BUT WAIT - THERE’S MORE!!
I’m going to skip the algebra this time and cut to the chase.
If we set the final probability - that of observing X at least once in N trials, equal to 1/N, what can we say about the probability of X in a single trial? Written out :
(1-(1-p)^N) = {1 \over N}
For clarity later, let’s assign a name, p_n = {1\over N}, so
(1-(1-p)^N) =p_n
It’s pretty easy to solve for p, giving,
p= 1-({{N-1}\over N})^{1\over N} = 1-\sqrt[N]{{{N-1}\over N}}
So if we have N=100 trials, and we assume the probability we will observe X at least once is p_n={1\over N} = 0.01, then p=0.0001005. Note that 0.01 is almost 100 times greater then 0.0001005. both probabilities are still pretty small, but multiple trials make a big difference!
Next let’s solve for N. This is much harder, but there is another trick! (Calculus!) For values of a close to one, the logarithm function is nearly linear, so that,
log(a) ~= a-1
For 0.9< a <1.1 this approximation is very good (and calculus is useful!)
Using this, and the fact that {{N-1}\over N} is close to one for large N and the probability p is very small (close to zero), I can give the following approximation:
N\approx{ 1\over\sqrt{p}}
So if p is …let’s say a* one-in-a-billion probability* (p=0.000000001) then N \approx 31,623, and p_n \approx {1 \over 32623} = 0.000031623
which is a LOT more than one-in-a-billion.
This is a good place to stop. It should be obvious there are serious implications for ID probability arguments, but I’ll let people figure this out for themselves.
Next I’ll work this out to find the number of trials needed so the probability of observing X (with probability p) at least once is 0.5.