All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
All 10 digits (Posted on 2012-07-06) Difficulty: 3 of 5
Pi takes 33 digits to get all 10 digits.
However, e only takes 21 digits.
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?

See The Solution Submitted by Math Man    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Possible solution | Comment 2 of 5 |
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 2012-07-06 11:34:42

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information