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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

•  Submitted by levik Rating: 4.1818 (11 votes) Solution: (Hide) After a long tenure on the "Unsolved" list, there have been some users who posted solutions agreeing with one another in their results. Please refer to the comments on this problem to read them.

 Subject Author Date hmm Joe C 2003-11-05 14:13:39 my methods and Charlie's corrections Steve Hutton 2003-03-09 02:07:57 re(2): a method of solving problem A Charlie 2003-02-06 09:07:50 re: a method of solving problem A Charlie 2003-02-06 09:01:32 re: Another analytic solution to problem B Charlie 2003-02-06 05:55:44 Another analytic solution to problem B Rick 2003-02-05 12:06:52 re(2): a method of solving problem B Charlie 2003-02-05 08:57:29 re: a method of solving problem B Charlie 2003-02-05 08:53:56 re(2): Analytic solution, A and B missing parens added Charlie 2003-02-05 08:36:48 re: Analytic solution, A and B missing parens added Charlie 2003-01-31 09:17:10 Analytic solution, A and B Charlie 2003-01-31 08:33:22 Analytic Solution to Part B(ctd.--part 3) Charlie 2003-01-30 10:01:13 Analytic Solution to Part B (ctd.) Charlie 2003-01-30 09:59:15 Analytic Solution to Part B Charlie 2003-01-30 09:56:57 re(3): Simulation results Charlie 2003-01-29 17:01:55 re(3): Simulation results Charlie 2003-01-29 17:00:45 re(2): Simulation results Tony 2003-01-29 13:25:16 re: Simulation results Charlie 2003-01-29 10:15:58 Simulation results Tony 2003-01-24 13:39:11 cards terry 2003-01-19 19:17:29 trading cards terry 2002-12-25 06:51:56 re: Trading cards terry 2002-12-20 02:47:19 Trading cards terry 2002-12-19 21:18:36 re: This should be the answer Cory Taylor 2002-11-29 08:58:07 This should be the answer Sophia 2002-11-29 03:19:56 attempt at case B Keedom 2002-11-19 00:26:36 attempt ar case A Keedom 2002-11-18 23:59:16 a method of solving problem A Steve Hutton 2002-07-31 13:55:51 a method of solving problem B Steve Hutton 2002-07-31 13:51:49 re: my result Steve Hutton 2002-07-28 06:05:44 my result Cheradenine 2002-06-11 08:22:50 No Subject Cheradenine 2002-06-11 08:21:18 re: it's more complicated levik 2002-05-26 12:27:41 My thoughts TomM 2002-05-10 08:32:37 its more complicated Eamon 2002-05-10 00:01:49 Possible solution Half-Mad 2002-05-07 23:33:12

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