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Record highs in dice

Expected valueHard

A fair $6$ -sided die is rolled $5$ times. A roll is a record if it is strictly larger than every earlier roll (the first roll counts). What is the expected number of records?

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Hint

Compute the chance that roll k beats all k − $1$ earlier rolls, allowing for ties.

Solution

Roll number $k$ is a record with probability $\sum (\frac{1}{6}) \times (\frac{v - 1}{6})^{k - 1}$ over faces $v = 1, \dots, 6$ , because all earlier rolls must be below $v$ .
This gives $1$ , $\frac{15}{36}$ , $\frac{55}{216}$ , $\frac{225}{1296}$ and $\frac{979}{7776}$ for $k = 1, \dots, 5$ .
On the denominator $7776$ the numerators are $7776$ , $3240$ , $1980$ , $1350$ and $979$ , which sum to $15325$ .
The expectation is $\frac{15325}{7776}$ .

Answer: $\frac{15325}{7776}$

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