For a randomly chosen real number x on the interval (0,10) find the exact probability of each:
(1) That x and 2^{x} have the same first digit
(2) That x and x^{2} have the same first digit
(3) That x^{2} and 2^{x} have the same first digit.
(4) That x, x^{2} and 2^{x} all have the same first digit.
First digit refers to the first nonzero digit of the number written in decimal form.
x and 2^x:
See correction at bottom, per Jer's later comment:
From 0 to 1, 2^x has a first digit of 1. In 1/10 of this interval, the first digit of x is also 1. So when x is between 0 and 1, the conditional probability of a firstdigit match is 1/10. The overall probability contribution of this portion is therefore 1/100.
There's a lack of matches until the numbers start with 5. The initial 5's match from log(50)/log(2) (that is, the base2 log of 50) to log(60)/log(2).
The 6's match from 6 to log(70)/log(2).
Neither the 7's nor the 8's have matches.
The 9's match from log(900)/log(2) to log(1000)/log(2).
In sum, the probability is (using notation in the Wikipedia article of lb for the binary (base2) logarithm):
(1/10 + lb(60)  lb(50) + lb(70)  6 + lb(1000)  lb(900)) / 10
=
(1/10 + log(60)/log(2)  log(50)/log(2) + log(70)/log(2)  6 + log(1000)/log(2)  log(900)/log(2)) / 10 ~= .0644320516223839
The correction per Jer's comment:
There are actually only 9 possible first digitsnot 10, so the answer should be
(1/9 + lb(60)  lb(50) + lb(70)  6 + lb(1000)  lb(900)) / 10
=
(1/9 + log(60)/log(2)  log(50)/log(2) + log(70)/log(2)  6 + log(1000)/log(2)  log(900)/log(2)) / 10 ~= .065543162733494
Edited on March 30, 2011, 10:53 am

Posted by Charlie
on 20110329 17:31:49 