There are incredibly usefull shorthands. What is the probability for 2n attempts?
In the infinite limit (or just large-ish x), the probability of at least one success, from nx attempts with 1/x odds on each attempt, will be 1 - ( 1 / e^n )
For x attempts, 1 − 1/e = 0.63212
For 2x attempts 1 − 1/e^2 = 0.86466
For 3x attempts 1 − 1/e^3 = 0.95021
And so on
You have the tools necessary to figure this out
There are incredibly usefull shorthands. What is the probability for 2n attempts?
In the infinite limit (or just large-ish x), the probability of at least one success, from nx attempts with 1/x odds on each attempt, will be 1 - ( 1 / e^n )
For x attempts, 1 − 1/e = 0.63212
For 2x attempts 1 − 1/e^2 = 0.86466
For 3x attempts 1 − 1/e^3 = 0.95021
And so on
You have the tools necessary to figure this out