20071129, 14:33  #1 
Feb 2007
1B0_{16} Posts 
toys, cereals and statistics
This may be a trivial one but well:
Yesterday morning at breakfast my son was deceived since opennig a new cereal box he found one of the 4 CD's to collect which he already got before. Now for the question: Given that in the boxes of some cereal brand there is hidden one of N different objects to collect, and each time you buy a box you get any one of the N with the same probability 1/N, how many boxes to you have to buy "in the mean" to have a complete collection ? 
20071129, 15:08  #2 
(loop (#_fork))
Feb 2006
Cambridge, England
2×29×109 Posts 
This is a standard problem (usually called couponcollecting)
To get from a collection of 0 to a collection of 1 takes 1 box To get from 1 to 2 takes N/(N1) boxes, since the chance of getting a new toy is (N1)/N To get from 2 to 3 takes N/(N2) ... So it's N * sum(i=1 .. N) 1/i, which is roughly N log N. 
20071129, 15:35  #3  
Feb 2007
2^{4}×3^{3} Posts 
Quote:


20071129, 17:03  #4 
"Lucan"
Dec 2006
England
6,451 Posts 

20071130, 09:04  #5 
"Lucan"
Dec 2006
England
6451_{10} Posts 

20071201, 14:29  #6 
"Lucan"
Dec 2006
England
1100100110011_{2} Posts 
I'm thinking that:
p + 2p(1p) + 3p(1p)^2 + 4p(1p)^3 +......... must equal 1/p. Is this obvious? I guess we say 1/p = (1(1p))^(1) and use the binomial theorem. Last fiddled with by davieddy on 20071201 at 14:39 
20071203, 00:44  #7  
"Lucan"
Dec 2006
England
6451_{10} Posts 
Quote:
S(1p)S=p(1+(1p)+(1p)^2+...) pS=p/(1(1p)) S=1/p The point being that this practically follows from the definition of probability. 

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