This post reminds me of another estimate. Using the same mistake as a starting point it can be phrased like: “It’s a p chance which I did n times, so it should be np if np<<1.” This is because (1-p)^n = 1 - np + n(n-1)/2 p^2 - …, and since np<<1 this is approximately 1-np.
However, I find this linearity more useful when combining small changes: a 1% increase followed by a 2% increase is approximately a 3% increase, since (1+p)(1+q)=1+p+q+pq and pq can be ignored in an approximation.
This post reminds me of another estimate. Using the same mistake as a starting point it can be phrased like: “It’s a p chance which I did n times, so it should be np if np<<1.” This is because (1-p)^n = 1 - np + n(n-1)/2 p^2 - …, and since np<<1 this is approximately 1-np.
However, I find this linearity more useful when combining small changes: a 1% increase followed by a 2% increase is approximately a 3% increase, since (1+p)(1+q)=1+p+q+pq and pq can be ignored in an approximation.