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If there is a die with n sides (1,2,..k,..n) and it is rolled n times. What is the probability of the side labeled 'k', not occurring in n rolls?

CMg
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  • Welcome to CV. Since you’re new here, you may want to take our tour, which has information for new users. Please add the [self-study] tag and read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. – jbowman Apr 08 '18 at 22:54
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    [(n-1)/n]$^n$ since (n-1)/n is the probability that a specific side will not occur on any given roll and the rolls are assumed to be independent. – Michael R. Chernick Apr 08 '18 at 22:55
  • @jbowman I'll make sure next time. – CMg Apr 08 '18 at 22:56
  • @MichaelChernick I got the same till (n-1)/n. But, I do not get why there is a power of n on it? – CMg Apr 08 '18 at 22:58
  • It has to happen n consecutive times. – Michael R. Chernick Apr 08 '18 at 23:09
  • See https://stats.stackexchange.com/questions/209104/what-is-the-difference-between-1-pk-and-1-pk and https://stats.stackexchange.com/questions/88980/why-on-average-does-each-bootstrap-sample-contain-roughly-two-thirds-of-observat – Glen_b Apr 09 '18 at 01:46

1 Answers1

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In one roll:

$\text{P}(\text{side = }k)=\frac{1}{n}$

$\text{P}(\text{side $\neq$ }k)=1 - \frac{1}{n}$

In $n$ rolls:

$\text{P}(\text{side $\neq$ }k \text{ in n rolls})=\left(1-\frac{1}{n}\right)^n$

sjw
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  • I have a follow up question, my homework says that you need to evaluate this expression for an increasing sequence of values. I don't need the answer for it, rather I just want to know what does the question mean/context of it is. It'd be great if you could use a different example to help explain this. Thanks! – CMg Apr 08 '18 at 23:41
  • I think that means that you should evaluate this expression for n=1, n=2, n=3, and so on. – sjw Apr 09 '18 at 00:15