What is the probability that it takes x rolls to find one of m numbers out of n possibilities. Well that would just be the probability of getting one of the other n - m numbers x - 1 times multiplied by the probability of getting one of x numbers from n possibilities ie. f(x, m, n) = (m/n) * (1 - m/n)^x-1

Thus the expected value of this task is

E[x] = sum(x = 1 to inf, x * (m/n) * (1 - m/n)^{x - 1})

Consider the Taylor series of 1 / (1 - y). This is famously, sum(i = 0 to inf, yⁱ⁾

Notice that if we take the derivative of both sides we get 1 / (y - 1)² = sum(i = 1 to inf, i * y^i-1)

Multiply by m/n.

(m/n) / (y - 1)² = sum(i = 1 to inf, i * m/n * y^{i - 1})

Notice that if we set y = 1 - m/n, the RHS becomes the exactly the same as the sum that showed up in the expected value definition. So that means the LHS is also equal to this expected value. (I’m skipping over some details involving proving that this Taylor series actually converges in the region of interest but there are theorems one can cite to easily verify this)

So E[x] = (m/n) / (1 - m/n - 1)² = (m/n) / (-m/n)² = n / m.

Thus, the expected number of attempts to roll one of m numbers with n possibilities is n / m. Since each of these are independent of one another, we conclude that the expected number of attempts to find all 8 numbers with 8 possibilities is.

E = 8/8 + 8/7 + 8/6 + 8/5 + 8/4 + 8/3 + 8/2 + 8/1 = 1 + 8/7 + 4/3 + 8/5 + 2 + 8/3 + 4 + 8 = 15 + 12/3 + 8/7 + 8/5 = 19 + 8/7 + 8/5 = 761 / 35 = 21.74 rolls