# What is the average number of dice rolls to obtain 2 specific numbers?

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 have you tried? Can you, say, determine the expected number of rolls it takes to get a $9$?
– lulu
May 2 '19 at 13:33
• "disregarding repeat numbers" can you elaborate on this part a bit further? Does this mean we can assume that our dice is magical so that we can't ever roll something we previously rolled? May 2 '19 at 13:39
• @Iulu I know the expected number of rolls to get a specific number on an n-sided dice is simply n. May 2 '19 at 14:29
• @WaveX I more meant that I wasn't asking for the chance of getting a 9 and a 10 without repeats, I put that poorly, I just meant that any 10 rolled after the first 10 should be treated like any other number. May 2 '19 at 14:30

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.
• And the other one (for finding all four rare items) would be $$\dfrac{100}{4}+\dfrac{100}{3}+\dfrac{100}{2}+\dfrac{100}{1}$$ correct? May 2 '19 at 13:54