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

Home > Probability
On Average (Posted on 2004-01-26) Difficulty: 4 of 5
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.)
re: solution | Comment 7 of 11 |
(In reply to re: The other answers by Penny)

(Please excuse the following if the math is inexcusable...)

The probability of NOT having seen a specific number after 14.7 times, if I've understood this correctly, should be

(5/6 ^ 14.7)

Thus the probability of not having seen ANY one number after 14.7 throws should be

(5/6 ^ 14.7) * 6 = 41.13%

So what makes this the expected result? It's not the point where the p of seeing all six number rises above 50%, for instance. Indeed, it hardly seems related at all to the probability of seeing all six.

I understand Federico's answer, I'm just wondering how, if at all, this could be worked out just by the probabilities of seeing (or not seeing) all six.
  Posted by Sam on 2004-01-26 21:19:23

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 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information