Forgive me if this has been answered, but I couldn't quite find what I'm looking for. Maybe I'm not used to the mathematical terms.
Say there is an event with a chance of 1 in 20 of being successful. This chance doesn't change in each trial.
If there were N trials (let's say 2 for a concrete example), then what is the chance of the next trial (N+1) being successful?
Trial 1: Fail Trial 2: Fail Trial 3: ?
As Arthur said, the probability 'resets' after each individual trial (that might almost be considered the definition of independent trials). So although the probability of waiting for large N before success decreases as N grows, the probability of success in the next M trials after N failures is completely independent of N. This can be illustrated using your own simulated data. The black line is the probability of success after N trials/attempts. The red, blue, and green lines are the probability of success of the remaining trials in the case that the first 50, 100, 150 (respectively) failed. Note that the lines are superimposed on top of the black line.
This demonstrates that even after (for example) 150 failures, the likelihood of success in the following trials is no different than it is when starting 'fresh' with no failures. In other words, no matter how lucky or unlucky or improbable you consider the past events - the probability of success in the future remains unchanged.
So if, for example, you had the extraordinary luck to toss heads 100 times in a row when tossing a provably fair coin (chance = 1 in 2^100), the likely wait before tossing tails from that point in time onward is no different to what it would be if you had just tossed 100 tails in a row, or a combination of 47 Heads and 53 tails or whatever...
And thus, to answer your original question after two failed trials (your concrete example), the probability of success in the next trial is still 1/20.