Solution
Leading Digits in Hexadecimal
Show the problem again
A large dataset of positive quantities spans many orders of magnitude and obeys Benford's law in every base: in base $b$ the leading digit $d \in \{1,\dots,b-1\}$ occurs with probability $\log_b(1 + 1/d)$. The values are written in hexadecimal (base 16). What is the probability that the leading hexadecimal digit is 1, 2 or 3? Give your answer as a reduced fraction.
Worked solution
The probabilities telescope: $\sum_{d=1}^{3}\log_{16}\frac{d+1}{d} = \log_{16}\left(\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\right) = \log_{16}4$. Since $16^{1/2} = 4$, this equals **1/2**. More generally, the probability that the leading digit is at most $k$ is $\log_b(k+1)$. In base 16, half of all numbers (in the log-uniform sense) lead with 1, 2 or 3, simply because $[1,4)$ covers half of the logarithmic range $[1,16)$.
Source: Benford's law in a general base b, a standard generalization; T. P. Hill (1995) showed that base-invariance characterizes the law. Statement written for AxiomIQ.