13Symmetry, Order Statistics, the LLN & the CLT

Chapter 12 grew a whole family of distributions out of a single coin flip, and every branch arrived with a formula you could sit down and evaluate. That is a real skill, and it has a ceiling. Two kinds of question walk straight past it. The first is the tangled one, where the pieces of a problem lean on each other so hard that no clean pmf is available. The second is worse, because nobody hands you the shape at all and you are still expected to say how wrong your answer is. This chapter builds the two engines that handle both. The first engine is symmetry, and it says that interchangeable pieces sharing a known total must each get the same share, with no independence assumed anywhere. The second engine is averaging, and it says that independent pieces, once averaged, are forced onto the mean at a rate you can compute, with a leftover wobble that can only be one shape. Between them they answer the questions where computing is either impossible or beside the point. And they leave you holding one small expression, σ/√n, that quietly prices every research decision a quant will ever make.

Look at what this page is standing on. Chapter 11 handed us three tools that carry most of the weight here: E[X] as a probability-weighted average, the identity Var = E[X²] − E[X]², and linearity of expectation, which splits a sum without ever asking how the pieces are related. Chapter 12 gave us the Normal, the Uniform, and one unpaid debt. Chapter 9 gave us the axioms, and the third one, that probabilities of disjoint events add, does more work on this page than anywhere else in the course.

That is the seam. Chapter 12 kept saying that sums of many independent things go Normal, and you watched it happen without ever being told why it must. This chapter pays that debt, states the conditions honestly, and then shows you the case where the machine breaks.

01The answer you refuse to compute

I want to start with a question, and I want you to commit to a number before reading on. No calculating and no hedging, just a number you would defend.

A plane has 100 seats and 100 passengers, and each passenger holds a ticket for one specific seat. The first passenger has lost his boarding pass, so he picks a seat completely at random and sits down. Every passenger after him takes his own seat if it is free. If it is taken, he picks uniformly at random from whatever seats remain. What is the probability that the 100th passenger ends up in his own seat?

Most people say 1 in 100, and a few insist it must obviously depend on how big the plane is. Both answers feel reasonable and both are wrong.

The answer is exactly 1/2, and it does not depend on the 100 at all. Change the plane to 3 seats or to 3 million and the answer is still one half, and I am not going to prove that yet.

Here is what I want you to sit with instead. You have spent four chapters learning one model of how a probability gets computed. Build the sample space, count the favourable outcomes, divide. Or write the pmf and grind out the sum. Both are honest, and both are the wrong tool here, because the branching in this problem is genuinely horrible.

There is another way to pin a number, and it is not approximation. You find a constraint on the whole thing that forces the answer, and then you never look at the parts at all. The result is exact — there is no arithmetic anywhere in it, and that combination is what makes people distrust it the first time they meet it.

This chapter builds two engines of that kind. The first is symmetry, which asks what interchangeable pieces are forced to share. The second is averaging, which asks what many independent pieces are forced to become. The plane is now a debt we owe you, and it gets paid before this chapter is half done.

02★ The licence: exchangeability

Every source I have read writes the phrase "by symmetry" and then moves on. That phrase is doing an enormous amount of work, and what it hides is the condition actually being used. Let's dig it out.

Take the cleanest possible setup: a deck of two cards, one ace and one blank, shuffled so that either order is equally likely. Ask for the chance that the first card is the ace and the answer is plainly 0.5000. Ask for the chance that the second card is the ace and it is 0.5000 as well, because relabelling which card you look at first cannot change anything.

Now push on it with Chapter 10's conditioning. Given that the first card is the ace, what is the chance the second card is the ace?

It is 0.0000. There is only one ace in the deck, so knowing where that ace landed tells you everything about the other card. Two cards, identical marginal probabilities, and about as much dependence as it is possible to have.

So here is the sentence that has to be said out loud, because the whole first half of this chapter rests on it. Exchangeable means the pieces are interchangeable. It does not mean they ignore each other.

The formal statement is short: a collection of random variables is exchangeable when relabelling them leaves the joint law unchanged, so you can swap any two names and every joint probability comes out identical. The joint law is the full description of how the pieces move together, not just how each one behaves alone.

That distinction is the tear worth guarding for the rest of the course. The marginal is what one piece looks like on its own, and the joint is how they move together. Equal marginals can sit on top of completely different joints, which is exactly what the two tables in that figure show, and nothing in Chapters 9 through 12 ever forced you to hold them apart. Almost every distribution there was assembled out of independent atoms.

One more thing to keep straight, since the words all sound alike. Independent is a strong condition, and it implies exchangeable whenever the pieces also share a distribution. Exchangeable is far weaker and far more common, which is precisely why it is worth having as a separate idea. Cards from a shuffled deck, balls drawn without replacement, and the pieces of a broken stick are all exchangeable, and not one of them is independent.

03★ Equal shares from a fixed total

Now the payoff for that care: there is one lemma underneath every symmetry argument in this chapter, and it is three lines long. Almost nobody writes those lines down, which is why "by symmetry, each is T/n" feels like a convention you are expected to nod at.

You have n pieces X₁, …, Xₙ which are exchangeable, and they are known to add up to a fixed total T. Fixed means a genuine constant here, the same number in every outcome, not a value that merely happens to be the average.

Line one: since T = X₁ + … + Xₙ is a constant, its expectation is itself, so E[T] = T. Nothing has happened yet.

Line two, and this is the line doing the heavy lifting. Linearity of expectation splits that sum into E[X₁] + … + E[Xₙ] = T. Stop and notice what was not required. Chapter 11 proved linearity with no independence assumption anywhere, so it does not care that these pieces may be violently dependent, and that is the whole power of the lemma.

Line three: exchangeability says all n of those expectations are the same number, and n equal numbers summing to T are each T/n. That is arithmetic, and the lemma is finished.

Read the receipt carefully, because knowing which line paid for what is the difference between owning a method and repeating a trick. Linearity did the splitting and is blind to dependence, while exchangeability did the equalising. Neither one needed a distribution and neither one needed independence.

The condition that can break is the fixed total. If T is itself random, line one collapses and you get E[Xᵢ] = E[T]/n instead, which is a weaker statement about an average rather than about a constant you actually know. That is the failure mode to watch for when you aim this at a new problem.

The same three lines work anywhere the two hypotheses hold, and that is worth seeing outside probability once. Identical resistors across one supply split a fixed current equally. Symmetric desks running the same strategy split a fixed daily volume equally. The argument never mentions what the pieces are made of.

04Symmetry on events: the plane and the circle

So far we ran symmetry on expectations, and the reason it worked is that expectations add. Chapter 9 handed us another quantity that adds, which is the probability of disjoint events. Run the identical argument with events in place of numbers. The move itself does not change at all.

The event version. If a relabelling maps the whole setup onto itself, any two events it swaps must have equal probability. And if those equal events are disjoint and cover everything between them, each one's probability falls out by division.

Now the plane, and we are going to attack it the cowardly way. Shrink it until the answer is undeniable, then hunt for the thing that did not depend on the size.

With 2 seats it is a coin flip, so the answer is 1/2. With 3 seats you can enumerate the whole thing by hand in about a minute. Passenger 1 takes his own seat with chance 1/3 and everything then runs clean. He takes seat 3 with chance 1/3 and passenger 3 is sunk. He takes seat 2 with chance 1/3, and passenger 2 then picks between seats 1 and 3 with chance 1/2 each. Adding the two winning roads gives 1/3 + 1/6 = 0.5000.

At that point you know it is not a coincidence, and you can stop hunting for the 100. So what is the invariant?

Watch any displaced passenger make a choice. From his point of view, seat 1 and the last seat are perfectly interchangeable. Both are simply free seats he has no claim on, and nothing in the rules ever treats one of them differently from the other. The whole process ends the instant one of those two is taken, because taking seat 1 unblocks everybody afterwards and taking the last seat ends the story. Two interchangeable seats, and exactly one of them gets hit first. That is a coin flip, and it is why the 100 never appears in the answer.

Now the second classic, which leans on the disjointness half of the move. Drop 3 points independently and uniformly on a circle. What is the chance that all three lie within some single semicircle?

The trouble with that sentence is the word "some". A floating semicircle is genuinely ambiguous, and people who try to count it directly end up double-counting configurations. The fix is to anchor it to one of the points.

For each point, define the event that the other two both lie in the half-circle running clockwise from it. Each of those events has probability (1/2)² = 0.2500, since the two other points are independent and each has an even chance of landing in that half.

Here is the step no textbook writes down, and it is the one that turns this into a theorem instead of a card trick. Those three anchored events are disjoint. Only one point can be the clockwise-most of a clustered trio, so two of the events cannot both hold. And if the three points do fit inside a semicircle, one of them must hold. Axiom 3 then adds them: 3 × 0.25 = 0.7500.

One honesty note, in plain words. Configurations where two points land at exactly the same spot would break the "only one can be clockwise-most" claim. For continuous points those have probability zero, so they cost us nothing. The general answer, by the identical argument with n anchors, is n/2ⁿ⁻¹.

Flip the question over and you get the version a trader actually cares about. The chance that the three points surround the centre is 1 − 0.75 = 0.2500, which is the same question as whether three independent bets leave you exposed in every direction at once.

05Where symmetry pays: gaps and order statistics

Now we point the lemma at something that looks like it needs real machinery. Shuffle a standard 52-card deck, and ask how far down you have to go on average before you hit the first ace.

The instinct is to write the distribution of that position and sum it out. You can, and it is unpleasant. Instead, look at the deck the other way around.

There are 4 aces and 48 other cards, so forget the aces for a second and lay the 48 blanks out in a row. The four aces then sit in the spaces between those blanks. Count the spaces, not the cards.

Count them carefully, because this is where the classic off-by-one lives. Four aces cut the row of blanks into five gaps, not four. There is a gap before the first ace and a gap after the last one, and both of those end gaps are frequently empty. Empty is exactly why the eye skips them. In general, k markers make k+1 gaps.

Those five gaps are exchangeable: relabel which ace is which, or mirror the whole deck end to end, and the joint law of the gap sizes does not move. They are also obviously dependent, since the five of them must add to exactly 48. That dependence is the precise thing the last section built the lemma to survive.

So apply it unchanged — five exchangeable pieces sharing a fixed total of 48 give E[gap] = 48/5 = 9.6000 blanks each.

The first ace sits one card past the first gap, so its expected position is 9.6 + 1 = 10.6000. Write that in general and it is (N−k)/(k+1) + 1, which tidies to (N+1)/(k+1), and for the deck that is 53/5 = 10.6.

Now separate the two +1s in that formula, because memorised together they mean nothing. The +1 in the denominator is the number of gaps, and it came straight from the picture. The +1 in the numerator is the ace itself, sitting one card past its gap. Completely different origins.

Take a sample and sort it. The sorted values are new random variables called the order statistics, written X₍₁₎ ≤ X₍₂₎ ≤ … ≤ X₍ₙ₎, and the first ace's position was one of them. Those parentheses matter. X₍₃₎ is the third smallest value, while X₃ is the third value you happened to draw, so the subscript has quietly changed meaning from identity to rank.

And they really are random. Sorting feels like a tidying-up operation that removes randomness, which is a comfortable picture and a wrong one. Shuffle again and X₍₁₎ lands somewhere else.

# this really ran ↓
random.seed(20260813)
for _ in range(100_000):
  shuffle(deck)
  pos.append(first_ace(deck))
print(mean(pos))  # 10.6042

That figure is a genuinely executed run rather than a drawing. It shuffles 100,000 decks and reports the mean position of the first ace, and the empirical answer lands on 10.6. The formula stops being a claim and becomes a measurement you took.

The continuous version is the same picture with the dots replaced by a stick. Drop k points uniformly on the interval from 0 to 1. They cut it into k+1 pieces, the pieces are exchangeable, and they share a fixed total length of 1.

import random
k, N = 4, 100_000
s = [0.0]*k
for _ in range(N):
  u = sorted(random.random()
             for _ in range(k))
  for j in range(k): s[j] += u[j]

So each piece has expected length 1/(k+1), and the j-th smallest point sits at E[U₍ⱼ₎] = j/(k+1). With k = 4 that gives 0.2000, 0.4000, 0.6000 and 0.8000, which is evenly spaced. Your gut wanted that answer, and now it has a reason for it.

06Counting the stories: a bound for free

Before we leave the first engine, one more move in the same spirit. Constrain the whole thing again, but with a different conserved quantity. Not a total this time, but a count of outcomes.

The classic setup: you have 12 balls that look identical, and exactly one of them is a different weight. Heavier or lighter, and you do not know which. You have a balance scale and 3 weighings, and you must find the odd ball and say whether it is heavy or light.

Almost everybody starts hunting for a clever first weighing. That search has no stopping rule, so before searching, ask a completely different question. Could three weighings ever be enough?

Count both sides of the ledger. Each weighing has three possible results, which are left heavy, right heavy, and balanced, so three weighings produce at most 3³ = 27 distinguishable outcomes. Meanwhile the number of states you must tell apart is 12 balls times 2 possibilities each, which is 24. That is the whole ledger: 24 states against 27 answers.

Two weighings would give only 3² = 9 outcomes against those 24 states, so two are provably impossible. No cleverness rescues them, because a procedure that cannot produce 24 different answers cannot always name the right one.

Now use how narrow that margin is, because there are only 3 spare outcomes. Almost no weighing can afford to waste one, and that alone forces the famous opening. Weigh k against k. If it tips, the live states drop to 2k. If it balances, they drop to 2(12 − 2k). Each branch has two weighings left, so each branch must fit inside 9.

Those two requirements are 2k ≤ 9 and 24 − 4k ≤ 9, giving k ≤ 4 and k ≥ 3.75. So k = 4 exactly. The four-against-four opening is not a flash of insight at all, it is the only number the budget permits.

And now the honest half, which matters far more than the puzzle. A counting bound proves impossibility. It never proves possibility. Having found that 24 fits inside 27, you have built nothing, and you still owe an actual strategy.

The case that settles it is 13 balls. The budget says 26 ≤ 27, so the count is perfectly happy. No strategy exists that both finds the odd ball and names it heavy or light. Room in the budget is a permit, not a construction, and reading it as one is a habit that will misfire for years.

07★★ The turn: where the √n is born

Everything so far deliberately refused to assume independence, and that refusal bought us a lot. It also has a hard ceiling. Symmetry can pin a mean and it can pin a probability, and it can never say one word about spread.

Independence is the one assumption we have not spent yet, so let's spend it and see what it buys. The first thing it touches is precisely the quantity symmetry could not reach, because Chapter 12 showed that variances add for independent pieces.

Here is where nearly everyone slips, so predict before you read on. Suppose X₁, …, Xₙ are independent with variance σ² each, and you form the average X̄ = (X₁ + … + Xₙ)/n. What happens to the variance?

The gut says divide by n, because that is what dividing does. It divides by n², and that single squared factor is the entire √n. The reason the gut cannot correct itself is that we quietly picture variance as living in the same units as the variable. It does not, and its own definition E[(X−μ)²] says so.

Measure a set of heights in centimetres and the standard deviation comes out at 7 cm, which makes the variance 49 cm². Now measure exactly the same people in metres. Every number shrinks by a factor of 100, so the deviations shrink by 100 while their squares shrink by 10,000.

The variance is therefore 0.0049 m² and the standard deviation is 0.07 m. Nothing about the people changed. The square is bookkeeping, not convention.

One line of algebra confirms it using Chapter 11's identity: Var(aX) = E[a²X²] − (aE[X])² = a²Var(X), and taking the square root gives SD(aX) = |a|·SD(X). That is the rung the rest of this chapter stands on, and it appears in neither Chapter 11 nor Chapter 12.

It also explains a habit you have probably noticed. Finance quotes standard deviation and essentially never quotes variance. A sigma is in percent, which you can feel. A variance is in percent squared, which nobody can.

Now put the two rules together. This is the centre of the chapter.

You average 100 noisy readings instead of taking one. How many times more accurate is your answer? Almost everyone writes 100. It is 10.

The reason is a picture, and it is one you already own from Chapter 6. Draw the first measurement's error as an arrow, then attach the second error to it. If the two errors were locked together, so that knowing one told you the other, they would point the same way and their lengths would simply add. That is the gut's picture, and it is what dependence looks like.

Independence means the second error carries no information whatsoever about the first. Geometrically that is a right angle. So the two errors do not queue up end to end, they brace against each other, and the combined length is the hypotenuse √(σ² + σ²) = 1.4142σ rather than 2σ.

Stack n of them, each coming in perpendicular to all the others, and the total error length is σ√n. Then divide by n to turn that sum into an average, and the scaling rule from the units panel carries it to σ/√n.

The algebra is now confirmation rather than argument: variance adds, so Var(X₁ + … + Xₙ) = nσ². Scaling by 1/n multiplies variance by 1/n², so Var(X̄) = nσ²/n² = σ²/n, and therefore SD(X̄) = σ/√n.

The mean, meanwhile, does not move: E[X̄] = μ. That is the equal-shares lemma read in the independent case, since n interchangeable pieces are sharing a total of nμ.

σ/√n is the standard error of the mean, and it is how wrong your average is likely to be. From Chapter 16 onward we use it constantly.

One word about noise before we move on, because it is two different things wearing one coat.

The spread of the sum grows like σ√n. The spread of the average shrinks like σ/√n. Both are true, both come from the same √n, and people who never see them on one panel spend years quietly unsure whether more data makes things noisier or cleaner.

Two consequences are worth carrying out of here. Doubling your data buys a factor of 1.4142, not 2. And to halve an error bar you need four times the data, which is the arithmetic that prices every research decision you will ever make.

A footnote on what independence was really needed for. In the variance addition, the cross term died because E[XY] = E[X]E[Y], and that condition is called uncorrelated, which is weaker than independent. The cross term itself has a name, covariance, and Chapter 18 does the work.

08The centre locks: the law of large numbers

We now know the average's spread heads to zero. It is tempting to declare victory and say the average converges to μ, but there is a join here that most treatments slide across without comment.

"The standard deviation goes to zero" and "the probability of being far from μ goes to zero" are different statements. The second one is what you actually want, and crossing from the first to the second takes one honest tool.

Start with something that looks almost too simple to be useful: let X be a random variable that is never negative, and pick a level a. The definition of expectation says E[X] = Σ x·p(x), and every term in that sum is non-negative.

Throw away all the terms with x < a and the sum can only shrink, so E[X] ≥ Σ x·p(x) over the values at least a. Every surviving x is at least a, so replace each one by a and it shrinks again to E[X] ≥ a·P(X ≥ a). Rearranged, that is Markov's inequality, P(X ≥ a) ≤ E[X]/a.

In words, mass far out to the right has to be paid for out of the mean. Now aim it at the squared deviation (X̄ − μ)², which is never negative and has expectation σ²/n.

Being at least ε away from μ is the same event as that squared deviation being at least ε², so Markov gives P(|X̄ − μ| ≥ ε) ≤ σ²/(nε²). That is Chebyshev's inequality, and it is the entire law of large numbers.

Put numbers on it with σ = 1 and ε = 0.1. At n = 100 the bound is 1.0000, which says nothing at all. At n = 400 it is 0.2500, at n = 2500 it is 0.0400, and at n = 10,000 it is 0.0100.

Notice that the bound is loose, meaning far more pessimistic than the truth, and notice that it does not matter. It heads to zero for any ε you care to name, and that statement is the weak law of large numbers.

Now for the thing this law is most often accused of saying. Chapter 10 named the gambler's fallacy. This is where we execute it, because we finally have the machinery to show what happens instead.

The phrase "the law of averages" makes people picture a correcting force. Ten extra heads must somehow be repaid by ten extra tails later. Nothing pays them back.

Flip a fair coin and watch two quantities at once. The absolute lead, meaning heads minus tails, has standard deviation √n. It is 10 at a hundred flips and 100 at ten thousand, so it grows.

The proportion gap, meaning the head fraction minus one half, has standard deviation 0.5/√n. It is 0.0500 at a hundred flips and 0.0050 at ten thousand, so it shrinks.

Same √n, two curves, opposite directions. The excess heads never come home. They simply get divided by a bigger and bigger n, which is why deviations are diluted, never repaid.

09★ The wobble's shape: standardizing and the CLT

The law of large numbers has done something slightly inconvenient. It squashed the average onto a single point, so if you plot the distribution of X̄ for growing n on a fixed axis, you get a spike and nothing left to look at.

A spike is not the end of the story. Structure is hiding inside it. To see structure in something that is shrinking you have to zoom in, and the only question is by how much.

Try three settings and let the picture decide. With no zoom at all you get a dot. Multiply by n and the picture explodes off the screen, because you magnified faster than the thing was shrinking. Multiply by √n and the image settles and stops changing as n grows.

That is the whole justification, and it is worth stating plainly. To photograph something that is shrinking, you zoom by exactly its shrink rate, no more and no less. We already know that rate. It is σ/√n.

So define the standardized average by dividing the deviation by precisely that: Zₙ = (X̄ − μ)/(σ/√n). Clear the fraction and it reads √n(X̄ − μ)/σ, which is the form textbooks print. It is the same object in less frightening clothing, and the √n on top is not fighting the one underneath. It is the one underneath, moved.

The zoom finally makes the real question askable. What does the picture actually look like?

Before the answer, kill the corruption. The most common false sentence in all of statistics is that the central limit theorem says data becomes Normal if you collect enough of it. It says nothing of the kind.

The theorem is about the average, or equivalently about the sum. It is never about the data. The left panel of that figure is a histogram of raw draws from a deliberately ugly parent, and sliding n does not move it one pixel. The right panel is a histogram of averages, and it is the only thing that changes.

With that separation held, here is the statement. Take any parent distribution with a finite mean μ and a finite variance σ², draw n independent values, and standardize the average. As n grows, the distribution of Zₙ approaches the standard Normal, whatever the parent happened to be.

Read what that costs the parent. Its shape is forgotten entirely, and only μ and σ survive the trip. That is why two completely different businesses can end up with identically shaped uncertainty about their averages.

While we are here, the n > 30 rule you may have met is folklore and not a theorem. There is no proof anywhere behind it. A symmetric parent is close to Normal by n = 5, and a badly skewed one is still visibly skewed at n = 100. The skew of the average falls like skew/√n, so for an exponential parent at n = 30 it is still 0.3651.

The last question is the one usually answered with a shrug. Why the Normal, of all possible shapes?

There are two honest parts to the answer. The first is that adding an independent piece convolves the shape against itself, and convolution is a smoothing operation. Every bump gets averaged against a shifted copy of the whole shape, and it can never sharpen back. Start with a coin, which is two bare spikes, and three convolutions later a bell is visibly assembling out of nothing but spikes.

The second part pins which smooth shape it has to be. Whatever the standardized average settles into must reproduce itself. Add two independent copies, rescale by √2, and you have to land back on the shape you started with. The Normal is that shape.

Now the honesty note. That argument identifies the destination and assumes a destination exists. Proving one exists needs machinery we have not built, namely characteristic functions, the Fourier transforms of the distribution. That debt is named, not hidden.

10The fine print

No new concept arrives in this section. What arrives is the difference between somebody who has heard of the CLT and somebody who can be trusted with money.

Every statement in the last section arrived with hypotheses attached: independent pieces, a finite variance, and a limit in n. So walk back through those three and ask what happens when each one fails.

Start with the limit, because it fails in the most dangerous place. The CLT converges in the middle first and in the tails last, and that is worth seeing with real numbers.

Take the average of 30 independent Exponential(1) draws, where the exact answer is computable. At one sigma out, the true tail probability is 0.157465 against the Normal's 0.158655, so the error is invisible. At three sigma it is 0.004164 against 0.001350, already off by a factor of 3.08. At four sigma it is 0.000386 against 0.000032, wrong by a factor of 12.20.

Read that ledger, then read a risk report. A 99% VaR is a question about the far tail, and the far tail is the last thing to converge. So the one place you most want the CLT is the one place it helps you least.

Now kill the finite variance. A Cauchy distribution looks like a perfectly ordinary bell-ish curve, and its tails are heavy enough that the variance does not exist. Take a running average of Cauchy draws and it never settles. It jumps forever, no matter how much data you feed it, because the hypothesis that failed is one the whole derivation quietly needed.

Then kill independence. Market returns cluster in a crisis: a bad day makes another bad day more likely. So the pieces stop being independent exactly when you need the machinery most. Your effective n is far smaller than the n you counted, and your error bars are too narrow.

That crack is not a footnote, it is a door. Chapter 34 walks through it with fat tails and volatility clustering, and Chapter 31 builds risk measures that do not lean on a Normal tail.

Now cash the chapter out in plain numbers. That one expression σ/√n is the standard error of Chapter 16, the width of every confidence interval in Chapter 17, and the √252 = 15.8745 that turns a daily volatility into an annual one.

It is also the reason a Sharpe ratio needs years of data before anybody should believe it, and the reason doubling your sample buys you 1.4142 rather than 2.

Carry three things out of here. First, "by symmetry" is a method with two named parts: linearity splits a fixed total without asking about dependence, and exchangeability makes the shares equal. Second, an average is wrong by about σ/√n, and that is a hypotenuse rather than a formula to recite.

Third, and this is the sentence the whole of statistics rests on. Averaging does not remove uncertainty. It shrinks it, at a known and provable and disappointingly slow rate.

Chapter 14 changes the question. Everything on this page described a static pile of draws, and the moment you ask how long until the first two heads in a row, or who goes broke first, a static distribution has nothing to say. The answer is to write the expectation as one step plus the same expectation again, and then solve.