# 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.

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 $$n\geq x$$.

• Thank you for the answer, but that formula is way beyond me, unfortunately my grasp on mathematical symbology is rudimentary at best. I can just about read and follow the logic of the various parts, which the best I can make out is. If 0.1 where successful number of outcomes is greater than 50t = 100t Sum of values 0.1 where successful outcomes = 50t by 50tr = 100t Sum of values 100t ? to n something P to the power of 1 - p) 100t - n but I couldn't use it to create calculate a value if my life depended on it. The effort is highly appreciated though. Sep 24 '20 at 18:15
• The pair of numbers in a bracketed column is called a Binomial coefficient, if that helps. The $\sum$ does indeeed mean a sum, of the term to its right with each $n$ replaced with $x, x+1,...$ and so on up to $N$ Sep 24 '20 at 18:21
• TBH I was expecting if had the value of probability for rolling at least 5 dice with a value of 1 from 10 dice rolled = 0.001635, there would be some simple power value that could be applied for each factor of 10 lol Sep 24 '20 at 18:29
• Thanks Redriod, Is there any apps that would allow me to input this formula and assign the variables numeric values, I've always been better at understanding things when I can see how each component changes the results for myself if that makes sense? Sep 24 '20 at 18:42
• Well, I tried computing it in MATLAB for you, but it seems that the result is so tiny that it shows up as zero (i.e., it is smaller than MATLAB's numerical precision). Even if you only want $5$ or more successes out of $10$ trials, with a success probability of $0.1$, the probability is $1.469\cdot 10^{-4}$, which is rather small. Sep 24 '20 at 19:15

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 $$\sigma^2 = Np(1-p) = 100\ \text{trillion}\cdot 0.1\cdot 0.9 = 9\cdot 10^{12}$$