Expected number of record highs in five draws

Question

You reveal 5 independent values one at a time, each drawn from the same continuous distribution (no ties). The first value is a record by default, and each later value is a record when it beats all before it.

What is the expected number of records among the 5 values?

Accepted Answer

The k-th value is a record exactly when it is the largest of the first k values, which happens with probability \frac{1}{k} by symmetry. Summing these chances gives the expected count: E[	ext{records}] = \sum{k=1}^{5} \frac{1}{k} = 1 + 	frac12 + 	frac13 + 	frac14 + 	frac15 = \frac{137}{60} \approx 2.283. Takeaway. Expected records equal the harmonic number Hn; five draws already average only about 2.3 records.

Expected number of record highs in five draws

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