nth Time's The Charm So, a question came up recently regarding an image macro shared in my work HipChat: 

Considering the above statement, each Time i has an equal probability P(i) of being The Charm, but we want The Charm to occur at least once in the set of n Times.
Following this, how would we model the equation to determine the probability of at least one Time i between 1 and a given n being The Charm? I'd assume it's a series with a limit of 1 as n approaches infinity, but it's been too long since college, and I've forgotten how to solve this.
 A: I hope I understood your question.
I also didn't understand why you were thinking about infinity. Do you want to calculate the probability that you have at least one "successful time" out of "$n$ times", each time independent with probability $p$?
If so - this is a binomial variable. The complement to getting at least one successful time is not getting any successful times at all, therefore you get:
$$P(\text{getting the charm at least once out of n tries}) =1-P(\text{not getting the charm at all}) = \ \ \ 1 - (1-p)^n$$
A: As I understand it, you are looking for the probability that the event $i$ occurs at least one time in $n$ consecutive events. This probability equals 1 minus the probability that it does not happen in $n$ consecutive events, and since this probability is always the same, we have:
$$1-(1-P[i])^n$$
A: In the abscence of real numbers I am going to model this akin to a dice roll:
Let 6 = "the charm", the odds of this happening for any given dice roll is of course 1/6, but over time we can add these up:
roll 2: the odds are 5/6 * 1/6 + 1/6, this is because it relies on the first roll not being a 6, and then getting a 6, plus the original possibility of getting a six.
Now this is just a matter of series: with each dice roll the odds is 5/6 * (odds of previous dice roll) which is a convergent series, emaning we can use the sum to infinity equation: a/(1-r) where a is 1/6 and r is 5/6, which gives 1, as expected.
