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

 On Average (Posted on 2004-01-26)
What is the expected number of rolls of a fair, normal 6-sided die, one is required to make, so that each of the 6 numbers comes up at least once?

Hint: this is not necessarily an integer answer
_____________________

As an aside, it would be interesting to see the computer program simulation of this, but this would not be proof of the solution (merely evidence supporting the proof).

 See The Solution Submitted by SilverKnight Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 1 of 11
This can be seen as a state machine, with probabilistic state changes. You start at the state "No numbers seen".

With probability 6/6, you change to the state "One number seen".

At this state, with probability 1/6, you remain there one more throw; with probability 5/6 you change to state "Two numbers seen".

At this state, with probability 2/6, you remain there one more throw; with probability 4/6 you change to "Three numbers seen", and so on.

The expected number of throws is 6/6+ 6/5+ 6/4+ 6/3+ 6/2+ 6/1 = 14.7.
 Posted by Federico Kereki on 2004-01-26 13:16:08

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

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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