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

Home > Probability
Trading cards (Posted on 2002-05-03) Difficulty: 3 of 5
A trading card series has 200 different cards in it, which are sold in 5-card packages.

Each package has a random sampling of the cards (assume that any card of the 200 has an equal chance of being in a package).

On the average, how many packages will need to be bought to collect the complete series if...

  • A: all the cards in a package will always be different
  • B: a package can have repeats

  • See The Solution Submitted by levik    
    Rating: 4.1818 (11 votes)

    Comments: ( Back to comment list | You must be logged in to post comments.)
    Analytic Solution to Part B | Comment 23 of 39 |
    This problem is to find the expected value of the number of packets (groups of 5) to reach 200 different cards (all the 200 that are available). To get an expected value, you multiply each possible number of packet draws by the probability that that draw will complete the set. Each of these products is added up to get the expected value (the average number of draws required). Each probability can be gotten by taking the probability that a set will have been achieved by the given draw minus the probability that it will have been achieved by the previous draw, turning a cumulative probability into a probability distribution.

    The cumulative probability of having completed a set at a certain number of draws is an ANDed probability (card 1 achieved AND card 2 achieved, etc.). There is an inclusion/exclusion formula that can convert a number of ORed probabilities into each ANDed probability you need. It follows along the lines of P(A and B and C) = P(A) + P(B) + P(C) – P(A or B) – P(A or C) – P(B or C) + P(A or B or C), with the sign reversing as you go from singles to pairwise to triples, etc. (the etc. needed only for larger numbers than 3). Note that all the possible singles, pairs, triples, etc. are included.

    (... more in next post ...)

      Posted by Charlie on 2003-01-30 09:56:57
    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 (9)
    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