Problem:
A non-zero digit is chosen in such a way that the probability of choosing digit d is log10β(d+1)βlog10βd. The probability that the digit 2 is chosen is exactly 1/2 the probability that the digit chosen is in the set
Answer Choices:
A. {2,3}
B. {3,4}
C. {4,5,6,7,8}
D. {5,6,7,8,9}
E. {4,5,6,7,8,9}
Solution:
Let Pr{d1β,d2β,β¦} be the probability that the digit chosen is one of d1β,d2β,β¦. Note that
Pr{d}=logdd+1β and Pr{d,d+1}=logdd+1β+logd+1d+2β=logdd+2β
We seek a set of digits with probability 2Pr{2}. Thus
2Pr{2}=2log23ββ=log49β=log45β+log56β+β―+log89β.=Pr{4,5,6,7,8}.β
Note 1. In this solution, we have not used the fact that the logs are base 10. However, this fact is necessary for the problem to make sense; otherwise the union of all possibilities (i.e, picking some digit from 1 to 9) does not have probability 1.
Note 2. If one collects a lot of measurements from nature (say, the lengths of American rivers in miles), the fraction of the time that the first significant (i.e., nonzero) digit is d is approximately log10β(d+1)βlog10βd. In particular, the distribution is not uniform, e.g., 1 is the first digit about 30% of the time. This counterintuitive fact, sometimes called Benford's Law, can be explained if one assumes that the distribution of natural constants is independent of our units of measurement.