We start with one die. We end with a tree. The question is the same for both: what can many readings tell us about the thing that produced them?
Probability is a number between 0 and 1. It says how likely a result is. 0 means "never". 1 means "always". 1/2 means "half of the time". When every result has the same chance, the rule is simple:
Probability of a result = number of ways to get that result ÷ total number of possible results
Rolling…
Probability of any one number, for example 4
= 1 ÷ 6 = 1/6 ≈ 0.167 ≈ 16.7 %
Each face has the same probability, 1/6. The six probabilities add up to 6/6 = 1, because some face must come up.
Flipping…
Probability of head (H) = 1 ÷ 2 = 1/2 = 0.5 = 50 %
Probability of tail (T) = 1 ÷ 2 = 1/2 = 0.5 = 50 %
Head or tail: 1/2 + 1/2 = 1. One of the two must come up.
Careful: 1/6 does not mean "4 comes up exactly once in every 6 rolls". In 6 rolls, 4 may come up 0 times, or 1 time, or 3 times. The probability 1/6 tells you what happens over many rolls: about one roll in six shows a 4. The next sections show this with many rolls.
Read this once now, and come back whenever a word is unclear. Every word is explained with the same tiny example: three readings, 5, 6 and 7. A later section, "Simple calculations", works out all the numbers by hand.
| These two | How they differ |
|---|---|
| Spread and SD | Spread is the idea: how far apart the readings are. SD is the number that measures it. "Large spread" and "large SD" mean the same thing. You can see spread; you calculate SD. |
| SD and SE | SD is the spread of the readings. SE is the spread of the mean: how much the mean would move if you measured everything again. SE is always smaller: SE = SD ÷ √n. For 5, 6, 7: SD = 1 but SE = 0.58. Note that an SE is itself an SD: "SE" is just the special name for the SD of a mean. |
| SD and CV | Both measure spread. SD has a unit (for example 1 m). CV has no unit, it is a percentage (for example 16.7 %). Use CV to compare two things measured in different units. |
| Mean and running mean | The mean uses all the readings. The running mean uses the readings so far and keeps changing. When all readings are in, the two are the same. |
| SE and 95 % range | SE is one number, a distance. The 95 % range is two numbers, a low end and a high end, about 2 × SE on each side of the mean. |
| Probability and share so far | Probability (1/6) is what theory says before rolling. The share so far (for example 21 fours in 100 rolls = 21 %) is what actually happened. With many rolls the share gets close to the probability. You will see this below, when a die is rolled thousands of times. |
A fair die shows each face (1, 2, 3, 4, 5 or 6) with probability 1/6. You cannot see "1/6" in one roll. One roll gives one face, nothing more. After thousands of rolls, each face has come up close to 1 time in 6. Then every bar in the left chart is close to the dashed line, and the running mean is close to 3.5 (the average of 1, 2, 3, 4, 5, 6).
The SD of one die roll is 1.708. This is the "typical distance" of a roll from 3.5: a roll of 2 is 1.5 below, a roll of 6 is 2.5 above, and so on. As you roll more, the "SD of the rolls so far" moves close to 1.708. How 1.708 is calculated from the six faces is shown later on this page, in the section on expected value.
In Panel A we could see the die. We knew it had six faces, 1 to 6, so we could calculate the probability 1/6 before rolling even once. After many rolls, the share of each face matched that probability, and the running mean settled at 3.5.
Now suppose you cannot see the die. Somebody rolls it behind a screen and only tells you the number. Can you work out what the die looks like? How many faces does it have? Which numbers are on them? Is every number equally likely? Below are three mystery dice. One may be an ordinary die. One may have faces 1, 2, 3, 4, 5, 5. One may have eight faces, 1 to 8. Choose a die, roll it many times, study the shares, and make your guess. Then press "Reveal the die".
Three questions to answer from the rolls: (1) How many different numbers appear? (2) Do all the bars become equally high? (3) Is one bar about twice as high as the others?
For the mystery die, the hidden thing is the set of faces and how likely each one is. The rolls are what we get to see. For a tree, the hidden thing is its true height. Each measurement is one "roll": it comes out a little different every time, because of the person measuring, the instrument, the angle and the ground. In both cases we never see the hidden thing directly. We only see readings.
From the readings we make three guesses: a middle (the mean), a spread (the SD), and a range inside which the hidden value very probably lies. And in both cases more readings make the guesses better: the shares settle down, the running mean settles down. The rest of this page builds exactly these three guesses, first with dice, then with a real tree.
In Panel A the running mean settled at 3.5. In Panel B each mystery die settled at its own number. These numbers can be calculated in advance, from the faces and their probabilities, without rolling at all. The result is called the expected value. It is the long-run average: the number the running mean settles at after many rolls.
Expected value = (value × probability of that value), added up over all the possible values
Every possible value takes part. A value with a high probability pulls the expected value toward itself more strongly. A picture helps: draw the probabilities as bars standing on the number line. The expected value is the balance point, the place where the bars would balance on a pivot (the small triangle).
Faces 1, 2, 3, 4, 5, 6. Each value has probability 1/6.
1 × 1/6 + 2 × 1/6 + 3 × 1/6 + 4 × 1/6 + 5 × 1/6 + 6 × 1/6
= (1 + 2 + 3 + 4 + 5 + 6) ÷ 6 = 21 ÷ 6 = 3.5
When all values are equally likely, the expected value is simply the plain average of the faces.
Values 1 to 4 have probability 1/6 each. Value 5 is on two faces, so its probability is 2/6. Value 6 has probability 0.
1 × 1/6 + 2 × 1/6 + 3 × 1/6 + 4 × 1/6 + 5 × 2/6 + 6 × 0
= (1 + 2 + 3 + 4 + 10) ÷ 6 = 20 ÷ 6 = 3.33
The second 5 and the missing 6 pull the balance point down from 3.5 to 3.33. The tall bar at 5 pulls harder than any other bar.
Eight values, each with probability 1/8.
1 × 1/8 + 2 × 1/8 + … + 8 × 1/8
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) ÷ 8 = 36 ÷ 8 = 4.5
More faces with bigger numbers move the balance point to the right.
Two things to notice. First, "expected" does not mean "the value you expect on the next roll". No face shows 3.5 or 4.5. The expected value is about the average of many rolls, not about one roll. Second, the mean of the rolls (from data) and the expected value (from the probabilities) are two routes to the same number. With few rolls they differ. With thousands of rolls they almost agree. Go back to Panel B and compare "Running mean" with the numbers above.
For a die we know the faces, so we can calculate the expected value exactly. For a tree we do not know the "faces": we do not know which readings are possible or how likely each one is. So we cannot calculate the expected value. We can only estimate it, with the mean of the readings we have. If the measuring method is fair, pulling neither up nor down, the expected value of a reading is the true height of the tree. The mean of our readings is our best guess for it, and the more readings, the better the guess.
The SD of the rolls so far in Panel A settles at 1.71. That number can also be calculated in advance, with the same recipe as the expected value. Instead of "value × probability" use "(distance from the expected value)² × probability", add up, and take the square root.
SD from theory = √[ (value − expected value)² × probability, added up over all the possible values ]
| Step, for an ordinary die (expected value 3.5) | Working |
|---|---|
| 1. Distance of each value from 3.5 | 1 − 3.5 = −2.5, 2 − 3.5 = −1.5, 3 − 3.5 = −0.5, 4 − 3.5 = +0.5, 5 − 3.5 = +1.5, 6 − 3.5 = +2.5 |
| 2. Square each distance | 6.25, 2.25, 0.25, 0.25, 2.25, 6.25 |
| 3. Multiply each square by its probability, 1/6, and add | (6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25) ÷ 6 = 17.5 ÷ 6 = 2.917 |
| 4. Square root. This is the SD of one roll | √2.917 = 1.708, written 1.71 |
Here we divide by 6, not by n − 1. The n − 1 correction is for data, where the mean had to be estimated from the readings. Here the expected value 3.5 is exact, so no correction is needed.
The same steps for the mystery dice. Die 1 (expected value 3.33): distances −2.33, −1.33, −0.33, +0.67, +1.67, +1.67; squares 5.44, 1.78, 0.11, 0.44, 2.78, 2.78; add and divide by 6: 13.33 ÷ 6 = 2.22; square root: 1.49. Die 2 (expected value 4.5): distances ±0.5, ±1.5, ±2.5, ±3.5; squares 0.25, 2.25, 6.25, 12.25, each twice; add and divide by 8: 42 ÷ 8 = 5.25; square root: 2.29. Compare these three numbers with "SD of the rolls so far" in Panel B after many rolls.
Two tiny data sets with the same mean, 6, but different spread. Follow the steps with a pen and a calculator. Then the same recipe is applied to ten rolls of an ordinary die.
| Step | Set 1: 5, 6, 7 | Set 2: 2, 6, 10 |
|---|---|---|
| 1. Count the readings, n | n = 3 | n = 3 |
| 2. Mean = sum ÷ n | (5 + 6 + 7) ÷ 3 = 18 ÷ 3 = 6 | (2 + 6 + 10) ÷ 3 = 18 ÷ 3 = 6 |
| 3. Distance of each reading from the mean | 5 − 6 = −1, 6 − 6 = 0, 7 − 6 = +1 | 2 − 6 = −4, 6 − 6 = 0, 10 − 6 = +4 |
| 4. Square each distance (the minus signs disappear) | 1, 0, 1 | 16, 0, 16 |
| 5. Add the squares | 1 + 0 + 1 = 2 | 16 + 0 + 16 = 32 |
| 6. Divide by n − 1 | 2 ÷ 2 = 1 | 32 ÷ 2 = 16 |
| 7. Square root. This is the SD | √1 = 1 | √16 = 4 |
| 8. CV = SD ÷ mean × 100 | 1 ÷ 6 × 100 = 16.7 % | 4 ÷ 6 × 100 = 66.7 % |
| 9. SE = SD ÷ √n | 1 ÷ √3 = 1 ÷ 1.73 = 0.58 | 4 ÷ √3 = 4 ÷ 1.73 = 2.31 |
| 10. 95 % range for the mean ≈ mean ± 2 × SE | 6 ± 1.16 → 4.84 to 7.16 | 6 ± 4.62 → 1.38 to 10.62 |
Why divide by n − 1 and not n in step 6? With n readings, the mean was calculated from those same readings, so the distances are a little too small. Dividing by n − 1 corrects for that. For large n it makes almost no difference.
The rolls were 2, 5, 3, 6, 1, 4, 4, 2, 6, 3. n = 10. Sum = 36. Mean = 36 ÷ 10 = 3.6.
Distances from the mean: −1.6, +1.4, −0.6, +2.4, −2.6, +0.4, +0.4, −1.6, +2.4, −0.6. Squares: 2.56, 1.96, 0.36, 5.76, 6.76, 0.16, 0.16, 2.56, 5.76, 0.36. Sum of the squares = 26.4. Divide by n − 1 = 9: 2.93. Square root: SD = 1.71.
CV = 1.71 ÷ 3.6 × 100 = 48 %. √10 = 3.16, so SE = 1.71 ÷ 3.16 = 0.54.
95 % range for the mean ≈ 3.6 ± 2 × 0.54 = 3.6 ± 1.08, so 2.52 to 4.68.
Now compare with what we know about an ordinary die. Its true mean is 3.5 and the SD of one roll is 1.71. Ten rolls gave a mean of 3.6, not exactly 3.5, but the 95 % range 2.52 to 4.68 contains 3.5. The ten rolls did not tell us the exact answer. They gave us a range, and the true value is inside it. That is what readings can do, for a die and for a tree.
So far: one die and its probabilities, a mystery die guessed from its rolls, and the recipe for mean, SD, SE and the 95 % range. Four experiments remain. Each has its own Play, Step and Reset buttons.
Panel C rolls several dice at once and keeps only their average. It shows that an average has less spread than a single roll, and by how much. This is the reason why more readings make a guess better.
Panel D moves to a tree. We are 250 students in the class. Let us assume that, hypothetically, every one of us measured the height and the DBH of a single tree. (DBH is the diameter at breast height: the diameter of the trunk measured 1.3 m above the ground.) These readings are the "rolls" of the tree. The numbers are made up for this page, but they are the kind a class would get. Panel D is a revision: every idea from the dice, applied to the tree, ending with a 95 % confidence interval for its height.
Panel E finds the same 95 % interval a second way, without the formula, by "rolling the bag" of readings. This is called the bootstrap. Panel F asks what to do when a tree has only one reading, as in our project. The answer is the Monte Carlo method: build the bag yourself from an assumed error, and then roll it.
Press Play at slow speed first. Press Fast-forward only after you have understood one step.
Now roll k dice at the same time and write down only their average. Compare two cases. One die (k = 1): the result can be anything from 1 to 6, and every value is equally likely. The average of 25 dice (k = 25): the result is usually between 3 and 4, and almost always between 2.8 and 4.2. It practically never goes below 2.5 or above 4.5. That is what "less spread" means: the values stay closer to the middle.
Why does the average stay in the middle? For the average of 25 dice to be 1, every one of the 25 dice must show 1 at the same time. The chance of that is 1/6 × 1/6 × … × 1/6, twenty-five times, which is about 1 in 3 × 1019, practically zero. For the average to be 6, all 25 dice must show 6, equally impossible. Even an average of 2 needs almost every die to show a low number, which is very unlikely. But there are very many ways to get an average near 3.5: some dice show high numbers, some show low numbers, and the high and low ones cancel each other out. So the average of many dice is pulled toward the middle, and its spread is small.
How much less spread? The SD of the averages is the SD of one die divided by √k. The SD of one ordinary die is 1.71, calculated from its six faces in the expected value section above. So with 25 dice (try it below) the SD of the averages is one fifth of 1.71, and with 100 dice it would be one tenth. This rule is the most useful fact in this course.
One number, two names. The SD of the averages has a special name: the standard error, SE. An SE is simply the SD of a mean. Here we can find it in two ways: measure it from the histogram of many averages, or calculate it from the formula SE = SD of one value ÷ √k. The two green boxes below show both. They give the same number.
How to read the two green boxes. The left one is measured: the SD of all the averages in the histogram, which is the SE found by experiment. The right one is the SE predicted by the formula: 1.71 divided by √k. Try each k with Fast-forward. The two numbers agree to about two decimal places every time.
The rule: when you average k independent values, the SD of the average (its SE) is the SD of one value divided by √k. "Independent" means one roll does not affect another.
95 % range: 95 out of 100 averages fall inside it. For k = 1 it is almost the whole range 1 to 6. For k = 25 it is roughly 2.8 to 4.2.
Let us assume that all 250 of us measured the same tree, tree #406: one height reading and one DBH reading per student. The 250 numbers in the bag below are made up for this page. They are not real measurements. They are the kind of numbers a class would get, and they let us practise every idea from the dice on a tree.
Nothing new is introduced in this panel. The tree is the mystery die of Panel B. Its true height is the hidden thing. Each student's reading is one roll. The table shows the analogy line by line.
| With a die | With the tree |
|---|---|
| The hidden thing: the faces and their probabilities | The hidden thing: the true height (or DBH) of the tree |
| One roll | One student's reading |
| Many rolls | 250 readings |
| Expected value, calculated from the faces (3.5 for an ordinary die) | The true height. It cannot be calculated, because we do not know the "faces". It can only be estimated. |
| The running mean settles at the expected value | The mean of the readings settles at the true height, if the measuring is fair (no instrument that reads too high or too low for everyone) |
| SD of one roll, 1.71 for an ordinary die, calculated from the faces | SD of the readings, calculated from the data: how far one reading typically is from the mean |
| SE of the average of k dice = 1.71 ÷ √k (Panel C) | SE of the mean of n readings = SD ÷ √n |
| 95 % range of the averages (Panel C) | 95 % confidence interval for the true height: mean ± 2 × SE |
Press Play to take the 250 readings out of the bag in the order the students handed them in. Watch the running mean and the 95 % confidence interval settle, exactly as the running mean of the die settled in Panel A.
A chip turns orange once its reading has been taken out and counted. The chip being read now is highlighted.
The interval is for the true height, not for one reading. It answers "where is the hidden value?", not "what will the next student write down?".
"95 %" describes the method, not this one interval. Imagine many classes, each of 250 students, each measuring this tree and each calculating its own interval. The intervals would differ a little, because the readings differ. About 95 out of 100 of those intervals would contain the true height; about 5 would miss it. We have one class and one interval. We cannot know whether ours is one of the 95 or one of the 5. That is why we say "95 % confident", not "certain".
Height is measured from a distance with an angle reader (a clinometer) and some trigonometry. DBH is measured with a tape touching the trunk. The CV lets us compare the two even though one is in metres and the other in centimetres.
One thing the dice cannot show, and 250 readings cannot fix: if every student's instrument reads 0.5 m too high, the mean is also 0.5 m too high. All the readings would be "rolls" of a die whose faces are shifted. More readings make the interval narrower, but around the wrong value. Only a better instrument, or a check against a tree of known height, can find this kind of error. It is called a systematic error, and we will come back to it when we look at biomass equations.
In Panel D we used a formula: SE = SD ÷ √n, then interval = mean ± 2 × SE. In Panel C we found the SE a different way: by experiment. We rolled the dice thousands of times, made a histogram of the averages, and measured their spread. Can we do the same for the tree? Not directly: we cannot send 2,000 new classes to measure tree #406. We have one bag of 250 readings and nothing else.
The trick. Treat the bag of 250 chips as if it were the tree itself. Drawing one chip from the bag plays the role of measuring the tree once. This is the same idea as in Panel B, where many rolls were a good picture of the die: here, 250 readings are a good picture of what a reading of this tree can look like. So we can "measure the tree" as often as we like, by drawing chips from the bag:
1. The formula only works for the mean. Press "SD" above. We often want to know how sure we are about the SD, or the CV, or the median, or a biomass calculated from the readings. There is no simple "÷ √n" rule for those. The bootstrap uses the same four steps for any quantity: calculate it for each pretend class and look at the spread.
2. The formula makes an assumption. "Mean ± 2 × SE" assumes that the means of many classes would pile up in a bell shape. Panel C showed that averages do this when the readings are spread evenly around the middle. But if the readings are not symmetric, with a long tail on one side, for example because a few students made very large mistakes, the bell shape is only approximate. The bootstrap makes no such assumption. It uses the actual shape of the readings in the bag.
3. It is one recipe for everything. Draw with replacement, calculate, repeat, sort, cut 2.5 % at each end. The computer does the repeating. We will use exactly this recipe again for the biomass of a whole plot.
What if we did not put the chips back? Then every pretend class would contain exactly the same 250 chips, only in a different order. Every pretend mean would be identical. The histogram would be one single bar, and it would tell us nothing. Putting the chip back is what lets the bag behave like the tree: it can give the same reading twice.
In our real project nobody measures one tree 250 times. Each tree is measured once: one height, one DBH. So for a single tree there is no bag of readings to roll. The bootstrap cannot start. A bag with one chip always gives the same answer.
The idea. If we cannot fill the bag with readings, we fill it with an assumption. From Panel D we know how far a height reading can be from the truth: the CV of height readings was about 10 %. (A flat die with limits ± 10 % is not quite the same as a bell die with CV 10 %; see the note at the end of this panel.) So we say: "the true height is somewhere between the reading − 10 % and the reading + 10 %, and we have no reason to prefer any value in that range." That is a die with very many faces, all equally likely, spread evenly from 5.4 m to 6.6 m for a reading of 6.0 m. This shape is called the uniform distribution: drawn as a picture, it has a flat top. The shape is our choice, and it is part of the assumption. The die could instead have the bell shape that the readings in Panel D made, with most values near the reading and fewer far away; this shape is called the normal distribution, and pages 3 and 4 use it. Whatever the shape, the method is the same. Then we roll the die, exactly as before. This is the Monte Carlo method, named after the casino town, because it works by rolling dice.
| Bootstrap (Panel E) | Monte Carlo (this panel) | |
|---|---|---|
| What is in the bag | Real readings | Values made from an assumption: a shape and an error size |
| Where the spread comes from | The data itself | The assumed error, for example ± 10 % |
| Needs | Many readings of the same thing | One reading plus a known or assumed error |
| The rolling and counting | Identical: draw, calculate, repeat thousands of times, sort, cut 2.5 % at each end | |
For one tree alone, Monte Carlo on the height gives nothing new: the mean is the reading and the range is ± 10 %. It becomes useful when the reading goes into a formula. Take the simplest tree formula: treat the trunk as a cylinder. Its volume is V = π × r² × H, where r is the radius, half the DBH.
One tree was measured once: H = 6.0 m and DBH = 8.0 cm, so r = 4.0 cm = 0.04 m. From the readings alone: V = 3.1416 × 0.04 × 0.04 × 6.0 = 0.0302 m³ = 30.2 litres (1 m³ = 1,000 litres). But how sure are we of 30.2? Height can be off by ± 10 %, DBH by ± 3 % (about the size of the DBH spread seen in Panel D). Two errors go into one formula, and the DBH is squared. Working this out with algebra is messy. Rolling is easy.
For a hand calculation, make the two dice with coarse faces. Height die: 13 faces, 5.4, 5.5, 5.6, …, 6.6 m (6.0 ± 10 %). DBH die: 5 faces, 7.8, 7.9, 8.0, 8.1, 8.2 cm (8.0 ± 3 %, rounded to 0.1 cm). Roll both, calculate V, repeat:
| Roll | Height die, m | DBH die, cm | r = DBH ÷ 2, m | V = 3.1416 × r² × H | V, litres |
|---|---|---|---|---|---|
| 1 | 5.7 | 8.1 | 0.0405 | 3.1416 × 0.00164 × 5.7 = 0.0294 m³ | 29.4 |
| 2 | 6.5 | 7.9 | 0.0395 | 3.1416 × 0.00156 × 6.5 = 0.0319 m³ | 31.9 |
| 3 | 5.5 | 8.2 | 0.0410 | 3.1416 × 0.00168 × 5.5 = 0.0290 m³ | 29.0 |
| 4 | 6.2 | 8.0 | 0.0400 | 3.1416 × 0.00160 × 6.2 = 0.0312 m³ | 31.2 |
| 5 | 5.9 | 7.8 | 0.0390 | 3.1416 × 0.00152 × 5.9 = 0.0282 m³ | 28.2 |
After 5 rolls the mean is (29.4 + 31.9 + 29.0 + 31.2 + 28.2) ÷ 5 = 29.9 litres. Five rolls are too few, just as 5 rolls of a die were too few in Panel A. Let the computer roll 2,000 times below. The pile of 2,000 volumes gives the mean, the SD and the 95 % interval of V. You will find: mean about 30.2 litres, SD about 2.0 litres, 95 % interval about 26.5 to 34.1 litres. So we report: V = 30.2 litres, 95 % interval 26.5 to 34.1 litres, or "30 litres, plus or minus about 13 %".
| Roll | H, m | DBH, cm | V, litres |
|---|
The answer is only as good as the assumption. The bootstrap gets its spread from real readings. Monte Carlo gets it from the ± 10 % we put in. Where does the 10 % come from? From an exercise like Panel D, where many people measure one tree; from the instrument's manual; or from a small test. If we assume ± 5 % instead, the interval for V becomes narrower, but not because we measured better. Try it above.
Uniform is the honest choice when you only know the limits. In Panel D the readings piled up in a bell shape, with most readings near the middle. A uniform die says "anywhere between the limits, equally likely". It is the simplest shape, and it is a safe choice when all you know is "about ± 10 %". When more is known, other shapes can be used, for example the bell shape. One care is needed: a flat die with limits ± 10 % is not the same as a bell die with CV 10 %. The flat die from 5.4 m to 6.6 m has an SD of only 0.35 m, a CV of about 6 %, because its values never go beyond the limits. A bell die with CV 10 % has SD = 0.6 m, and about 1 draw in 20 lands more than 1.2 m from the reading. Panel D showed a bell shape with CV about 10 %, so the later pages use the bell die with CV 10 %. This page uses the flat die only because it is easy to roll by hand.
The same dice, a bigger formula. Here the formula was a cylinder with two inputs. For the project, the formula will be the real biomass equation, which has five uncertain inputs instead of two. The method does not change: build a die for each uncertain input, roll them all, calculate, repeat, sort, cut.