Pi takes 33 digits to get all 10 digits.
3.14159265358979323846264338327950
However, e only takes 21 digits.
2.71828182845904523536
Suppose you pick random digits from 0 to 9 with probability 1/10 of each digit. What is the average number of digits required to get all 10 digits?
Assume that we are allowed to collect one number each day.
On the first day we are bound to get a number we do not already have.
On the second day we are almost certain; in fact there is a 9 in 10 chance.
So we allow 10/9 of a day for the second number.
Then on the third day, the chance is 8 in 10, so we allow 10/8 of a day and so on, until for the last number we need 10 days, giving the solution:
1*1+1*10/9+1*10/8+1*10/7+1*10/6+1*10/5+1*10/4+1*10/3+1*10/2+1*10/1, or about 29 days, i.e. digits.
Mathematically: Sigma (1 to n) X/n where X is the number of items being collected (e.g. here, 10) and n=X, giving n* Hn, or n times the nth harmonic number.
Edited on July 7, 2012, 2:56 am

Posted by broll
on 20120706 11:34:42 