If there is a 1% probability (of getting an item in a loot box for example) of getting a thing, and you do this 100 times, what is the resulting probability that you get what you want?

Question

e.g

A loot box has an item that has a 1% drop rate. This will neither increase nor decrease from the 1st to the last draw. If I draw this 100 times, what is the overall probability of getting what I want?

This isn't 1% or 100% and I do not know how to work it out, so I'm wandering if any of you experts can help...

Thanks for helping!

Answer 1

For every loot box opened, we have the probability Pr(desired item) =0.01. This is equivalent to saying Pr(not desired item)=0.99, and each loot box can be regarded as something called a Bernoulli trial.

Since each loot box is independent, opening multiple boxes (repeated i.i.d. Bernoulli trials) forms what's called a Binomial Distribution. We can use then use its probability mass function to determine the probability of receiving exactly k items from n loot boxes given a fixed probability p per box.

More explicitly, Pr(X=k) = C(n,k) * p^k * (1-p)^{n-k}, where C(n,k) is the binomial coefficient defined as n!/(k!(n-k)!.

However, since you are only interested in the probability of receiving the item (possibly multiple ones), there is a much simpler approach. We are interested in finding Pr(X>=1), which is equivalent to 1-Pr(X=0). The term Pr(X=0) can be thought of as the probability of not receiving the item across 100 loot boxes, which is 0.99*0.99*...*0.99 = 0.99^100. Thus the probability of receiving at least one desired item in those 100 loot boxes is 1-0.99^100 ≅ 63.40%.