My Math Forum Odds of seeing all possible values

 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 14th, 2008, 07:06 AM #1 Newbie   Joined: May 2008 Posts: 3 Thanks: 0 Odds of seeing all possible values Is there a way to compute the odds of seeing all possible values in a set of values? For example, on average, how many times would you have to throw a 6 sided dice to see all 6 possible numbers? Or, if you have a 1000 items in a database and you are shown one randomly, on average, how many times would you be shown items before you have seen each item at least once? For the dice problem, I simulated it in Excel and after 5000 runs I had an average of 15 rolls before you have seen all 6 possibilities. I'm just looking for the mathematics to verify this answer.
 May 14th, 2008, 08:51 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I find an average (expected number of flips before all sides show up) of 14.7, exactly, for a six-sided die. Other averages: 3 for a fair coin, 25/3 for a three-sided die, and 761/35 for an eight-sided die. I'll see if I can explain my method.
 May 14th, 2008, 11:57 AM #3 Newbie   Joined: May 2008 Posts: 3 Thanks: 0 This is what I came up with: For the dice example N = 6 1 + 1/((N-1)/N) + 1/((N-2)/N) + 1/((N-3)/N) ... + 1/((N-5)/N) I just don't know how to express this or solve it more concisely. Using the above I would still have to do 1000 calculations if N = 1000.
 May 19th, 2008, 06:15 AM #4 Newbie   Joined: May 2008 Posts: 3 Thanks: 0 For anyone else that may happen upon this... This problem is more famously known as the "Coupon Collector Problem". Just do a search for that and you will find lots of information. An approximation for the solution is: N*ln(N).

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