs in four draws

Question

Four independent values are revealed one at a time, each from the same continuous distribution (no ties). Day 1 is a record by default, and each later day is a record when it beats all before it.

What is the expected number of records among the 4 values?

Accepted Answer

The k-th value is a record with probability \frac{1}{k}, because it is a record exactly when it is the largest of the first k exchangeable values. Adding these chances: E[	ext{records}] = \sum{k=1}^{4} \frac{1}{k} = 1 + 	frac12 + 	frac13 + 	frac14 = \frac{25}{12} \approx 2.083. Takeaway. Even a short sequence of four draws averages only about two records; the count climbs slowly as the harmonic number Hn.

Expected number of record highs in four draws

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