Say I have a 10 sided dice, and I want to know the average number of rolls it would take me to roll both a 9 and a 10, disregarding repeat numbers, what would the maths look like for this? For context, imagine in a game an opponent has a 1/100 chance of dropping 4 specific rare items. How many kills on average would it take to obtain all 4?
What is the average number of dice rolls to obtain 2 specific numbers?
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A general solution makes use of the Coupon Collector's Problem. https://en.wikipedia.org/wiki/Coupon_collector%27s_problem
In terms of the dice, we'll call a successful roll one where you roll either of the 9 or 10. The coupon collector's problem will tell you the expected number of successful rolls needed until both the 9 and the 10 are rolled: when n=2, expect to require 3 successful rolls. Now we can say a successful roll happens 20% of the time (the chances of rolling a 9 or 10 on a 10-sided die = 2/10). This means you should expect 1 successful roll for every 5 rolls. To get 3 successful rolls you should expect 5*3 = 15 total rolls.
In terms of the game opponent, let's say a "successful drop" is when one of the four items is dropped (regardless of whether it was dropped previously). The coupon collector's problem tells us to expect to need 9 successful drops to get all 4 items. We know the odds of getting a successful drop are 4/100 = 1/25 = 1 successful drop in 25 drops. This means we should expect 9*25 = 225 total drops before all 4 rare items are received. I was looking into this for Runescape drops so thought I'd come back to answer the question for you after I understood it.
I apologize for not deriving the Coupon Collector's Problem here and for not formatting the math elements. This is my first post here.
If you have a sequence of Bernoulli trials with probability of success $p$, then the average number of trials until the first success is $\frac1p.$ The trick here is to consider two events. The first is that either a $9$ or a $10$ shows up. the probability of success is $\frac15$ so the expected number of trials is $5$. After that you have to wait for the number that didn't show up to occur. That takes $10$ rolls on average, so in all the expected waiting time is $15$ rolls.