0

For my problem, a game drop table that has 8 individual items with 0.2% drop rate , 6 other individual items with 0.1% drop rate, and a last item with a 97.8% drop rate. Every time a monster is killed, 5 loot bags of the same drop table as aforementioned drops. Let's say the 14 items are group A and the last item is group B, then group A has a 2.2% chance of dropping, and group B has a 97.8% chance of dropping.

Here's the question, if you kill that type of monster 100 times, is the probability of getting group A 2.2% or is it ( 1 - probability of not getting it ) which is (1 - (0.978)^100 ) or ~89.2%?

I read the closest question, but the answer didn't really answer the question, or maybe I didn't understand the answer.

Nobody
  • 3
  • 1
    Are you looking for the probability of getting group A at least one time in 100 drops? Or precisely one time? It does matter. Or maybe you mean to the probability of getting group A just in the last time? You explained well everything except what you are looking for.. or at least I didn't get it – Shaq Jan 23 '19 at 19:52
  • Just to find the probability of getting group A at all. Doesn't matter how many times group A actually show up. – Nobody Jan 23 '19 at 20:52
  • 1
    @Nobody the probability of getting at least one item from A group is equal to (as you wrote) $1-(0.987)^{100}$. – vermator Jan 24 '19 at 06:53

1 Answers1

0

Asking what is the probability to get group A in any of the attempts, is equal to 1 - the probability not to get any, which is way easier to calculate.

So not getting any group A means you get B every time, 100 times in a row not related one to each other, so $P(X_1=X_2=...=X_{100}=B)=P(X_1=B)\cdot...\cdot P(X_{100}=B)=0.987^{100}$ So $P(\exists n<101: X_n=A)=1-P(\forall n<101:X_n=B) = 1-0.987^{100} $

Shaq
  • 468