Cost of rounding in a probability computation

You’re making some quick computations and you don’t want to fuss around with a calculator or computer. You’ve got some fraction \(\frac{x}{y + \epsilon}\) but you’d much rather round it to \(\frac{x}{y}\). What’s the error, \(\Epsilon\)?

\[ \Epsilon = \frac{x}{y+\epsilon} - \frac{x}{y} = x\left ( \frac{1}{y + \epsilon} - \frac{1}{y} \right ) = x\left ( \frac{y}{y(y + \epsilon)} - \frac{y + \epsilon}{y(y + \epsilon)} \right ) = x\left ( \frac{-\epsilon}{y(y + \epsilon)} \right ). \]

It’s reasonable to assume \(\epsilon\) much smaller than \(y\) so that

\[ \Epsilon \approx \frac{-x}{y^2}. \]

So if \(y\) is substantially larger than \(x\) this error is quite small.

Example

You’ve got 8 balls marked “1” to “8”. You draw three balls, what is the probability that “1” “2” and “3” aren’t drawn?

\[ P(\text{none of 1 2 or 3}) = 5/8\cdot 4/7 \cdot 3/6 = 5/42. \]

Now \(5/42 \approx 5/40 = 1/8 = 0.125\) with an approximate error of \(-5/40^2 = -0.003125\). The true answer is \(5/42 = 0.119047619\), so the actual error is more like \(-0.00595\). If you’re busy with a poker game or some other game of chance, “one in eight” with “error roughly in the third digit” should be good enough.