Probability Mass Function of 50 collectibles

[Verdict]

Homework Statement

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)

Homework Equations

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

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?

 

[Ray Vickson]

Homework Statement

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)

Homework Equations

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

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 N₄₉ is where you start at, if I understand correctly.
Now, the pmf has to do with the chance of 'being at a point Ni.' I'd say N₄₉ is the most likely, and N₀ 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.
None of them really seem to fit the situation in my opinion, so maybe someone could provide a hint as to what to do?

Google "Coupon Collector's Problem".

 

[Verdict]

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

 

[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

 

[Ray Vickson]

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

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.