Mathematicians! Please help settle a question about rounding numbers

[therealthom]

A colleague of mine posted a query about rounding numbers and I am not at all sure about my reply which I've spoilered below. Am I even correct? And can anyone provide a concise easily understandable explanation? I am dissatisfied with my argument being based on an example rather than a principle.

Thanks in advance for your help.

"Questioning Colleague" wrote:

I’m no mathematician, but I don’t understand why there needs to be this augmentation to the basic rounding rules that we all learned in first or second grade. If I’m given the instruction to round a given number to two significant figures, how does rounding a “5” up or down 50 % of the time making the rounding process more even?

Example:
Using the “original” method, I would round the value 0.220 to 0.22, 0.221 to 0.22, 0.222 to 0.22, 0.223 to 0.22, 0.224 to 0.22, 0.225 to 0.23, 0.226 to 0.23, 0.227 to 0.23, 0.228 to 0.23, and 0.229 to 0.23. That gives my five values that are round to 0.22 and five values rounded to 0.23. If I look at all of the possible values of 0.22 out to the forth decimal place, I would have 50 values that would be rounded to 0.22 and 50 values that would be rounded to 0.22, and this goes on and on in powers of ten for infinity. Are we forgetting that “0” is a number too? In the field of measurement science, the value 0.220 is not the same as 0.22.

So my question is, why is this method created in the first place and why should anyone feel it is necessary to use it?

My reply:

Hmm. I've never thought to question this, perhaps because I'm not a mathematician either. While I agree that .220 is not the same as .22 for metrology purposes, I am not sure that your conclusion necessarily follows in purely mathematical terms.

If we look at this question in terms of the 'distance moved' in rounding the number then we might get a very different result : Assume a flat distribution of results.

For the 5 results that would be rounded down:

.22 -.220 = 0
.22 - .221 = -.001
.22 - .222 = -.002
.22 - .223 = - .003
.22 - .224 = - .004

And potentially the total distance moved while rounding down is -.010.

Similarly for rounding up :

.23 - .225 = .005 \\ .23 - 226 = .004 \\ .23 -.227 = .003 \\ .23-.228 = .002 \\ .23 -.229 = .001

And the total distance moved while rounding up would be +.015. This would bias the average of your total population higher.

But I'm no expert. It might be nice if someone better qualified weighed in with an argument.

[Orfamay Quest]

You are correct. The issue is not the number of numbers, but the effect of the rounding process. Zero is a number, but it's a number that doesn't need to be rounded as it introduces no bias and no error when you round the number 0.32000 to 0.32. On the other hand, it introduces the maximum possible error when you round 0.32500 to 0.32 or to 0.33.

Basically, for every real number x over the relevant interval [0,1), there is another number in that range y = 1-x that is different from x, with two exceptions:

* if x == 0, then y == 1 and y is not in that range.
* if x == 0.5, then y == 0.5 and y == x

We now assume a uniform distribution (technically, a uniform distribution of terminating decimals of fixed maximal length) on that range.

In the general case, there is no problem: if you round x up and y down, or vice versa, you will introduce equal and opposite biases with equal probability, and hence a net zero bias.

The first special case is not a problem; rounding 0 down to 0 introduces no bias.

The second special case however, is problematic. If you always treat it the same, you will introduce a systematic bias of 0.50 divided by the probability of that specific number arising. By randomly rounding either up or down, the net bias will again be zero.

You can confirm this for yourself. Roll a die and write the number of pips shown on one piece of paper. Then write HALF the number of pips shown (rounded) on another piece of paper. Repeat until you're bored.

If you always round 0.5 up (for example, rolling a 3 becomes a 1.5 rounded to 2), then the total number of pips on the second paper will be more than half of the total on the frist. If you always round down, the total number on the paper will be less than half.

If, however, you always round every other 0.5 up or down, you will find that the total on the second piece of paper is always exactly half the number on the first paper (when the total on the first is even) or within 0.5 of half that number (when the total on the first paper is odd).

[therealthom]

Thanks, Orfamay! Brilliant.