1

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?

RobPratt
  • 50,938
Kahel
  • 187
  • What have you tried? Also, I interpret the scenario as each common/uncommon/epic character having a 7%/4%/5% chance respectively. Each uncommon character is actually rarer than each epic one, which doesn't seem right. – Parcly Taxel Oct 09 '20 at 12:58
  • I think that the info from the pop up of the game means that for each time you open a gacha you have 70% of getting an Uncommon. In total you can collect 10 Uncommon. Epic you have only 10% chance and there are 2 of them in total – Kahel Oct 09 '20 at 13:36
  • Yes, but what are the probabilities of getting any one common character, say a Nickit or a Yamper? Are they equal among all characters in a tier or not? – Parcly Taxel Oct 09 '20 at 13:38
  • yes chances in the tier are the same – Kahel Oct 09 '20 at 13:41

2 Answers2

0

Hint: Look into hypergeometric distributions.

Noah Solomon
  • 1,922
  • I have looked into that but I don't understand how to incorporate the fact that technically there are no failures: you maybe be after an epic and get an uncommon but if you don't have that uncommon you are still good toward collecting them all? – Kahel Oct 09 '20 at 13:40
0

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 $100(1/20+1/16+1/12)=\frac{235}{12}=19.5833\dots$

Parcly Taxel
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