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 2007-11-29, 14:33 #1 m_f_h     Feb 2007 1B016 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 ?
 2007-11-29, 15:08 #2 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 2×29×109 Posts This is a standard problem (usually called coupon-collecting) To get from a collection of 0 to a collection of 1 takes 1 box To get from 1 to 2 takes N/(N-1) boxes, since the chance of getting a new toy is (N-1)/N To get from 2 to 3 takes N/(N-2) ... So it's N * sum(i=1 .. N) 1/i, which is roughly N log N.
2007-11-29, 15:35   #3
m_f_h

Feb 2007

24×33 Posts

Quote:
 Originally Posted by fivemack This is a standard problem (usually called coupon-collecting) To get from a collection of 0 to a collection of 1 takes 1 box To get from 1 to 2 takes N/(N-1) boxes, since the chance of getting a new toy is (N-1)/N To get from 2 to 3 takes N/(N-2) ... So it's N * sum(i=1 .. N) 1/i, which is roughly N log N.
sorry, so I was right concerning triviality... mea culpa... and thanks nevertheless.

 2007-11-29, 17:03 #4 davieddy     "Lucan" Dec 2006 England 6,451 Posts
2007-11-30, 09:04   #5
davieddy

"Lucan"
Dec 2006
England

645110 Posts

Quote:
 Originally Posted by m_f_h sorry, so I was right concerning triviality... mea culpa... and thanks nevertheless.
I would suggest that this solution is simple but that
its validity not trivial.

 2007-12-01, 14:29 #6 davieddy     "Lucan" Dec 2006 England 11001001100112 Posts I'm thinking that: p + 2p(1-p) + 3p(1-p)^2 + 4p(1-p)^3 +......... must equal 1/p. Is this obvious? I guess we say 1/p = (1-(1-p))^(-1) and use the binomial theorem. Last fiddled with by davieddy on 2007-12-01 at 14:39
2007-12-03, 00:44   #7
davieddy

"Lucan"
Dec 2006
England

645110 Posts

Quote:
 Originally Posted by davieddy I'm thinking that: p + 2p(1-p) + 3p(1-p)^2 + 4p(1-p)^3 +......... must equal 1/p. Is this obvious? I guess we say 1/p = (1-(1-p))^(-1) and use the binomial theorem.
S=p + 2p(1-p) + 3p(1-p)^2 + 4p(1-p)^3 +.........
S-(1-p)S=p(1+(1-p)+(1-p)^2+...)
pS=p/(1-(1-p))
S=1/p

The point being that this practically follows from
the definition of probability.

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