Indeed, I got to the same result.
First car has a probability of 1 to be the head of a cluster.
The second car has a probability of 1/2 to be the head of a cluster. It's a head of a cluster only if it's smaller then car 1.
The third car has a probability of 1/3 to be the head of a cluster. It's a head of a cluster only if it's smaller then car 1 and car 2.
...
so the result is 1+1/2+1/3+...1/N, since all clusters need heads (they are not zombies, are they).

[abia]

E(k) sa-i zicem.

E(k) = suma cu z de la 1 la infinit din z*P(z)
unde P(z) = probabilitatea ca abia dupa z extrageri sa obtinem un pokemon diferit.
P(z) = R(k)^(z-1) * (1-R(k))
unde R(k) este probabilitatea sa nu extragi un pokemon diferit daca ai deja k pokemoni diferiti
R(k) = k/n
=> P(z) = (k/n)^(z-1) * (n-k / n)
=> E(k) = (n-k)/n * suma cu i de la 1 la infinit din (k/n)^(i-1)*i
=> (via Wolfram) E(k) = n/n-k

Fiind 4 culori si 5 carti, e foarte probabil sa conteze ca 2 carti sunt de aceeasi culoare

alta problema?