Five dice, one equation, twenty thousand rolls: how sure are we about the weight of one tree?
On page 2 the equation gave 13.6 kg for our tree #406 (D = 8.0 cm, H = 6.0 m, ρ = 0.60 g/cm³). But each of those three numbers was measured only once, and the equation itself has its two doubts. So the true weight is not exactly 13.6 kg. How far from 13.6 could it be?
On page 1 we answered this kind of question with Monte Carlo: give each uncertain number a die, roll all the dice, calculate, repeat thousands of times, and look at the pile of answers. There we had two dice and a cylinder. Here we have five dice and the real equation. Panel A introduces the five dice. Panel B rolls them, one draw at a time. Panel C reads the pile. Panel D asks which die matters most. Panel E shows how to write the result.
An adding die says "+ something" or "− something". The height die is one: reading 6.0 m, die says + 0.6 m or − 0.6 m, result 6.6 m or 5.4 m. The two results are the same distance from 6.0 on both sides. This is the ordinary bell curve.
A multiplying die says "× something" or "÷ something". The scatter die is one: point value 13.6 kg, die says × 1.43 or ÷ 1.43, result 19.4 kg or 9.5 kg. Going up adds 5.8 kg, going down takes away only 4.1 kg. The two results are the same distance from 13.6 only in "times", not in kg. A pile made by such a die is squeezed on the left and stretched on the right. It is called log-normal because taking the logarithm turns "times" into "plus", so on the log scale it is an ordinary bell curve again.
Why not use an adding die for everything? Try it for the scatter: an adding die with CV 40 % has SD = 40 % of 13.6 kg = 5.4 kg. (Panel C shows 5.8 kg, because there the 40 % is taken of the mean, 14.6 kg.) Three SDs down gives 13.6 − 16.3 = −2.7 kg, a tree with negative weight. Impossible. The multiplying die never does this: ÷ 1.43 three times gives 13.6 ÷ 2.9 = 4.7 kg, small but still positive. For errors that are naturally "a percentage of the value", and for values that cannot go below zero, the multiplying die is the honest one.
Why not use a multiplying die for everything? For small errors the two dice give nearly the same answer. Height: adding ± 10 % gives 5.4 and 6.6 m; multiplying by 1.1 and dividing by 1.1 gives 5.45 and 6.6 m. The difference is 5 cm, not worth the extra idea. So the height and diameter dice stay simple adding dice. The wood density and scatter dice multiply by a log-normal factor, because those errors are naturally a percentage and the value can never be negative. The a, b die (Panel A) also multiplies the answer, but its 1 % is so small that it is written as a plain adding die on the number 1.
Every number that goes into the equation can be a little wrong. We give each one a die. The table says what each die stands for, how big its error is, and where that size comes from. Nothing here is new: the sizes were all found on pages 1 and 2.
| Die | What can be wrong | Size of the error, as CV (the SD as % of the value) | Where the size comes from | Same for every tree, or different? |
|---|---|---|---|---|
| Height H | The height reading | CV 10 % (for H = 6.0 m, SD = 0.6 m) | Page 1, Panel D: 250 readings of one tree had a CV of about 10 % | Different for each tree |
| Diameter D | The tape reading | CV 2.5 % (for D = 8.0 cm, SD = 0.2 cm) | Page 1, Panel D: the DBH (diameter at breast height) readings had a CV of about 3 %; we use a slightly smaller 2.5 % here so that the error of D² is a round 5 % | Different for each tree |
| Wood density ρ | We did not measure it. We took it from a table, and published values for one species often spread over a range like 0.54 to 0.64 (hypothetical numbers) | CV 10 %, log-normal (for ρ = 0.60 g/cm³, SD ≈ 0.06 g/cm³) | The spread of the published values | Different between species, but the same for every tree of one species. For one tree it behaves like an ordinary die; for a whole plot it is a shared error. |
| a and b | Doubt 1 the line is a little off | CV 1 % | Page 2, Panel D | Same for every tree |
| Scatter ε | Doubt 2 this tree is not on the line | σ = 0.357 on the log scale, about 40 % | Page 2, Panel E | Different for each tree |
On page 1 the height die was flat: every value between −10 % and +10 % was equally likely. Here each die is bell-shaped: values near the reading are more likely than values at the edge. We choose this shape because it is the shape the 250 readings made on page 1.
"CV 10 %" means the SD is 10 % of the reading: 0.6 m for a 6.0 m tree. About 2 draws in 3 land within ± 1 SD, that is within ± 10 % of the reading, and about 19 in 20 land within ± 20 %. The cards in Panel B show this curve, with the shaded part covering the middle 95 %.
The density die and the scatter die are the multiplying kind described in the words box. The scatter die draws ε from a bell curve with SD 0.357, and the weight is multiplied by eε: ε = +0.357 gives × 1.43, ε = −0.357 gives ÷ 1.43 (that is × 0.70). The die is fair in "times", not in "plus".
This is why a tree can come out twice as heavy as the point value but never below zero. The pile in Panel B will show it: a long tail to the right, a steep wall on the left.
On this page the a, b die is one multiplying factor on the whole answer: draw one bell number z, set factor = 1 + 0.01 × z, and multiply. This is the same as shaking a by about 1 % and keeping b fixed at 0.976.
The more correct way is to treat a and b as a pair with a joint probability distribution (one die for the pair): a two-dimensional bell curve with a mean for a, a mean for b, an SD for each, and a correlation between them (when one comes out high, the other tends to come out low). Each draw would then pick one (a, b) pair from this curve. The correlation matters: in a least-squares fit (the method page 2 used to find the line), a refit that gives a steeper b also gives a smaller a, so that the line still passes through the centre of the data (the fan in Panel D of page 2 pivots, it does not shift). Two separate, independent dice for a and b would ignore this and would exaggerate the doubt.
We use the simple form for two reasons. Chave and others (2014) do not give the SDs and the correlation of a and b in the main paper, so the joint curve cannot be written down from it. And Doubt 1 is about 1 % while the scatter die is about 40 %, so under the "square, add, square root" rule the way we split it changes nothing visible. The one thing the simple form hides: a shake of b changes big trees more than small trees, because b is a power of the size value. For tree #406 (size value 230), a 1 % shake of b would move the weight by about 5 %.
Point value: AGB = 13.6 kg · carbon = 6.4 kg · CO₂ = 23.5 kg
Untick a box to switch that die off. Changing any number empties the pile and starts again.
Press Step. An orange marker lands on each die at the value drawn. The five drawn values go into the equation. The answer drops into the pile. Press Play to repeat slowly, or Fast-forward to do 20,000 draws at once.
Under each curve: the value 2 SD below the middle, the middle, and 2 SD above. The shaded part holds about 95 % of the draws.
The pile is lopsided. It has a long tail on the right and a steep wall on the left, because the dice multiply the answer instead of adding to it (Panel A). A tree sometimes comes out at twice the point value, 27 kg (about 1 draw in 25), but never below zero.
The same numbers as on page 1, now read from the pile of draws. They need a few hundred draws before they settle. By 20,000 draws they hardly move.
| Quantity | Point value | Median | Mean | SD | CV % | 95 % interval |
|---|---|---|---|---|---|---|
| AGB (kg) | 13.6 | – | – | – | – | – |
| Carbon = 0.47 × AGB (kg) | 6.4 | – | – | – | – | – |
| CO₂ = 3.67 × carbon (kg) | 23.5 | – | – | – | – | – |
Only the point value comes from a formula. Every other column comes from the pile: the list of all draws made so far, sorted from smallest to largest. The numbers below are for the starting tree after 20,000 draws. Your run gives the same numbers, because the page uses the same random numbers every time, unless you press New random numbers.
| Column | How it is calculated | Result |
|---|---|---|
| Point value | The plain equation with the readings as they are and every die at its centre: 0.0673 × (0.60 × 8.0² × 6.0)0.976. No draws are used. | 13.6 kg |
| Median | Sort the draws. Take the one in the middle: half the draws are below it, half above. | 13.6 kg |
| Mean | Add up all the draws and divide by the number of draws. | 14.6 kg |
| SD | As on page 1: subtract the mean from each draw, square, add, divide by n − 1, take the square root. In kg. | 5.8 kg |
| CV % | SD ÷ mean × 100. The same spread, as a percentage. | 5.8 ÷ 14.6 = 40 % |
| 95 % interval | Sort the draws. Cut off the lowest 2.5 % and the highest 2.5 %. The interval runs from the first draw kept to the last draw kept. | 6.4 to 28.9 kg |
Why the median equals the point value. Every die is centred on its reading, and the multiplying dice are fair in "times": × 1.43 is exactly as likely as ÷ 1.43. So half the draws land above the point value and half below, and the middle draw is the point value itself. (The adding dice on H and D push it up or down by about a tenth of a percent; the gap is too small to see.)
Why the mean is higher. Multiplying dice make a lopsided pile. Going up by × 1.43 adds 5.8 kg; going down by ÷ 1.43 removes only 4.1 kg. The ups are bigger than the downs in kg, so when you add all the draws and divide, the mean lands above the middle. For the scatter die alone the mean is eσ²/2 = e0.064 = 1.07 times the median, about 7 % above. The median ignores the tail, which is why we report it.
Why the 95 % interval is not symmetric. The point value sits 7.2 kg above the lower end but 15.3 kg below the upper end. The same lopsided pile again: "13.6 ± 11" would be wrong on both sides. On the log scale the interval is symmetric: 13.6 ÷ 6.4 = 2.1 and 28.9 ÷ 13.6 = 2.1. That is the signature of multiplying dice.
Why SD and CV are only a rough summary here. On page 1 the 68–95 rule said "2 draws in 3 within ± 1 SD". That rule is for a bell-shaped pile. This pile is not bell-shaped, so 13.6 ± 5.8 kg does not hold 2 draws in 3 on each side equally. The CV of 40 % is still useful as one number to compare dice (Panel D uses it), but the interval is the honest report.
Three rows, one uncertainty. Carbon and CO₂ are the weight multiplied by a fixed number, so their piles have the same shape and the same CV as the weight. Only the units change. (The carbon fraction 0.47 is itself a little uncertain, about 5 %, but that is small next to the scatter die and we leave it out here.)
The button runs the experiment five more times, each time with only one die switched on (5,000 draws each), then once with all five together. The bars show the CV each die causes on its own.
Press the button to see the numbers. The rule: square each CV, add the squares, take the square root. The biggest die decides the answer almost alone.
A single number hides the uncertainty. Write the middle value, the range, how many draws, and which die matters most.
Writing "13.6" suggests we know the answer to within 0.1 kg. We do not: the 95 % interval runs from about 6 kg to 29 kg. Round the number to what you actually know.
The pile is lopsided. "13.6 ± 5.8 kg" (the SD from Panel C) would suggest that the pile reaches equally far below and above 13.6 kg, to 8 kg and to 19 kg. It does not: the spread runs much further up than down. The 95 % interval shows this. "± SD" hides it.