How can I calculate these highly improbably odds? If I repeated an experiment 100 trillion times, and the chances of seeing the desired outcome is 10% per repetition, what are the odds of that 1 in 10 chance occurring 50 trillion times or more.
I've tried finding a percent calculator that did the type of sum I wanted, the closest thing I could find was a dice probability by selecting a 10 sided dice with 50 dice rolling 25 dice with the same value of 1 that gave me the odds 0.000000000000907% I tried it with 100 dice and 50 dice of the same value but the value was too low for it to work. https://www.omnicalculator.com/statistics/dice#how-to-calculate-dice-roll-probability
I read the formula they used but couldn't make sense of it.
 A: The number $X$ of successful outcomes follows a binomial distribution with $N$ being 100 trillion, and $p = 0.1$. You want $P\left(X \geq x\right)$ where $x$ is 50 trillion. This evaluates to
$$P\left(X \geq x\right) = \sum^N_{n=x} P\left(X = x\right) = \sum^N_{n=x} {N\choose n}p^n\left(1-p\right)^{N-n}.$$
For each possible $n$ value, you want exactly $n$ out of $N$ successes. If you look at a sequence of experiments containing exactly $n$ successes, any such sequence occurs with probability $p^n\left(1-p\right)^{N-n}$. But there are ${N \choose n}$ such sequences, and you don't care about exactly when you got the successes, only how many there were, so you have to add up the probabilities of all such sequences. You then repeat this for every $n\geq x$.
A: While sven svenson's answer is correct that this has a binomial distribution, it could be impractical to calculate binomial coefficients for numbers this large. Instead, we can use the Central Limit Theorem to say that this binomial distribution is well approximated by a Normal distribution with the same mean and variance given by $\mu = Np = 0.1\cdot 100\ \text{trillion} = 10^{13}$ and  $\sigma^2 = Np(1-p) = 100\ \text{trillion}\cdot 0.1\cdot 0.9 = 9\cdot 10^{12}$
