Bankers’ rounding means preferring even digits in edge cases: both 3.5 and 4.5 are rounded to 4. When rounding a random amount of money to the nearest dollar in this way, the expected value added or subtracted is zero. Indeed: 0 cents don’t get rounded, 1 through 49 get rounded down, 51 through 99 get rounded up (and the amounts match 1-49), so 50 has to be split between the two directions.

Python follows this convention: rounding the numbers 1/2, 3/2, 5/2, … via

`[round((2*k+1)/2) for k in range(10)]`

results in [0, 2, 2, 4, 4, 6, 6, 8, 8, 10].

However, when rounding 0.05 = 1/20, 0.15 = 3/20, … to 1 digit after the decimal dot, we get a surprise:

`[round((2*k+1)/20, 1) for k in range(10)]`

returns [0.1, 0.1, 0.2, 0.3, 0.5, 0.6, 0.7, 0.8, 0.8, 0.9]. An irregularly spaced array, in which odd digits win 6 to 4…

This is because the numbers like 0.05 cannot be precisely represented in the double-precision format used for floating point numbers in Python (and other languages): the denominator of 1/20 is not a power of 2. And so, 0.05 gets represented by 3602879701896397/2^{56}. Since 0.05*2^{56} = 3602879701896396.8, this representation is greater than the real number 0.05. And this pushes the rounding routine over the edge, toward 0.1 rather than 0.0.

Out of the numbers tested above, only 0.25 = 1/4 and 0.75 = 3/4 get a precise binary representation and are therefore rounded predictably. For the other 8 the direction of rounding has nothing to do with bankers… oddly enough, 6 out of these 8 are rounded toward an odd digit.

This isn’t the end of surprises, though. Recalling that numerical manipulations on arrays are best done with NumPy library, we may want to import it and try

`numpy.around([(2*k+1)/20 for k in range(10)], 1)`

The output ought to make any banker happy: [0., 0.2, 0.2, 0.4, 0.4, 0.6, 0.6, 0.8, 0.8, 1.]

Why? NumPy takes a different approach to rounding: when asked to round to 1 digit after the dot, it multiplies the numbers by 10, then rounds to the nearest integer, and divides by 10. Multiplication by 10 restores sanity: 0.05 * 10 = 0.5 is represented exactly and gets rounded to 0; similarly for the others.

This algorithm isn’t perfect, however. When rounding 0.005, 0.015, …, 0.995 to two decimal digits, it misplaces 0.545 (to 0.55) and 0.575 (to 0.57). The other 98 numbers are rounded as expected; tested with

`numpy.around([(2*k+1)/200 for k in range(100)], 2)`

Out of 1000 numbers of the form , , there are six deviations from the bankers’ rule. They all appear together, in increments of 0.002, from 0.5015 to 0.5115.

Is there a rounding algorithm that follows bankers’ convention for all (not too long) finite decimal expansions?