# How do you round numbers? **URL:** \ **Category:** Factual Questions\ **Created:** [November 5, 2009, 5:03pm UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440 "2009-11-05T17:03:04Z")\ **Posts on this page:** 11\ **Page:** 4
**Author:** ![The\_Hamster\_King](https://avatars.discourse-cdn.com/v4/letter/t/8edcca/32.png) [@The\_Hamster\_King](https://boards.straightdope.com/u/The_Hamster_King)\ **Post date:** [November 10, 2009, 1:21am UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/61 "2009-11-10T01:21:01Z")
> [@robby](#): > > In this instance, however, there would be no reason to round your measurement. You only round when you combine numbers in calculations. I agree with you. I was trying to keep my example in line with the situation described in the OP where the rounding was done (for whatever reason) on the raw data. But my main point remains unchanged. Whenever you’re rounding, if you always round 5’s up, you’re introducing a slight upward bias to your results. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [November 12, 2009, 11:37pm UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/62 "2009-11-12T23:37:57Z")
**Chronos** said: > [@](#): > > One argument for rounding fives up (as opposed to rounding them down, at least) is that you don’t know what might have been done to the numbers before you got them. If someone else tells you that a measurement is 13.5 cm, it’s possible that they’ve already lost some digits. In particular, it’s possible that your source has (incorrectly) truncated instead of rounding, in which case rounding up will get you closer to the true value than rounding down will. How likely is that, vs the person already rounded up? 13.47 just became 13.5 and you want to make it 14 instead of 13. Vs. the measurement being 13.56 and they truncated to 13.5? **ElvisL1ves** said: > [@](#): > > I’m not sure you get the point there. Half of all digits go to the ten on the low side, the other half go the ten on the high side. One of the ten digits doesn’t have far to go, though. No, you fail to get the point. There are an odd number of digits that round. Zero doesn’t round. It is the demarcation point. Zeros truncate, because they are the even line. The next zero is the next even line. 5 is exactly the midpoint, which is why there are numerous systems for rounding that address different things to do with 5. For what you are saying to be true, zero also rounds up. It simultaneously rounds both directions. Look at a metric ruler. Look at the hash marks on a centimeter. Which value is halfway between hash marks? 5. Right at the midpoint. **robby** said: > [@](#): > > Again, wrong. Accuracy is unchanged. Rounding reduces the indicated precision of the data. There is a difference between these two terms (accuracy vs. precision). **Alex\_Dubinsky** said: > [@](#): > > Wait, don’t you mean it reduces accuracy (correspondence with ‘truth’) but preserves precision (repeatability / lack of randomness)? Accuracy is how close to the target. Precision is how tightly clustered the hits. Both rounding and truncating reduce precision - the clustering is opened up. But rounding tries to keep things as close to the same target point as possible. Truncating drifts off target. **robby** said: > [@](#): > > However, what **ElvisL1ves** is trying to explain to you is that **all of these figures are referred to as “significant figures,” which, by definition, includes all of the certain digits, plus ONE uncertain digit.** > > In your example, all three digits of your measurement, including the 3, are significant. Your measurement, in other words, has three significant figures. So if you have an additional digit beyond your uncertain digit (i.e. the last significant figure), it is indeed meaningless. If my scale reading has the last digit at zero, and I’m trying to estimate the portion below, and I want to estimate at 1/4, that would be .25. But the value 13.25 would have too many significant digits, because 13 was the last certain value. Now, how do I pick that estimate value? Is it 13.2 or 13.3? Why? Again, I was estimating 1/4, which is actually a larger block than tenths. Certainly anything like 13.21 or 13.29 would be too many digits. But the best I can read is 13 and 1/4. How do you convert that to digital? ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [November 13, 2009, 2:17am UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/63 "2009-11-13T02:17:00Z")
By the way, we’re not taught to round 5 down in Britain. Always up. (I’d never come across the “round to even” method until a couple of years ago, either.) ---
**Author:** ![robby](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/robby/32/11048_2.png) [@robby](https://boards.straightdope.com/u/robby)\ **Post date:** [November 13, 2009, 2:39am UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/64 "2009-11-13T02:39:13Z")
> [@Irishman](#): > > …But the value 13.25 would have too many significant digits, because 13 was the last certain value. Now, how do I pick that estimate value? Is it 13.2 or 13.3? Why? Again, I was estimating 1/4, which is actually a larger block than tenths. Certainly anything like 13.21 or 13.29 would be too many digits. But the best I can read is 13 and 1/4. How do you convert that to digital? If your measuring device is delineated by unit increments (i.e. by “ones”) such that you can tell the measurement is between 13 and 14, but you are estimating the tenths digit, then there is no way that you can confidently state that the “correct” measurement is **exactly** 13 and 1/4 (i.e. 13.25). The best you can do is to estimate the tenths digit. If you are estimating the tenths digit (which is your uncertain digit), then the best you can do is to estimate whether the last digit is 2 or 3. If you are quite certain that the measurement you are trying to take really is 13.25, then this would actually be a case for rounding your measurement to even and recording 13.2. (Whereas if your best estimate was 13.75, you would record 13.8.) ---
**Author:** ![Arjuna34](https://avatars.discourse-cdn.com/v4/letter/a/eb8c5e/32.png) [@Arjuna34](https://boards.straightdope.com/u/Arjuna34)\ **Post date:** [November 13, 2009, 2:59am UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/65 "2009-11-13T02:59:30Z")
> [@Irishman](#): > > If my scale reading has the last digit at zero, and I’m trying to estimate the portion below, and I want to estimate at 1/4, that would be .25. But the value 13.25 would have too many significant digits, because 13 was the last certain value. Now, how do I pick that estimate value? Is it 13.2 or 13.3? Why? Again, I was estimating 1/4, which is actually a larger block than tenths. Certainly anything like 13.21 or 13.29 would be too many digits. But the best I can read is 13 and 1/4. How do you convert that to digital? If you’re estimating in 4 parts instead of 10, use a different base, where the number of parts is evenly divided by the base. Instead of choosing between 13.2 and 13.3, use binary: 1101.01 - two significant binary figures, exactly matching your ability to estimate the fractional part in fourths. ---
**Author:** ![Napier](https://avatars.discourse-cdn.com/v4/letter/n/ce73a5/32.png) [@Napier](https://boards.straightdope.com/u/Napier)\ **Post date:** [November 13, 2009, 5:54pm UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/66 "2009-11-13T17:54:28Z")
> [@Arjuna34](#): > > If you’re estimating in 4 parts instead of 10, use a different base, where the number of parts is evenly divided by the base. Wow… you just convinced me to blindly truncate. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [November 13, 2009, 10:56pm UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/67 "2009-11-13T22:56:08Z")
**robby** said: > [@](#): > > If your measuring device is delineated by unit increments (i.e. by “ones”) such that you can tell the measurement is between 13 and 14, but you are estimating the tenths digit, then there is no way that you can confidently state that the “correct” measurement is exactly 13 and 1/4 (i.e. 13.25). The best you can do is to estimate the tenths digit. No, I’m saying that I cannot even estimate the tenths digit, the best I can do is estimate whether it is a half or a quarter unit between. I could record my data as 13 and 1/4 13 and 1/2 13 and 3/4 etc. That would eliminate the fuss over 13.25 vs 13.2 vs 13.3, at the cost of having to write fractions on the paper and then try to work the math with fractions instead of convenient decimals. Had a real world example today. For various reasons, I was using a dial torque wrench with a scale that had delineations every 2 in-lbs in the 0 to 50 in-lb range. Because of the size of the dial and the spacing of the lines and the comparative size of the dial needle, the _best_ I could do was estimate to the half an in-lb. So its 12 or 12.5 or 13. Not 12.2 - can’t see that finely. Do I round the halfs to the nearest in-lb? **Arjuna34** said: > [@](#): > > If you’re estimating in 4 parts instead of 10, use a different base, where the number of parts is evenly divided by the base. Instead of choosing between 13.2 and 13.3, use binary: 1101.01 - two significant binary figures, exactly matching your ability to estimate the fractional part in fourths. Yes, because _that_’s easier. :rolleyes: Why don’t I just record the data in Sanskrit while I’m at it. I know, I’ll just use tick marks. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [November 13, 2009, 11:00pm UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/68 "2009-11-13T23:00:28Z")
By the way, yes I could record with 1/2, or 1/4 instead of decimals. But here is the key: in base 10, 1/4 = 0.25 exactly. So writing 13.25 is precisely and exactly equivalent to 13 1/4. Now as long as I am crunching the numbers, and I’m aware that I’m actually working with quarter units, not tenths, I can round the final value to whatever. But annotating the process so someone else can follow the work, I need to be clear. ---
**Author:** ![Arnold\_Winkelried](https://avatars.discourse-cdn.com/v4/letter/a/3d9bf3/32.png) [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)\ **Post date:** [November 14, 2009, 12:09am UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/69 "2009-11-14T00:09:12Z")
> [@Irishman](#): > > > [@toadspittle](#): > > > > Please expand on this concept. What would you round to in the following cases: > > > > 3.390 > > 3.393 > > 3.387 > > 3.383 > > 3.39 > 3.39 > 3.38 > 3.38 Irishman, I believe you meant to say: if one were to “round to even” keeping three significant digits, then the rounding would be like this: 3.390 is rounded to 3.39 3.393 is rounded to 3.39 3.387 is rounded to **3.39** 3.383 is rounded to 3.38 In general round to even works like this (I am rounding to two significant digits). In the example below, the round to even rule only needs to be evoked for the three values that have a five in the last significant digit. (I put in bold the values where I needed to apply the “round to even” rule) 1.10 -\> 1.1 1.11 -\> 1.1 1.12 -\> 1.1 1.13 -\> 1.1 1.14 -\> 1.1 **1.15 -\> 1.2** 1.16 -\> 1.2 1.17 -\> 1.2 1.18 -\> 1.2 1.19 -\> 1.2 1.20 -\> 1.2 1.21 -\> 1.2 1.22 -\> 1.2 1.23 -\> 1.2 1.24 -\> 1.2 **1.25 -\> 1.2** 1.26 -\> 1.3 1.27 -\> 1.3 1.28 -\> 1.3 1.29 -\> 1.3 1.30 -\> 1.3 1.31 -\> 1.3 1.32 -\> 1.3 1.33 -\> 1.3 1.34 -\> 1.3 **1.35 -\> 1.4** 1.36 -\> 1.4 1.37 -\> 1.4 1.38 -\> 1.4 1.39 -\> 1.4 etc… In my example above, I was only rounding and eliminating the last digit. If I were to eliminate more than one digit in my rounding, I consider all the digits when applying my rules e.g. rounding to two digits after the decimal point, using “round to even”. the “round to even” rule only needs to be evoked for numbers that have, immediately following the last significant digit that I’m rounding to, a “5” followed by one or more zeroes. (I put in bold the values where I needed to apply the “round to even” rule) 1.3475 -\> 1.35 1.3549 -\> 1.35 **1.3550 -\> 1.36** 1.3551 -\> 1.36 1.3649 -\> 1.36 **1.3650 -\> 1.36** 1.3651 -\> 1.37 1.3749 -\> 1.37 **1.3750 -\> 1.38** 1.3751 -\> 1.38 (notice that, since I am using “round to even”, 1.3650 and 1.3651 round to different values, whereas 1.3750 and 1.3751 round to the same value) ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [November 14, 2009, 9:11am UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/70 "2009-11-14T09:11:12Z")
Um, yeah, what he said I said. 😉 ---
**Author:** ![Alex\_Dubinsky](https://avatars.discourse-cdn.com/v4/letter/a/dbc845/32.png) [@Alex\_Dubinsky](https://boards.straightdope.com/u/Alex_Dubinsky)\ **Post date:** [November 15, 2009, 1:32am UTC](https://boards.straightdope.com/t/how-do-you-round-numbers/516440/71 "2009-11-15T01:32:47Z")
> [@Arjuna34](#): > > If you’re estimating in 4 parts instead of 10, use a different base, where the number of parts is evenly divided by the base. Instead of choosing between 13.2 and 13.3, use binary: 1101.01 - two significant binary figures, exactly matching your ability to estimate the fractional part in fourths. Yes! YES! Another advantage of my harebrained Binary Inches scheme. [Previous page](https://boards.straightdope.com/t/how-do-you-round-numbers/516440.md?page=3)