TITLE: What is the chance of rolling a specific number after a certain amount of rolls?
QUESTION [8 upvotes]: Say I roll a $6$ sided dice and I want to roll a $6$. what is the probability that I will have rolled the number I want after $6$ rolls?
I have been using this: $\displaystyle1-\left(1-\frac{1}{x}\right)^y$
where $x$ is the number of sides and $y$ is the amount of rolls, so it would be $\displaystyle1-\left(1-\frac{1}{6}\right)^6$ for rolling a specific number in $6$ rolls, which is $\approx66.5\%$ is this the correct way of calculating the probability of something like this, if not what is the proper way?
i'm not really sure why that formula works(if it does) so some elaboration on that would be nice.
sorry for lack of technical language
thanks in advance

REPLY [2 votes]: I would say that it's easier to calculate the chance that you never roll this number.
So that all 6 rolls are not the number you want.
This chance is (5/6)^6 = 33.49%.
Now to calculate the chance that you roll your number at least oncentre within 6 rolls you can take 
1-(5/6)^6= 66.51%. (I don't have an approximate sign)
This is exactly the same as your equation. I just used 5/6 instead of 1-1/6.
As for the explanation.
All the possibilities that exist can be added up to 100%.
What we do here is we calculate the chance for all the possibilities that you won't get what you are looking for. 
Now if you subtract these you will be left with all the possibilities you are looking for.
The reason we sometimes use this method is because the conjugate is much easier to calculate, like in this situation.