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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?

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  • $\begingroup$ What have you tried? Can you, say, determine the expected number of rolls it takes to get a $9$? $\endgroup$
    – lulu
    May 2 '19 at 13:33
  • $\begingroup$ "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? $\endgroup$
    – WaveX
    May 2 '19 at 13:39
  • $\begingroup$ @Iulu I know the expected number of rolls to get a specific number on an n-sided dice is simply n. $\endgroup$ May 2 '19 at 14:29
  • $\begingroup$ @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. $\endgroup$ May 2 '19 at 14:30
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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.

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  • $\begingroup$ 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? $\endgroup$ May 2 '19 at 13:54
  • $\begingroup$ @saulspatz That makes a lot of sense, thank you. $\endgroup$ May 2 '19 at 14:28

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