I’ve found a proper approximation after some time and some searching.
Since the binomial distribution has a very large n, we can use the central limit theorem and treat it as a normal distribution. The mean would be obviously 500 billion, the standard deviation is √(n * p * (1-p)) which results in 500,000.
You still cannot plug that into WA unfortunately so we have to use a workaround.
You would calculate it manually through:
Φ(b) - Φ(a), with
b = (510 billion - mean) / (standard deviation) = 20,000
and
a = (490 billion - mean) / (standard deviation) = -20,000
and
Φ(x) = 0.5 * (1 + erf(x/√2))
erf(x) is the error function which has the neat property: erf(-x) = -erf(x)
You could replace erf(x) with an integral but this would be illegible without LaTeX.
Therefore:
Φ(20,000) - Φ(-20,000)
= 0.5 * [ erf(20,000/√2) - erf(-20,000/√2) ]
= erf(20,000/√2)
≈ erf(14,142)
WolframAlpha will unfortunately not calculate this either.
However, according to Wikipedia an approximation exists which shows that:
1 - erf(x) ≈ [(1 - e^(-Ax))e^(-x²)] / (Bx√π)
And apparently A = 1.98 and B = 1.135 give good approximations for all x≥0.
After failing to get a proper approximation from WA again and having to calculate every part by itself, the result is very roughly around 1 - 10^(-86,857,234).
So it is very safe to assume you will lose between 49% and 51% of your gut bacteria. For a more realistic 10 trillion you should replace a and b above with around ±63,200 but I don’t want to bother calculating the rest and having WolframAlpha tell me my intermediary steps are equal to zero.
Zweibel would be two Bel (or two Decibel, converted to the better known unit) if you forgot about your space key.
Surely you mean Zwiebel.