Hi, I am currently collecting coupons from cigarette packets, and I would
like to know how many packets of cigarettes I need to buy before I can be
95% percent confident that I will have collected the whole set (10
pictures of film stars).

Generally how can I represent the distribution of number of packets
purchaced versus the probability of completing my set?

If anyone has references to easy to understand material on the subject I
would be greatfull.

It seems that it should be relativly easy to put a statistical bound on
an estimate of the actual number of coupons given the behaviour of the
'wait time for a new coupon' in number of packets of cigarettes opened.

For example, how sure can I be that the true number of coupons is actually
10?

Also, as I am collecting coupons I can get some idea of the underlying
distribution of the coupons in circulation.

How accurately can I estimate the underlying distribution, and how does
this affect my estimate of the remaining coupons?

Finally, given a fixed cumulative probability of getting cancer with each
pack of cigarettes I open, can I possibly finish my investigation before I
die?

My maths is not good, so any clear / intuative / descriptive feedback is
most welcome - I cant promise to be able to understand technical
feedback.

Cheers,
Dan.

Dan,

let p(n,k) be the probability of getting k diffenrent coupons (out of c,
c = 10 in your case) in n packages.

Then we can set up the following recursion equation (derivation see below)

(1) p(n+1,k) = p(n,k-1) * (1 - (k-1)/c) + p(n,k) * k/c

The initial conditions are

(2) p(n=1,k=0) = 0; p(n=1,k=1)= 1

and the boundary conditions are

(3) p(n=0,k) = 0

I have put all this into my Excel and found that p(n,k=10) = 95% is
reached betweeen n=50 and n=51. Hence you need at least 51 packages to
have 95% confidence for getting all 10 coupons.

This solves only your first question, of course.

Derivation of (1):

We can have k coupons in step n+1 from two situations in step n: we can
either have k-1 coupons or k coupons. In the first case the probability
of increasing the number of coupons by 1 is given by considering the
still empty part of our coupon list. The length of that is c-(k-1), the
total length of the coupon list is c, hence the probability is
(c-(k-1))/c. In the second case we must hit the occupied part of the
list which has length k. The corresponding probability is k/c.Putting
things together gives (1).

By the way, the recursion relation can be shown to lead to the so called
Stirling numbers.

Wolfgang

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