Earlier today there was Youtube video attempting to solve a problem for a certain game. In it he tries to calculate the probability of certain events happening which narrows down to this equation:
$P(r) = 1-(1-10^{-1014})^{10^{155}}$
Unfortunately, there isn't a trivial way to calculate the percentage form of the answer due to very large exponents, even with a machine. This is why he was unable to come up with an percentage himself.
So how does one approximate or calculate a percent form of large exponent probabilities similar to this ?
Correction: The original value I posted was $P(r) = (1-10^{-1014})^{10^{155}}$, this was different the the actual formula used in the video which has been corrected to the one above, but doesn't effect the question.
You have $$ (1-10^{-1014})^{10^{155}} = e^{10^{155}\ln(1-10^{-1014})} \simeq e^{-10^{155}\cdot 10^{-1014}} = e^{-10^{-859}} \simeq 1-10^{-859} $$ using the low-order Taylor expansions $\ln(1+u) = u +o(u)$ and $e^u = 1+u+o(u)$ when $u\to 0$.
Another approach, using $(1+u)^n = 1+nu + o(u)$ for fixed $n$ and $u\to 0$ (which is a valid approximation as $"nu" = 10^{155}\cdot 10^{-1014} \ll 1$) gives directly $$ (1-10^{-1014})^{10^{155}} \simeq 1+ 10^{155}\cdot (-10^{-1014}) = 1-10^{-859}. $$ (the second approach is faster, but a bit more dangerous, as for the approximation you need to argue or "feel" that your $n$ can be considered as a constant w.r.t. your $u$).