15Paradoxes, Utility & Kelly

Chapter 14 handed us a machine that never refuses. Give it states and arrows and it returns a number, every time, with no complaint. Its very last result was a fair walk that hits zero with probability 1 and takes, on average, infinitely long to get there. That is a number nobody can live. So here is the question this chapter turns on, and it is the one most courses skip on the way to the formula: when an expected value comes back, what exactly have you been handed? The answer is that E[X] is never a fact about the world. It is the output of a model somebody chose, with three slots filled in — which outcomes can happen, with what chances, and what each outcome is worth to you. Every paradox in this chapter is that model telling you which slot broke. We will work through six of them, and the last one costs real money, because dollars add but wealth multiplies. By the end you will know why a bet that grows your money 5% per flip on average leaves you holding 51 cents. You will know why ln 0 = −∞ is the whole of risk management in three symbols, and why a 60/40 coin is a 20% bet and never an all-in one.

Look at what this page stands on. Chapter 9 gave us the axioms, and the one that does the most work here is the plainest of them: the total probability is exactly 1. From Chapter 10 came conditioning, and the fact that looking at something moves the odds. Chapter 11 gave us E[X], linearity, and Jensen's inequality. Chapter 12 handed over the Geometric waiting time and the Beta prior. Chapter 13 gave us the law of large numbers, and Chapter 14 the absorbing state. Then there is Chapter 5, and the identity that a log turns a product into a sum. Keep that one within reach. It is the hinge this whole chapter swings on.

That is the seam. Chapter 14 taught us to decompose a future we could not enumerate. This chapter teaches us to distrust the number that decomposition returns, until we can name the model it came from.

01One coat, on average, forever

Start somewhere with no money at stake at all. A coat check, closing time, and the attendant has lost every ticket, so n coats go back to n people in a completely random order. How many people, on average, get their own coat back?

Commit to a guess before you read on, and commit to three of them. What is the average number of matches when there are 3 coats, when there are 10, and when there are a million? Almost everybody's guesses shrink as the coats pile up, and that instinct is very reasonable. With more coats there are more ways to go wrong.

The instinct is wrong, and the tool that shows why is one you already own from Chapter 11 — the indicator trick. Instead of trying to count matches directly, define one little 0/1 variable per person: Xi = 1 if person i gets their own coat, and 0 otherwise. The total number of matches is just X1 + X2 + … + Xn.

Now take the expectation, and use the one property that does not care about dependence. Person i has exactly one coat that counts as theirs out of n coats, all equally likely, so E[Xi] = 1/n. There are n people, so linearity adds up n terms of size 1/n, and the n's cancel in front of you.

# coats.py · seed 20260813
P = range(n)
for p in permutations(P):
  k = sum(p[i]==i for i in P)
  c[k] += 1
N = factorial(n)
E = sum(k*v for k,v in c)/N

So E[matches] = 1, for every n, whether there are five coats or five million. Notice what the argument never used: it never once assumed the matches were independent, and they are emphatically not. If the first 99 of 100 people get their own coat, the hundredth is guaranteed to get one as well. Linearity survives dependence, which is exactly why it is the sharpest tool in Chapter 11's box.

Now for the part that matters more than the answer. You have a mean of 1, so picture the outcome: most people see something clustered around one match, maybe zero sometimes, maybe two. Stop imagining it and look at the real distribution.

With 4 coats, 9 of the 24 possible orders give nobody their own coat. That is 9/24 = 0.375, and the exact chance of a total shutout heads for 1/e ≈ 0.3679 as the coats pile up. An arrangement where nothing sits in its own place is called a derangement, and Chapter 9's inclusion–exclusion counts them for us. The fraction of orders that are derangements comes out as Σ(−1)k/k!, and if you look at that sum for a moment you will recognise it. It is Chapter 2's Maclaurin series for e−1, arriving here without being invited.

One honesty note, because the usual telling of this is slightly false. People say "the most likely outcome is that nobody matches," and that depends on the parity of n. At n = 4 the zero-match bar wins at 0.3750 against 0.3333. At n = 5 the one-match bar wins at 0.3750 against 0.3667. The two bars trade places by a margin of exactly 1/n!, and both walk onto 1/e. The robust statement is the useful one: about 73.6% of the time you get zero matches or one, and the mean is 1.

Hold both of those facts in your head at once, because the gap between them is the whole chapter. The mean is 1, while the single most likely count is 0 or 1, each around 36.8%. So the number 1 is not a typical outcome so much as a coincidence of the arithmetic, and nothing here is wrong. The machinery answered the question it was asked, and our gut asked a different question and heard the answer to that instead.

That is the first kind of failure, and the mildest one, and I will call it a paradox of the gut. The mathematics is fine and the intuition is not. The rest of this chapter is about the two failures that are worse, where the mathematics itself has a hole in it.

02What an expectation is actually made of

Here is the belief that makes paradoxes feel like tricks. Most readers arrive thinking an expected value is discovered, the way you discover the mass of a rock. Nothing in Chapters 9 through 14 forced anyone to think otherwise, because every exercise handed you the outcomes and the probabilities already filled in, and quietly assumed the payoff was measured in dollars.

So let me make the choosing visible on a sentence that could not be more harmless. Take the words "a coin comes up heads." Now build three completely legitimate models of that sentence, and watch them produce three different numbers, all of them correct.

The first model says the coin is fair, so P = 0.5. The second remembers Chapter 12: we saw 7 heads in 10 tosses, a uniform prior updates to a Beta(8,4) posterior, and its mean is 8/12 = 0.667. The third describes a physical spin on a real table and measures 0.510. Same six words, three models, and three different answers with no arithmetic error anywhere.

So an expectation is a functional of a model, not a property of the world. The model has exactly three slots, and every one of them is somebody's choice. Slot Ω is which outcomes get in at all. Slot P is what chance each one gets. The third slot is the value scale, what each outcome is actually worth to you. That is the slot everybody leaves on its default setting of "dollars" without ever noticing there was a setting.

Name the slots and you can name the failures. This taxonomy is the instrument the rest of the chapter runs on. A paradox of the gut is what we just met: the machinery is right and your intuition is wrong. A paradox of specification means a slot was left blank, so the question has no answer until you fill it. A paradox of criterion means the machinery is right, the model is complete, and you asked it the wrong question.

Three slots and three failure modes — that is the whole toolkit, and from here the chapter is just aiming it at one slot at a time.

03Slot Ω — "at random" names a machine, not a set

Before the famous one, here is the smallest example I know that forces the point. It has no geometry to hide behind and you can hold the whole thing in your head.

Pick a random square whose side is somewhere between 0 and 1. What is the chance its area is less than 1/4?

If "random square" means the side is uniform on (0,1), then area below 1/4 means side below 1/2, and the answer is 0.5. If it means the area is uniform on (0,1), the answer is 0.25, by definition. Both are honest readings of the same four words, and the phrase alone never chose between them.

That is the entire lesson, and everything else in this section is the same lesson wearing better clothes. Uniform in x is not uniform in x², so re-parametrise the space and the word "uniform" silently moves with you.

Now notice what just quietly retired. In Chapter 13, symmetry was a genuine method, and on six faces of a die or 52 cards in a deck it really did pin the probabilities down uniquely. That worked because the outcome set was countable and the symmetry group had one orbit. On a continuum there are many ways to stretch the space, and symmetry alone no longer picks one out.

So to Bertrand, who asks you to draw a circle, inscribe an equilateral triangle, and pick a chord at random. What is the chance the chord is longer than the triangle's side? Three procedures, all of them natural, give three different answers.

Pick a random point on a radius and take the chord perpendicular there, and you get 1/2. Pick two random points on the circumference and join them, and you get 1/3. Pick a random point anywhere in the disc and treat it as the chord's midpoint, and you get 1/4. None of these is the mistake, because the mistake was in the question, which never said what machine was drawing the chord.

Here is the way out, and it is refreshingly physical. Build a machine that actually throws straws across the circle and the question acquires an answer, because a real mechanism is a measure, and nothing else is.

Translate that into the language of the desk and it stops being a puzzle. A model is a stated data-generating process. If someone tells you their signal is "random" without telling you what generates it, they have not given you a model, and any expectation computed from it is a number without a source.

04Slot P — the two-envelope paradox

Two envelopes sit on the table, and one holds twice as much money as the other. You pick one, open it, and see an amount, which we will call A.

The other envelope holds either 2A or A/2, and it feels obvious that each is equally likely. So the value of swapping is ½(2A) + ½(A/2) = 1.25A, and you should trade. But the same argument applies to whoever holds the other envelope, so they should trade too. And once you have swapped, the argument tells you to swap back. Forever.

The algebra is genuinely fine, and you can feel that it is, which is exactly why this one is maddening rather than instructive. The break lives at one unmarked step. The symbol A is written as though it were an amount you have observed, and the ½/½ is then used as the probability given that observation. Those are two different probabilities, and telling them apart was Chapter 10's entire lesson.

Most treatments dissolve this by saying the words "improper prior" and moving on. That is a name for the gap rather than a bridge across it, so let us build the bridge with actual numbers. Here is the smallest legal prior that really exists: the pair is either ($10, $20) or ($20, $40), each with probability ½, and then one envelope of the pair is handed to you at random.

What happens next is not what most tellings claim. You open your envelope and see $20, and the ½/½ is genuinely correct here — you are equally likely to be the large half of the small pair or the small half of the large pair. So swapping really is worth ½($10) + ½($40) = $25. The naive rule survives at $20.

It dies at the ends of the ladder. Open the envelope and see $10 and you know exactly where you are, at the bottom of the small pair, so swapping gives you $20 with certainty. If you see $40, you are at the top, and swapping costs you $20 with certainty. The ½/½ was never true there, because the prior had run out.

Add up all four equally likely cells and the advantage is exactly zero: ¼(+10 − 10 + 20 − 20) = 0. It had to be. Swapping before you look is just choosing the other envelope, and the two were symmetric.

So the error was never the arithmetic at any single amount. The error was assuming the ½/½ holds at every amount at once. And now you can ask the sharp question: is there any prior at all that does hold it everywhere?

There is not, and the proof is short enough to carry. Line the possible pairs up as a ladder of scales, and let pk be the probability of the pair at rung k. Seeing amount A means you are either the top of rung k−1 or the bottom of rung k, so the ½/½ demands pk−1 = pk at every rung. That means infinitely many equal positive numbers, and Chapter 9's axiom says they must sum to 1. They cannot, so no such distribution exists.

That is a paradox of specification, and it has the pleasant property that it dissolves the instant you write a legal model down. It also leaves you with a portable test. If an argument recommends the same action from either side of a symmetric situation, it is not describing an edge at all — it is describing something that does not converge.

05The population you averaged over

Two treatments for kidney stones. Treatment A works on 81 of 87 men and 192 of 263 women. Treatment B works on 234 of 270 men and 55 of 80 women. Work the rates out before you continue, because the punchline needs you to have done it.

For men, A gets 93.1% against B's 86.7%, and for women A gets 73.0% against B's 68.8%. A wins both groups, so pool them: A is 273 of 350, which is 78.0%, and B is 289 of 350, which is 82.6%. B wins overall, nobody lied, and no number was invented.

The gut is running a rule it has never examined: better in every part means better in the whole. That rule is true for sums and false for ratios, and nobody has ever said which of the two a success rate is. When you pool a/b with c/d you get (a+c)/(b+d). That object has a name, the mediant, and it is a weighted average of the two rates whose weights are the group sizes.

Now make it geometric, because once you can see it the paradox stops being one. Draw each group's record as an arrow: trials to the right, successes up. The success rate is that arrow's slope, and pooling two groups is then literally vector addition, tip to tail, which is Chapter 6's very first move.

Drag the group sizes and watch the reversal appear and vanish. Treatment A is a short steep arrow plus a long shallow one, and treatment B is the opposite shape. The long arrow dominates the sum's slope, so whichever treatment was mostly tested on the harder group finishes with the worse pooled number. That is the entire mechanism, and you can now predict a reversal before you see the table.

Here is the honest and uncomfortable part. The numbers alone will never tell you which table to believe. If doctors gave the severe cases to A, then the split tables are the fair comparison and the pooled one is contaminated. If the split were caused by something downstream of the treatment itself, the pooled table might be the right one. That choice is causal, not statistical, and I am recording the debt here so Chapters 18 and 19 can pay it.

The version that will cost you money looks like this. A strategy can have a positive average return in every market regime and still lose money overall, if the size you traded differed by regime. Same mediant, same reversal, real dollars.

06Slot value — St. Petersburg, and the price of a ticket

Two slots audited, one to go. This one is forced on us: the next paradox has a completely specified Ω and a completely specified P, and it still returns nonsense.

The game: a pot starts at $1 and doubles every time a fair coin comes up tails. The first head ends it and you take the pot. So if the first head lands on flip n, you win 2n−1, which happens with probability 2−n. That is Chapter 12's Geometric waiting time with a payoff rule bolted on, and nothing about it is ambiguous.

Multiply the payoff by its probability and every single term is worth exactly 50 cents. There are infinitely many terms, so the fair price of a ticket is infinite, and you should hand over your house for one flip of this game. Nobody would pay $100.

Before we fix anything, notice something we have never had to notice. An expectation is a series, and Chapter 2 taught us that series come in two kinds. Every expectation in this course so far was assumed to exist. This one does not converge, so E[X] is not a large number here — it fails to be a number at all.

There are two separate repairs, and almost every telling blurs them together, which is why readers can never say which idea is doing the work. Keep them apart.

The first repair is truncation, and it needs no new theory at all. There is no counterparty with infinite money. If the house can pay at most $1,000,000, the game survives about twenty doublings, so the sum has about twenty terms of fifty cents plus a small capped tail.

Run the dial and the shock evaporates. A million-dollar bank makes the ticket worth $10.95, and a billion-dollar bank makes it worth $15.93. Making your counterparty a thousand times richer buys you about five extra dollars, because the value grows like the logarithm of the bank. The infinity was an artefact of a counterparty who does not exist.

Now hold the bank at infinity anyway, and ask what you would pay. You still say something small, so a second thing must also be doing work, and it lives in the third slot.

Bernoulli's move was to change what you are averaging. Do not average wealth. Average a function of wealth, u(w), called a utility. It is increasing because more is better, and concave because the ten-millionth dollar does less for you than the first. Concave is Chapter 4's word: u′ > 0 and u″ < 0.

Take u(w) = ln w and the St. Petersburg sum converges immediately. E[ln payoff] = ln 2 = 0.693, so the certainty equivalent is e0.693 = $2 exactly. Honesty note: that values the payoff on its own and ignores the wealth you already have. The full calculation uses ln(W + payoff − price), and the shape of the conclusion does not change.

Concavity also explains risk aversion without a single word about psychology, and this is Chapter 11's Jensen inequality arriving exactly on cue. For concave u, E[u(W)] ≤ u(E[W]). Say you own $100 and I offer a fair coin that pays $100 or nothing, so your expected wealth is $150. Expected log-wealth corresponds to a certainty equivalent of √(200 × 100) = $141.42. The $8.58 gap is the risk premium, and selling that gap is what an insurance company does for a living.

And now I have to raise an objection against my own argument, because if I do not, you will quietly file this whole topic under hand-waving. Why ln, and why not √w, which is also increasing and concave and also makes the sum converge? As it stands, log utility is a fit, not a derivation, and it was chosen because it works. That suspicion is completely correct, and it stays unresolved for exactly one more section.

07Wealth multiplies — the sentence this chapter exists for

Forget psychology entirely, because I want to ask a question with no taste in it at all: what does a repeated bet actually do to actual money?

Here is the bet, and it runs on a single fair coin. Heads, your wealth grows by 50%, and tails, it falls by 40%. So each flip multiplies what you have by 1.5 or by 0.6, and the expected multiplier is ½(1.5) + ½(0.6) = 1.05. Your money grows 5% per flip on average.

Write your answer down before you scroll. You start with $100 and you flip 100 times, so how much do you have at the end?

Nearly everyone computes 100 × 1.05100 and gets $13,150. That number is correct — it is the true expected value, and there is no error in it.

Now walk one honest path. Suppose you get 50 heads and 50 tails, which is the most likely split. Multiplication does not care about order, so pair each head with a tail: 1.5 × 0.6 = 0.9. Every pair of flips shrinks your pile by 10%. Fifty pairs gives 0.950 = 0.00515, so you are holding 51 cents.

That is not a bad path — it is the median path. Both numbers are exactly right, and the reason they can both be right is the thing this chapter was written to say.

E[·] averages across parallel worlds. It takes every one of the 2100 possible flip sequences, weights each by its probability, and adds. But you do not live in the ensemble. You live in one world, in sequence, and in sequence your wealth is a product of growth factors, not a sum of payoffs. Wn = W0 × G1 × G2 × … × Gn.

The mean is hauled upward by a handful of astronomically lucky paths. Only about 1% of the 2¹⁰⁰ paths end above $13,150. Meanwhile 86% of them end with less than the $100 you started with, and 54% end with under a single dollar. The mean is a real number that belongs to a world almost nobody lives in.

So ask the useful question. Chapter 13's law of large numbers is the only thing that pins a long-run average down. What quantity does it actually apply to here? It applies to sums of independent things, and what we have is a product. There is exactly one move in mathematics that converts one into the other, and Chapter 5 handed it to us: ln(1.5 × 0.6) = ln 1.5 + ln 0.6.

Take the log and the product becomes a sum. Now the LLN has something to bite on, and it says the average of ln G converges. So the quantity that governs your future is the growth rate g = E[ln G]. For our coin that is ½ln(1.5) + ½ln(0.6) = −0.0527 per flip, which is a per-flip factor of e−0.0527 = 0.9487 = √0.9. That is negative, so every flip, on the only scale that describes you, you are shrinking.

That is the keystone, and it settles the debt from the last section. The log is not a taste and it is not a patch. It is the unique move that makes the law of large numbers applicable to something that compounds. Log utility was never psychology — it was mechanics wearing a psychologist's coat.

The gap between the two averages has a name and a size. Jensen again: ln E[G] ≥ E[ln G], always, because ln is concave. That gap is called volatility drag, and Chapter 4's Taylor expansion puts it at roughly σ²/2. For our coin the exact drag is 0.1015 and σ²/2 = 0.45²/2 = 0.1013. Remember that term, because it walks back out of Itô's lemma in Chapter 26 wearing completely different clothes, and you will recognise it.

08Ruin, on the only scale that matters

Push the same logic to its edge. What happens to the growth rate of a strategy that can lose everything?

Substitute into the formula we just built. If any outcome has G = 0, then that term contributes ln 0 = −∞. A single minus infinity swamps every finite term beside it, however small its probability. So the long-run growth rate is minus infinity, no matter how beautiful the odds look in dollars.

Here is the concrete version, and the two curves are worth watching cross. A bet wins 90% of the time and doubles your money, and 10% of the time takes all of it. You stake your whole bankroll each round. The expected multiplier is 0.9 × 2 = 1.8, an enormous edge, and after 100 rounds E[W] has grown by a factor of 3.4 × 10²⁵.