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A fair die is rolled until a $6$ appears. What is the probability that is must be cast more than 5 times?

So this is $1- P(\text{dice has to be cast less than or equal to}\ 5 \ \text{times})$. So this probability is equal to $$ P =\frac{1}{6}+ \frac{5}{6} \frac{1}{6}+\left(\frac{5}{6} \right)^{2} \frac{1}{6} + \cdots + \left(\frac{5}{6} \right)^{4} \frac{1}{6}$$

So just add $-1$ to this?

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If you add $-1$ to $P$ you will get a negative number. – André Nicolas Dec 25 '11 at 17:52
@AndréNicolas: I meant I do the following: $-P+1$? – Thomas Dec 25 '11 at 17:55
That's fine. Your calculation is then longer than the simple $(5/6)^n$, but perfectly correct. – André Nicolas Dec 25 '11 at 17:59

The probability of getting a non-$6$ the first five times is $(5/6)^5$.

$$ \begin{align} & \Pr(\text{a non-}6\text{ on the 1st }5\text{ trials}) \\ \\ & = \Pr(\text{a non-}6\text{ on the first trial and a non-}6\text{ on the 2nd trial and a non-}6\text{ on the 3rd trial and }\ldots) \\ \\ & = \Pr(\text{a non-}6\text{ on the 1st trial})\cdot\Pr(\text{a non-}6\text{ on the 2nd trial})\cdot\Pr(\text{a non-}6\text{ on the 3rd trial})\cdots\cdots \\ \\ & = \frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56. \end{align} $$

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But would my method work (i.e using the complement)? – Thomas Dec 25 '11 at 17:55
As @DavidMitra points out, and your last step shows, you assume the independence of the rolls. You could add this to your answer to make it self-contained. – user21436 Dec 25 '11 at 17:56
@Thomas Yes, it will. But you'll have to manipulate carefully as Andre Nicolas points out in his comments. – user21436 Dec 25 '11 at 17:58

For your solution, you'd subtract what you have from 1. It is fine.

It can be done a bit more simply though: The probability that it has to be cast more than 5 times to obtain the first 6 is exactly the probability that each of the first 5 rolls do not result in a 6. This would be $ \bigl({5\over 6}\bigr)^5 $, assuming independence of the rolls.

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The easiest way to do this is simply to raise the probability that a six is not rolled to the fifth power, as these probabilities are independent, which gives $$P(\text{cast more than five times}) = P(\text{individual roll is not 6})^5 = \left(\frac{5}{6}\right)^5$$

This is equivalent to $1 - P$ where $P$ is what you wrote, as $$1 - P = 1 - \frac{1}{6}\left(\frac{5}{6} + \cdots + \left(\frac{5}{6}\right)^4\right) = 1 - \frac{1}{6}\frac{1-(5/6)^5}{1 - 5/6} = \frac{1}{6}\frac{(5/6)^5}{1/6} = \left(\frac{5}{6}\right)^5$$

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