# 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.

this file last touched 2024.08.13