When there is a competition to estimate some number (e.g. the amount of money in a jar full of coins) we want the person closest to the actual value to win. Since the distance between two numbers x and y is just x-y (or y-x if y happened to be bigger) the player with the lowest difference to the actual value will win. Say there were 9$ in the glass and player A estimated 4$ while B said 16$. Because 9-4 = 5 and 16-9 = 7, player A was better. Or was he?
To understand why this question makes sense we will look at a more extreme example. Suppose the question was what the average US citizen earns in a year with the correct answer being 50,000$ (I didn’t look it up but this will do here). This time B said 100,000$ and A went with 1$. Obviously B did the better job because 100,000$ was fairly reasonable while 1$ was complete nonsense. But as done above we obtain 100,000-50,000 = 50,000 and 50,000-1 = 49,999 so A won by a margin of 1$.
The reason B seemed much better at first is that humans intuitively think about most things not as sums or differences but rather as products. What that means is that we don’t perceive B as off by 50,000$ and A by 49,999$ but rather notice that B is “only” off by a factor of two while A missed by a factor of 50,000. A scale looking at products instead of sums/differences is called a logarithmic scale. You might have seen this in some graphs depicting vastly different numbers. To my knowledge all human senses work like this, the best known example being the decibel scale for how loud something is which is also logarithmic. This also extends to how we think about numbers (particularly big ones) explaining why we favoured B earlier.
The logarithmic scale is closely tied to something called the geometric mean. You will be familiar with the arithmetic mean of two numbers x and y which is (x+y)/2. Now we will again replace the sum by a product. This poses the problem that the result would be in dollars (or any other unit x and y had) squared which doesn’t make sense so the geometric mean of x and y will be the square root of x*y, keeping the original unit.
Looking at the first example there would have been another way of figuring out who was closer. We calculate the arithmetic mean of the two guesses and if it is bigger than the actual value then the lower guess (A) was better, otherwise B wins. This can quickly be obtained when rearranging the differences but it also makes sense intuitively. The mean is exactly the same distance from both guesses so if the correct answer is below that it has to be closer to the smaller guess. Indeed, (4+16)/2 = 20/2 = 10 is bigger than 9 (the real value) so A is closer. If we want to use the logarithmic scale we can do the exact same thing, replacing the arithmetic mean by the geometric one. 4*16 = 64 and the square root of that is 8. Because this is lower than 9, B wins this time.
Finally, I want to ask for your opinion. Is B the real winner or does the logarithmic scale not matter for numbers that are so close together? Another thing to factor in is that the traditional arithmetic method is certainly much easier to understand so everyone can clearly see who won without having to do difficult calculations like square roots.
