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 x² have the same first digit

(3) That x² and 2^x have the same first digit.

(4) That x, x² and 2^x all have the same first digit.

First digit refers to the first non-zero 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 first-digit 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 base-2 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 (base-2) 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 digits--not 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 2011-03-29 17:31:49