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So, I'm not the strongest with probability and I'm a bit confused how to approach this problem.

Let's say I have to roll a number from $1$ to $99$. How would I calculate the probability of someone rolling higher than the number I have just rolled, within $N$ rolls?

As an example:

There are $15$ people in a group, including me. Each person rolls once. I roll a $85$. What is the odds of someone else getting $> 85$?

I have read through this Q & A. But I'm not sure how to extend this to my use-case?

Rietty
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Assuming you mean strictly higher, there are $14$ rolls that beat yours, so the chance a given roll will not beat you is $\frac{85}{99}$. The chance that $N$ rolls in a row do not beat you is $(\frac{85}{99})^N$. The chance that at least one of them beats you is then $1-(\frac{85}{99})^N$.

Ross Millikan
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  • Yes, I mean strictly higher. – Rietty Nov 10 '19 at 00:35
  • Also I apologize, but I meant from 1-99. Does that affect anything? – Rietty Nov 10 '19 at 03:39
  • It changes the chance that one roll beats yours to $\frac {14}{99}$ Can you follow through the argument with that? – Ross Millikan Nov 10 '19 at 05:58
  • Okay so. If I understand the concept. 1 roll beats mine at 14/99 chance. Hence the chance that none of the rolls beat me is.. 1-(85/99)^N. If there are 14 other people then it would be. 1-(85/99)^14 = 88.17% any of them don't beat me? Is that right? It seems... kinda high? – Rietty Nov 10 '19 at 17:24
  • Are we sure we have to subtract that number from 1? – Rietty Nov 10 '19 at 17:33
  • Shouldn't it be "The chance that at least one of them beats you is $1-0.85^N$"? It was brought up in chat by the OP. I just want to confirm that I wasn't wrong. – stressed out Nov 10 '19 at 17:54
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    The subtraction from $1$ gives the chance that at least one beats you. I had confused myself and have fixed it – Ross Millikan Nov 10 '19 at 17:54
  • Accepted! Thank you! – Rietty Nov 10 '19 at 19:24