Aaron Kriegman /

Decibels

In our number system, it is not very easy to quickly judge the magnitude of a number. The first thing you look for to get a feel for the size of a number is how many digits it has, and the second thing you look for is the first digit. Now as it turns out, for this purpose, different digits give you different amounts of information. If the first digit is a $1$, then your number could be as large as, say, $1999$, nearly twice the smallest it can be, $1000$. So you really need to look at the second digit to get a feel for the size of your number. On the other hand, if the first digit is a $9$, then the largest possible value is only $11\%$ larger than the smallest, so you already have a decent idea of the magnitude. This idea is often formalized as Benford’s law, which states that in typical data sets the first digits of the numbers are not evenly distributed, with $1$ appearing about $30\%$ of the time, $2$ appearing the second most, etc.

Quickly judging multiplication is not easy either, and is also dependent on the first digit. If I asked you what’s $3950$ times $5050$, you could estimate that as $4000$ times $5000$, and tell me about $20$ million pretty easily. If I asked you what’s $1250$ times $1580$, it’s not so straightforward. Maybe you could estimate it as $1200 \cdot 1600$, but now you’re doing multiplication with two significant figures, which is already much harder. If you just wanted to quickly estimate the answer, you might not bother in this case. In both these cases, the proportions of our two numbers are the same. That is to say, in a world where $3.16$ centimeters made an inch, you’d be doing the first problem if you worked in centimeters and the second if you worked in inches. So these two multiplication problems are really the same problem with a different skin, yet one is harder for no good reason.

Enter decibels¹. The decibel scale was originally meant for measuring the volume of sound, but it can be combined with other units², or just used to represent plain numbers. The definition of the scale is that every $10$ decibels is a factor of $10$. So $20$ dB $= 100$, $-10$ dB $= 0.1$, $0$ dB $= 1$, etc. This means that $5$ dB is the square root of ten, $1$ dB is the tenth root of ten, etc. One incredibly useful coincidence is that $3$ dB is very close to $2$. This corresponds to the fact that $2^{10} = 1024 \approx 1000 = 10^3$.

Now let’s revisit our previous examples but with decibels, and see if they’re easier. $1000$ and $1999$ written in decibels are $30$ dB and $33$ dB. Before we needed to look at three pieces of information to get a feel for the magnitudes of these numbers: the number of digits, the first digit, and the second digit. But after expressing the numbers in decibels, we can get a good feel for their sizes with only two pieces of information: the tens place³ tells us the order of magnitude of the number, and the ones place tells us pretty well where the number falls within that order of magnitude. The numbers starting with $1$ get the $0$ to $3$ range of the ones place when written in decibels, so a proportional amount of our scale is dedicated to these numbers. Meanwhile $80,000$ is $49$ dB and $90,000$ is $49.5$ dB, so these numbers get squeezed into a smaller part of our scale, as they should since they are proportionally closer.

One of the key properties of decibels is that $x$ dB $\cdot\ y$ dB $= (x + y)$ dB. That is, decibels turn multiplication into addition (and division into subtraction). So our previous problem, $3950 \cdot 5050$, becomes $36$ dB $\cdot\ 37$ dB, which is just $73$ dB $= 20$ million. About as easy as before. On the other hand, $1250 \cdot 1580$ becomes $31$ dB $\cdot\ 32$ dB $= 63$ dB $= 2$ million, which is much easier than before. Importantly, it’s the same difficulty as the first problem, as it should be.

Decibel arithmetic tips

One time in a meeting in my first internship, we were discussing poker hand probabilities, when someone asked how many possible five card hands there are. Nobody knew the answer off the top of their head, so we just moved on. Sitting in that meeting, using decibels, I was able to calculate in my head that it was about $2$ million. I then refined my estimate to $2.5$ million, impressing all my coworkers when we looked it up and it was about $2.6$ million. Once you learn the tips in this section, you’ll be able to do these kinds of calculations in your head too.

Converting to and from decibels

As I mentioned before, the approximation $3$ dB $= 2$ is going to be our best friend. This tells us that $6$ dB $= 4$ and $9$ dB $= 8$. Working backwards from $10$ dB $= 10$, we get that $7$ dB $= 5,$ $4$ dB $= 2.5$, and $1$ dB $= 1.25$.

For $5$ dB, we can use the approximation $\sqrt{10} = \pi$ to get $5$ dB $= 3.14,$ $8$ dB $= 6.28$, and $2$ dB $= 1.57$. This approximation is not the most accurate or the easiest to use, but it is easy to remember. Slightly more accurate is $5$ dB = $3 \frac16$ (That’s an improper fraction! This is the first time I’ve used those since elementary school.) which gives $2$ dB $= 1 \frac7{12}$ and $8$ dB $= 6 \frac13$. A helpful way to remember the $2$ dB approximation is that $0$ dB, $1$ dB, $2$ dB, and $3$ dB break the interval from $1$ to $2$ into pieces of proportion $3$ to $4$ to $5$. The relative errors⁴ of all these estimates are all smaller than $-21$ dB, or $0.8\%$.

Then we can get an estimate for any whole number of decibels by just sliding over the decimal point. For example, $14$ dB $= 25$, and $-6$ dB $= 0.25$.

For values in between a whole number of decibels, the best way to convert is by interpolating linearly. For example, to convert $3$ into decibels, we recall that $4$ dB $= 2 \frac12$ and $5$ dB $= 3 \frac16$. $3$ is $3/4$ of the way from $2 \frac12$ to $3 \frac16$, so $3 = 4 \frac34$ dB.

Now as an example of how to convert the other way, say we want to find $2 \cdot 3$ using decibels. That’s $(3 + 4 \frac34)$ dB $= 7 \frac34$ dB. $7$ dB $= 5$ and $8$ dB $= 6 \frac13$, so the answer is three quarters of the way from $5$ to $6 \frac13$ which is approximately $6$. This is a silly example, but it’s cool that it gives us the right answer.

The relative error of this linear interpolation is also smaller than $-21$ dB, so as long as we’re using the reference points mentioned above there’s no reason to use a better interpolation method⁵.

An example

Now I’ll show you how to use this to estimate the number of poker hands. There is a well known formula for the number of ways to choose $5$ cards out of a $52$ card deck:

$$\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$$

This number is called “$52$ choose $5$”. We can approximate the numerator as $50^5$, especially since the factors are spread symmetrically around $50$, so the errors in this approximation will partially cancel⁶.

Now $50 = 17$ dB, so $50^5$ is $(17 \cdot 5)$ dB $= 85$ dB. The denominator comes out to $120$, which is close to $125 = 21$ dB. So our final answer is $(85 - 21)$ dB $= 64$ dB $= 2.5$ million. Pretty close to the actual answer of $2,598,960$! I originally estimated the denominator as $20$ dB, which is how I got my first estimate of $2$ million.

But we can do better. Instead of rounding $120$ to $125$, we can interpolate. $120$ is $4/5$ of the way from $100$ to $125$, so it’s closer to $20.8$ dB. So our final answer is $64.2$ dB, which is $1/5$ of the way from $2.5$ million to $3 \frac16$ million, so about $2.63$ million. This is slightly closer to the answer and has a relative error of just over $-20$ dB. If you round this estimate to $2.6$ million you’ll get lucky and get a lot closer.

Conclusion

The first takeaway from all this is that when you encounter numbers in the wild and you need to multiply, divide, or compare their proportions, you may find it useful to convert them to decibels first. I find that it’s often not worth it for one off multiplication or division, but for exponents, comparisons, and products of many numbers, this conversion can be very helpful.

But the bigger point I want to make is that I think we should start presenting many numbers as decibels in the first place. As discussed, our number system is biased towards certain digits, and makes common operations more difficult than they have to be. If we started using decibels in news articles, on nutrition labels, etc., it would be easier to intuitively judge numbers. Unfortunately, this would require getting our entire society to become literate in decibels, which would not be worth it for this marginal improvement. So it seems that this system of approximations that I’ve come up with will be relegated to the realm of party tricks and occasionally useful tools for the foreseeable future. Maybe one day this could be incorporated into a school curriculum, but that would probably be a bad choice because there’re more important things for kids to learn in math class.

I would like to point out that not all numbers are better in decibel form. Some numbers are meant to be multiplied, divided, and compared using ratios, and for those numbers we should use decibels. Other numbers are meant to be added, subtracted, or compared using differences, and for these numbers it would not be helpful to use decibels.

Lastly, I’ll mention that decibels are what is known as a logarithmic scale. I’ve avoided using this word because I don’t want to scare people away. I think that the content in this post is actually pretty accessible, but if I’m not careful then people will get intimidated and their eyes will glaze over.