Number of tries to get all characters

Question

If I have the following table

1. 10 Common characters 70% chance
2. 5 Uncommon characters 20% chance
3. 2 Epic characters 10% chance

How many tries to get all the characters (I can draw duplicates) How many tries to get 3 Uncommon?

Answer 1

Part 1 is a generalised coupon collector's problem. If we use the formula here: $$E=\int_0^\infty(1-(1-e^{-0.07t})^{10}(1-e^{-0.04t})^5(1-e^{-0.05t})^2)\,dt$$ $$=\frac{119545721298102103570778707488616661069}{1796916887363723638261180642662688236}=66.5282\dots$$ So all characters will be obtained in, on average, $66.5282\dots$ tries.

For part 2, the expectation is the sum of three other expectations:

• The number of tries to get the first uncommon character. There is a $\frac{20}{100}$ chance of getting a new uncommon character, so the expected number of tries needed is $\frac{100}{20}$ – it follows a geometric distribution.
• The number of tries to get the second uncommon character. The $\frac{20}{100}$ from before is now $\frac{16}{100}$, so the expected number of tries is $\frac{100}{16}$.
• For the third one, the expectation is, well, $\frac{100}{12}$.

Thus the expected number of tries to get three distinct uncommon characters is $100(1/20+1/16+1/12)=\frac{235}{12}=19.5833\dots$

Answer 2

Hint: Look into hypergeometric distributions.