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

Home > Probability
Roulette (Posted on 2005-03-23) Difficulty: 2 of 5
If I spin the roulette (numbers from 0 to 36) N consecutive times, how many different numbers should I expect to get?

(If N=1, your formula should give 1 as an answer, but for N=2, it should be a little under 2, and for all other N, the formula should be less than the minimum of N and 37.)

Another question: how many times should you spin the roulette, so chances are better than one in a million, that all 37 numbers will have appeared?

  Submitted by e.g.    
Rating: 4.5000 (2 votes)
Solution: (Hide)
Must check but... D(n)=37(1-(36/37)n) = (37n-36n)/37n-1. As a check, D(1)=1 and D(2)=73/37... almost 2.

For the second question, the odds that an specific number hasn't appeared after n spins are (36/37)n, so the odds that any number hasn't appeared after n spins are 37x(36/37)n ≤10-6, so n≥636.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Hints/TipsOne step needed to get to final solutionRupesh Khandelwal2005-04-16 10:22:59
SolutionMy take on this problem (different solution?, spoilers)ajosin2005-03-24 23:15:32
SolutionPart 2 spoilerCharlie2005-03-23 16:05:43
Hints/TipsHadn't noticed this while it was in the queue, but ... (part 1 spoiler)Charlie2005-03-23 15:05:14
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 (13)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information