Working out a neat probability rule from divisbyzero.com.

For small \(x\) (on the order of \(1/10000\)), what is \(1 - (1-x)^t\), where \(t\) is some large number.

Using the binomial theorem expansion

$$(1+x)^n = {n\choose 0}x^0 + {n\choose 1}x^1 + \ldots + {n\choose n}x^n$$

we have:

\begin{eqnarray} (1-x)^t &=& {t\choose 0}(-x)^0 + {t\choose 1}(-x)^1 + \ldots + {t\choose t}(-x)^t \\ &\approx& 1 - {t\choose 1}x + O(x^2) \\ &=& 1 - tx + O(x^2) \\ \end{eqnarray} Thus the original equation can be approximated by:

\begin{eqnarray} 1-(1-x)^t &\approx& 1 - (1 - tx + O(x^2)) \\ &=& tx - O(x^2) \end{eqnarray} The original post was trying to approximate \(1-(1-D/P)^L\), and with \(x=D/P\) and \(t=L\), we arrive at:

\begin{eqnarray} 1-(1-\frac{D}{P})^L &\approx& 1 - (1 - \frac{LD}{P} + O(\left(\frac{D}{P}\right)^2)) \\ &=& \frac{LD}{P} - O(\left(\frac{D}{P}\right)^2) \end{eqnarray}

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