# Probability Mass Function of 50 collectibles

1. Mar 11, 2013

### Verdict

1. The problem statement, all variables and given/known data
Some supermarket gives its customers a free ”collectible” (C) for every
15 Euro's spent. There are 50 different types of Cs. Assume that each time a
customer receives a C it is equally likely to be one of the 50 types. Define Ni,
i = 1, 2, . . . , 49, to be the number of additional Cs that need to be obtained after
i distinct types have been collected in order to obtain another distinct type, and let
N denote the number of Cs collected to attain a complete set of at least one of
each type.

Find the pmf (probability mass function) of Ni. What famous distribution is this?
(And later, after I figure this one out):
Determine the expectation value of N (so not Ni)

2. Relevant equations
I can't really think of any at this time, other than maybe pmf (probability mass function) for X is given by fx = P(X=x)

3. The attempt at a solution
Right, so, starting with the pmf of Ni. Ni is the number of collectibles that are still required to be obtained after already getting i of them, so this is 50-i and the pmf has to do with the chance of 'being at a point Ni ', so I'd say that i = 1, or N1 = 49 is the most probably and i = 49 the least.

Now, the (discrete) distributions that I know of are the point mass function, the bernoulli distribution, the binomial distribution, the geometric distribution and the Poisson distribution.
I don't really see which one fits exactly, but a geometric distribution sounds alright. Could anyone provide a hint?

Last edited: Mar 11, 2013
2. Mar 11, 2013

### Ray Vickson

3. Mar 11, 2013

### Verdict

So it's a geometric distribution, I already had that feeling. Alright, I'll try and go from there!

Last edited: Mar 11, 2013
4. Mar 11, 2013

### Verdict

Ok, so clearly I misread what Ni is. It is the number of C's that have to be collected before having i + 1 different C's.

Would I be correct in thinking that N = 'the sum from i=1 to 49' of Ni, plus 1? (As you have to collect 1 C to get to i = 1)? I googled the problem, but I don't see how this is a geometric distribution. Isn't there supposed to be a constant factor between the terms, like 1/2 + 1/4 + 1/8 etc for it to be a geometric series?

Alright, so again my mistake. A geometric series is not the same category as a geometric distribution. Knowing that, it makes perfect sense. I got it, thanks

Last edited: Mar 11, 2013
5. Mar 11, 2013

### Ray Vickson

No, no, no. It said very clearly that Ni is the number of *additional* Cs that must be collected to get a new type, after i distinct types have been collected already. So Ni has an ordinary geometric distribution, with 'success' probability dependent on i.

The (random) number of Cs that must be collected altogether until i different types are obtained is the sum $S_i = 1 + N_1 + N_2 + N_3 + \cdots + N_{i-1},$ where the $N_k$ are independent but not identically-distributed geometric random variables. Getting the actual probability distribution of $S_i$ is a challenging problem.