A statement made by an *ExcelTips* subscriber (Chuck) in answer to a Help Wanted question provoked some interesting comments from many other *ExcelTips* subscribers. Chuck made the statement that, statistically, the value .5 should be round up half the time and down the other half of the time because it is exactly in the middle of two whole values. The analogy was provided that if a tennis ball was balanced on a net, statistically the ball should fall left half of the time and right the other half of the time.

It seems that there are some strong feelings about such statements, even among other statisticians. (All disciplines seem to have their various religious wars where feelings run high.) As one correspondent mentioned, this is "the old 'fences vs. fence posts' problem in counting intervals between numbers." The argument is where something will "fall" when it is situated right on a fencepost. The problem with the tennis ball and net analogy (or fences and fence posts) is that the net in the middle of the court is not the only precise dividing line.

For instance, let's say that the left end of a tennis court has a line marked "4.0" and the other end has a line marked "5.0." This means that the net is marked "4.5." While a tennis ball could balance on the 4.5 mark and fall either way, theoretically the ball could also balance on the line at either end of the court (4.0 and 5.0) and fall either way off of them, as well.

One correspondent expressed the feeling that rounding .5 either up or down (half one way and half the other) is inappropriate because it introduces bias into the data. Consider the situation where you are dealing with one digit to the right of the decimal point: You have numbers 7.0, 7.1, 7.2, etc., all the way through 7.9. When rounding these figures, five values would always round down (7.0 through 7.4), one value could round either way (7.5), and four values would always round up (7.6 through 7.9). In other words, over time 5.5 values would round down and 4.5 values would round up. In a true even application of statistical probability, 5 values should round down and 5 up, but the "waffling" of the center value (7.5) makes a bias in favor of rounding down and against rounding up.

So, which theory of rounding is correct? Should 7.5 round up half the time and down half the time, or should it always round up? Microsoft has obviously made its mind up, as it always round 7.5 up (the tennis ball always falls to the right for positive values and to the left for negative values). Does Microsoft's decision mean that always rounding .5 up is correct? Your position in the rounding religious war will determine your answer.

Well, perhaps another data point will help. It appears that there is an ANSI standard on this whole issue. One subscriber indicated that he had always followed the standards ASTM E29 and ANSI Z25.1, both of which specify that an exact fractional value of .5 should be rounded to the nearest number ending in an even digit. If you need to do this type of rounding, then the proper formula to use is this:

=IF(A1-INT(A1)-0.5,EVEN(ROUNDDOWN(A1,0)),ROUND(A1,0))

To see how this can affect the outcome of rounding, I generated a series of 25,000 random numbers between 1 and 100, where each result had up to two decimal places. I then rounded the values to a whole value using the regular ROUND function in one column, and in the next column I rounded the numbers using the above formula. I then summed each column to see which method of rounding produced results closer to the original sum. In my test, the results were over 50% closer to original sum by using the above formula rather than Excel's ROUND function alone.

I then generated 25,000 random numbers with up to three decimal places, and the results were the same—the formula was closer than a generic ROUND. The same held true with numbers with four and five decimal places, as well.

One thing I did notice in my testing was that in the first set of test data (random numbers with up to two decimal places) there were 234 values that exactly matched the criteria of being exactly .5 (and thus eligible for rounding up or down). In the list with three decimal places the number of matches dropped to 14 values, with four decimal places it was 2 values, and with five decimal places it was 0 values. It stands to reason that the fewer values there are that meet the criteria of ending in .5, the less necessity there is to apply the "round up or down" logic. Thus, the rounding formula, above, loses its effectiveness when you start dealing with numbers having four, five, six, or more digits to the right of the decimal point by virtue of the fact that there are exact-center matches.

Any discussion of rounding, of course, needs to assume that you are rounding raw values, not previously rounded values. For instance, if a raw value is 14.46 and you round it to 14.5, it would be improper to later round the 14.5 to 15. The correct procedure would be to examine the original 14.46, which should round down, to 14. Thus, you should always use ROUND as one of your last steps in working with numbers, not as one of the first. This means that when using aggregative functions, such as SUM or AVERAGE, you should not apply them to values that have already been rounded. Instead, you should SUM or AVERAGE the raw values, and then do the rounding on the SUM or AVERAGE. You will get more precise results by remembering this tip.

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2020-07-17 07:04:00

Maarten Boers

Thanks for the clear explanation. I think Excel should provide a function that does this without the fairly complex formula you described.

I had a bit of trouble understanding it, until I noticed a typo:

The formula should read:

=IF(A1-INT(A1)=0.5,...

rather than:

=IF(A1-INT(A1)-0.5,...

Just before posting I noticed the discussion below, so the error has been corrected in ways similar to the above.

But you haven't corrected the formula yet...

Cheers,

M

2020-04-01 11:46:55

Rick H.

The formula should be "=IF(A1-INT(A1)-0.5=0,EVEN(ROUNDDOWN(A1,0)),ROUND(A1,0))"

For two decimal places: "=IF(A1*100-INT(A1*100)-0.5=0,EVEN(ROUNDDOWN(A1*100,0))/100,ROUND(A1,2))"

Dave - It seems to work fine with negative values. -0.5 rounds to zero, -1.5 rounds to -2.

2020-03-31 12:50:35

Chester Hood

Our calculation tool at that time was a slide rule and we were taught about significant digits and orders of magnitude. Four function calculators were introduced in this era and cost over $100, so almost all of us had a Keuffel & Esser Log Log Duplex Decitrig (recommended) or a Pickett. We also learned to do "back of the envelope" calculations (get the decimal point in the right place) which served me well throughout my career and I taught this technique to my children.

2020-03-31 12:07:28

Dave

I wonder whether the formula works seamlessly for numbers as we transition to negative numbers...

2017-11-02 12:35:17

Ron MVP

Thanks for the update.

2017-11-02 11:28:00

Chris C

2017-11-01 09:58:39

Allen

I just now moved it from the menu site to the ribbon site. I also updated the tip (above) with info that was in the older article you mentioned.

-Allen

2017-11-01 04:03:05

Ron MVP

Could you migrate this tip to the Ribbon site, so it does not get lost when you eventually retire the menu tips.

There used to be a second article to expand on this one. But I can't find it any more to share with other users (I have a copy in my archive disks)

http://excel.tips.net/T002835_Rounding_Religious_Wars_Take_Two.html

2016-05-19 16:26:30

Tom Bates

Thanks Rick Rothstein for that tip. That will come in very handy!

2016-05-18 05:32:08

Rick Rothstein

Num = 2.45

MsgBox Format(Num, "0.0")

2016-05-17 08:51:20

Tom Bates

tenths =ROUND() CLng()

1.2 1.0 1

1.3 1.0 1

1.4 1.0 1

1.5 2.0 2

1.6 2.0 2

1.7 2.0 2

1.8 2.0 2

1.9 2.0 2

2.0 2.0 2

2.1 2.0 2

2.2 2.0 2

2.3 2.0 2

2.4 2.0 2

2.5 3.0 2

2.6 3.0 3

2.7 3.0 3

2.8 3.0 3

2.9 3.0 3

3.0 3.0 3

3.1 3.0 3

3.2 3.0 3

3.3 3.0 3

3.4 3.0 3

3.5 4.0 4

3.6 4.0 4

3.7 4.0 4

3.8 4.0 4

From this list, we see that there are 10 2.0 values in column 2, and 10 3.0 values, because the ROUND function is correctly rounding .5 up.

In contrast, the CLng function rounds incorrectly, because the results are obviously very skewed towards even numbers: we see more 2's (11) than 3's (9).

Thanks to stas for pointing that out; henceforth, I will be using the spreadsheet function ROUND rather than allowing VBA to perform the rounding incorrectly.

2016-05-17 08:30:20

Tom Bates

If you consider rounding to a multiple of 10, 70 *remains* 70, while 71 *becomes* 70. It may be easier to recognize that 70 to 79 are "the seventies"; 70 to 74 are the lower half, 75 to 79 the upper half, exactly even. 80 is clearly not part of the 70's.

So .49999999999999 rounds down, .5 rounds up. Always (mathematically speaking).

2016-05-16 18:42:22

Scott Renz

2016-05-16 13:47:42

Rick H.

Trying to stay away from VBA, I worked up this formula to round cell C16 to two places and rounding even when the third and final digit was a 5:

=IF(AND(C16*1000=INT(C16*1000),C16*200=INT(C16*200),ISEVEN(C16*100)),INT(C16*50+0.5)*0.02,ROUND(C16,2))

If it matters, I'm still using Excel 2000, Win7.

2016-05-16 10:15:29

stas

worksheet round VBA round

0.5 1 0

1.5 2 2

2.5 3 2

3.5 4 4

4.5 5 4

5.5 6 6

6.5 7 6

7.5 8 8

8.5 9 8

9.5 10 10

10.5 11 10

11.5 12 12

12.5 13 12

13.5 14 14

14.5 15 14

15.5 16 16

16.5 17 16

VBA round was from

For i = 1 To 17

Cells(i, 3) = Round(Cells(i, 1), 0)

Next i

I Use Excel 2013 with Win7.

2016-05-16 09:24:33

Rick H.

7.0 is not rounded down. The decimal is simply dropped. 7.0 = 7

This leaves 7.1 through 7.4 (four values) always rounding down, and 7.6 through 7.9 (four values) always rounding up.

Are the values in this scenario only between 7.0 and 8.0?

I was taught to round even. This takes care of skewing the data, you don't have to keep track of which direction the last value was rounded, and subsequent halving isn't left with another .5 to deal with.

If our scenario includes values from a minimum of 0 to a maximum of 10, and we always round the .5 values up, the average is skewed. Rounding even corrects this issue.

Raw values: .5+1.5+...8.5+9.5=50, Avg.=5

Rounding up: 1+2+...9+10=55, Avg.=5.5 which rounds to 6.

Rounding even: 0+2+...8+10=50; Avg.=5

2016-05-14 08:17:15

Roger

2016-05-14 05:54:52

Rick Rothstein

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