Numbers, Chance, and Shape

The equipment is old. Human number sense counts four objects at a glance, feels the gap between twenty and forty but not the gap between twenty million and forty million, and files every probability under one of three headings: won't happen, might, will. That kit served foragers well. It still runs constantly — fast, fluent, and wrong in ways that carry no sensation of being wrong. The mistaken answer to every puzzle in this chapter arrives instantly, wearing total confidence.

The muscle trained here is the handoff: feeling the moment a question leaves intuition's jurisdiction and passing it to a stronger instrument — a notation, a computation, a proof. The instruments are not natural equipment. Positional arithmetic had to be smuggled into Europe inside a puzzle book (Chapter 1); probability theory is not yet four centuries old. Nobody perceives a conditional probability directly. The people who handle them correctly are the people who stopped trusting the feeling and learned to check.

Four movements: numbers, chance, estimation, shape — knowing exactly, knowing the odds, knowing roughly, and finally knowing that a thing is impossible. Along the way the gut will keep offering answers: 183 people, fifty-fifty, "the weekday obviously can't matter," "whichever treatment wins overall is better." Each is wrong. Notice what the certainty felt like just before each correction; that feeling is the syllabus. Where gut and derivation disagree, the derivation wins. Every time.

Numbers

Arithmetic first: three puzzles about letting symbols carry what attention drops.

★ 26 · The Seventeen Camels

A father dies leaving seventeen camels: half to the eldest child, a third to the middle, a ninth to the youngest. Camels do not divide. While the heirs argue, a stranger rides up and lends them her camel. Eighteen camels: the eldest takes nine, the middle six, the youngest two — seventeen in all — and the stranger takes her own camel back and rides off. Every heir received at least the promised share. What happened?

Pause here. Where is the trick hiding?

Solution. Add the will's fractions: 1/2 + 1/3 + 1/9 = 9/18 + 6/18 + 2/18 = 17/18. The will was never a partition of the estate. It under-allocates by exactly one eighteenth, and that gap is the whole mechanism. The borrowed camel completes the herd to a number the fractions bite on cleanly, and because the will only ever promised 17/18 of a herd, the integer shares 9 + 6 + 2 = 17 fit inside eighteen camels with one left over — the stranger's own. Each heir does better than the will's letter: nine camels against a promised 8.5, six against 5.67, two against 1.89. The trick is honest precisely because the original instruction was incoherent. The stranger didn't divide the estate; she repaired its arithmetic, and charged nothing. The tale is folklore across Arab, Indian, and European traditions — often told of the court wit Birbal — and Tartaglia was already discussing the seventeen-horse version in the sixteenth century.

The Move — Borrow a unit, return it. Temporarily extend a system to make its structure visible, then retract the extension. Mathematics does it constantly: add and subtract the same term, multiply by a cleverly written 1, adjoin a point at infinity, introduce a slack variable. The camel is a slack variable with legs.

★★★ 41 · SEND + MORE = MONEY

Each letter stands for a different digit; leading letters are not zero. Solve:

SEND + MORE = MONEY

Take a pencil. The carries are the whole game.

Solution. Name the carries c₁ (units into tens) up through c₄ (thousands into ten-thousands), and walk the columns from where the structure is stiffest.

MONEY has five digits, so the addition overflows: c₄ = 1, and since a carry never exceeds 1, M = 1.

Thousands column: S + M + c₃ = O + 10·c₄, so S + 1 + c₃ = O + 10. If c₃ = 0, then S = O + 9, forcing O = 0, S = 9. If c₃ = 1, then S = O + 8, so O is 0 or 1 — and 1 is taken. Either way O = 0, with S = 9 or 8 pending.

Hundreds column: E + O + c₂ = N + 10·c₃, which is E + c₂ = N + 10·c₃. If c₃ = 1, the left side must reach 10 + N; the only escape is E = 9, c₂ = 1, N = 0 — and 0 belongs to O. Dead. So c₃ = 0, which settles S = 9, and leaves N = E + c₂. Letters denote different digits, so N ≠ E, forcing c₂ = 1 and N = E + 1.

Tens column: N + R + c₁ = E + 10·c₂ = E + 10. Substitute N = E + 1 and it collapses to R + c₁ = 9. R = 9 is taken, so R = 8 and c₁ = 1.

Units column: D + E = Y + 10. The digits left are {2, 3, 4, 5, 6, 7}; E and N = E + 1 must be consecutive picks from them, and D + E must reach at least 12. Test the candidates: E = 5 (N = 6) demands D = Y + 5, and from the remaining {2, 3, 4, 7} that gives D = 7, Y = 2 — everything fits. Every other choice of E strands D above the digits that remain. The unique solution, confirmed by exhaustive search: 9567 + 1085 = 10652. Dudeney published the puzzle in the Strand Magazine, July 1924; the genre got its name — the cryptarithm — in 1931.

The Move — Squeeze the tightest constraint. Don't start where filling is easy; start where the structure is most rigid — here, the overflow into a fifth digit — and let the forcing cascade. In any constraint problem, most-constrained-variable-first is the professional's default. Amateurs start in the friendly corner.

★★★ 48 · The Self-Describing Number

Find the 10-digit number whose first digit says how many 0s the number contains, whose second digit says how many 1s, and so on, the tenth digit counting the 9s.

Build it digit by digit. The number polices itself.

Solution. Call the digits d₀ through d₉, where dᵢ counts the occurrences of digit i. Two facts discipline the search. First, the digits sum to 10: the census counts each of the ten positions exactly once. Second — counting another way — the sum of the number's digit values equals Σ i·dᵢ, and that is the same sum again: 10. Two totals, both 10, so the number is mostly zeros: a large d₀ and a light tail.

Try d₀ = 7. Then digit 7 appears, so d₇ = 1, and the tail beyond d₀ must sum to 3. That 1 sitting at position 7 means the number contains a 1, so d₁ ≥ 1. But d₁ = 1 is self-defeating — writing it creates a second 1 — and d₁ = 2 summons a 2 into existence, needing d₂ = 1, and the total hits 7 + 2 + 1 + 1 = 11. Dead.

Try d₀ = 6. Then d₆ = 1, tail sum 4. The 1 at position 6 forces d₁ ≥ 1; set d₁ = 2, which puts a 2 in the number, so d₂ = 1 — and the census closes: 6 + 2 + 1 + 1 = 10. Read it off: 6210001000. Check it against itself: six 0s, two 1s, one 2, one 6, nothing else. It holds, and machine search confirms it is the only such number in base 10. The family is called the autobiographical numbers — recreational folklore of the Gardner era.

The Move — Seek the fixed point. When the description is part of the described, demand x = f(x) and let consistency do the eliminating. Fixed points are where self-reference stops being a parlor trick and becomes mathematics: Gödel numbering, quines, Nash equilibria, and every "this sentence…" that actually works.

Chance

Probability is where confident people go to be wrong. The gut offers three settings — won't, might, will — and then backfills a story. The mathematics that corrects it is barely 370 years old, and it has never become instinct in anyone. For the four puzzles that follow, do one extra thing: before solving, say the instant answer out loud and write it down. The point of this section is not the right answers. It is the documented experience of the wrong ones.

★★ 32 · The Birthday Paradox

How many people must be in a room before it is more likely than not that two of them share a birthday?

Commit to a number before reading on.

Solution. The instinct says something near 183 — half of 365. The answer is 23. Compute the complement, the probability that all birthdays differ: the second person avoids the first with probability 364/365, the third avoids both with 363/365, and so on. For 23 people, 1 − (365/365)(364/365)⋯(343/365) = 0.5073; at 22 people it is 0.476, so 23 is the crossing. Intuition breaks because it counts your potential matches — 22 chances, which feels rightly small. But the event lives on pairs, and 23 people carry C(23,2) = 253 of them, each a lottery ticket nobody is watching. The growth is brutal: by 50 people the probability is 97 percent; by 70, it is 99.9. The result is usually credited to Richard von Mises (1939), carried into folklore by Feller's probability textbook and Gardner's columns.

The Move — Count the complement, and count pairs. "At least one" almost always means: compute none, subtract from 1. And when items interact, the scale of the problem is n², not n — the arithmetic behind hash collisions, DNA-database coincidences, and why "small world" stories are cheap.

★★★ 33 · Monty Hall

Three doors. Behind one, a car; behind the other two, goats. You choose a door. The host — who knows where the car is, must open one of the doors you didn't pick, and must show a goat — opens one: goat. He offers you a switch to the last closed door. Do you take it?

Decide now — switch or stay — and rate your confidence.

Solution. Staying wins 1/3 of the time; switching wins 2/3. Nearly everyone's gut says two closed doors, fifty-fifty — the mailbag below proves how strong the pull is. Enumerate, having picked door 1. Car at 1 (probability 1/3): the host opens 2 or 3; switching loses. Car at 2 (1/3): he is forced to open 3; switching wins. Car at 3 (1/3): forced to open 2; switching wins. Two worlds out of three, the switch takes the car.

The fifty-fifty feeling treats the host's reveal as scenery. It is not. Your first pick is wrong 2/3 of the time, and in every one of those worlds the host's hand is forced — his choice concentrates the entire remaining 2/3 onto the door he was not allowed to open. Amplify until the signal is deafening: a hundred doors, you pick one, and the host, knowing, swings open ninety-eight goats and leaves one door shut. Your door keeps its 1/100. The survivor carries 99/100. Nobody feels fifty-fifty at a hundred doors; three doors is the same machine with the volume turned down.

The constraint is everything. Contrast "Monty Fall": a host who slips, opens one of the other doors at random, and happens to reveal a goat. The stage looks identical — and now the answer genuinely is 1/2. Same picture, different procedure, different probability.

History, for consolation. Steve Selvin posed the problem in The American Statistician, 1975 (its grandfather is Bertrand's box paradox, 1889). When Marilyn vos Savant published the correct answer in Parade in September 1990, roughly ten thousand letters arrived telling her she was wrong — around a thousand from PhDs. She wasn't. Paul Erdős reportedly stayed unconvinced until shown a computer simulation. If the puzzle beat Erdős's intuition, it was allowed to beat yours. What matters is which instrument gets trusted next time.

The Move — Condition on the procedure, not the picture. Evidence is not what is seen; it is how likely that sight was under each hypothesis, given the process that produced it. Identical observations from different protocols carry different information. Ask what machine generated the reveal — in boardrooms and biopsy results as much as game shows. This may be the most practically valuable probability habit in the book.

★★★ 35 · Simpson's Paradox

Two treatments for kidney stones, real clinical data. Open surgery beats keyhole surgery on small stones and on large stones. Overall, keyhole wins. All six numbers are correct:

Which treatment is better — and how can the numbers even do this?

Check the arithmetic yourself. It all holds.

Solution. Nothing is wrong with the arithmetic; something is wrong with the question "which row wins?" asked of the last column. The reversal comes from the case mix. Open surgery was assigned mostly the hard cases — 263 of its 350 patients had large stones — while keyhole got mostly easy ones, 270 of 350 small. The "All patients" column therefore compares open-surgery-on-hard-cases against keyhole-on-easy-cases, and severity, not treatment, drives the totals. Severity is a confounder: it steered the choice of treatment and it worsens the outcome. Given that, the stratified columns are the honest comparison — open surgery wins — and the aggregate answers a different question, usually the wrong one. The data are from Charig et al., BMJ, 1986; the effect carries Simpson's name (1951) with Yule's priority (1903), and it walks the earth: the 1973 Berkeley admissions case, where women applied to the more competitive departments (Bickel et al., Science, 1975), batting averages, and every A/B test segmented after the fact.

The Move — Stratify before comparing. Check the denominators; hunt for the third variable steering cases into groups. And respect the converse discipline: aggregation is never neutral — someone chooses the level at which the truth gets read. When a summary and its parts disagree, ask which one answers the question actually being asked.

★★★★ 34 · The Tuesday Boy

A parent says: "I have two children. One of them is a boy born on a Tuesday." What is the probability both children are boys? And — seriously — does the Tuesday matter?

Your first answer is wrong. Probably your second, too.

Solution. The gut says the weekday is decoration: one is a boy, so the other is a boy with probability 1/2. A better-trained gut recalls the classic two-children puzzle and says 1/3. The enumerated answer is 13/27 ≈ 0.481 — and the Tuesday is doing real work.

Take the sample space seriously: each child is a sex-and-weekday pair, 14 equally likely kinds, so an ordered pair of children comes in 196 kinds. Families with at least one Tuesday-boy: 14 + 14 − 1 = 27 (first child a Tuesday-boy, or second, minus the one family counted twice — where both are). Among those 27, the both-boy families number 7 + 7 − 1 = 13. Hence 13/27: strictly between 1/3 and 1/2. The mechanism is that subtracted 1. The more specific the mentioned detail, the closer the answer creeps to 1/2, because a specific enough detail effectively names which child is being discussed — and once one child is pinned, the other is a plain coin flip. "One is a boy" leaves maximal ambiguity (1/3); "one is a boy born at 6:52 on a Tuesday morning" is nearly a proper name (≈1/2); Tuesday alone sits between.

Now the required honesty. 13/27 answers a census question: among all two-child families having at least one Tuesday-born boy, what fraction have two boys? If instead a parent simply volunteers a fact about one child, the selection procedure is different and so is the number — typically straight back to 1/2. The puzzle has no unique answer until the sampling procedure is stated ⚠. Gardner, who introduced the underlying boy-girl paradox in 1959, later retracted his own 1/3 as ambiguous for exactly this reason. Gary Foshee detonated the Tuesday version from the stage of the Gathering 4 Gardner in 2010, and the worldwide argument it started was, itself, the lesson.

The Move — The sampling defines the odds. Before computing any conditional probability, write down the procedure that produced the statement. Data never speaks alone; it speaks through its selection mechanism — the same lesson that returns as survivorship bias, Berkson's paradox, and half of all misread statistics.

Estimation

Past the edge of available data lies most of real life. What replaces exactness there is not guessing; it is controlled roughness.

★★ 46 · Fermi's Piano Tuners

How many piano tuners work in Chicago? No looking anything up — that restriction is the whole exercise.

One page, a pencil, no search bar.

Solution. Build a chain of factors, each defensible to within a factor of two or three. Chicago: about 3 million people. At roughly 3 per household, call it 1 million households. Perhaps 1 in 20 keeps a piano: 50,000 pianos, with schools, churches, and concert halls adding a few thousand more. A piano wants tuning about once a year, so ~50,000 tunings a year. One tuner manages maybe 4 a day across 250 working days: 1,000 a year. Demand over capacity: 50,000 ÷ 1,000 ≈ 50 tuners. The honest answer is a band, not a point — 30 to 150, order 10² — and the old Chicago phone directories agreed, listing dozens to a hundred-odd. Enrico Fermi ran his Chicago classroom on questions like this in the 1940s; the same reflex let him estimate the Trinity bomb's yield on 16 July 1945 from how far the blast wave pushed scraps of falling paper.

Why it works: each factor is soft by a factor of ~2, but the errors are independent — some high, some low — and they partially cancel in the product. The order of magnitude survives even though every input is a guess. And the discipline pays twice: an estimate made before a serious calculation is the alarm that later catches the calculation's slipped decimal point.

The Move — Decompose, estimate, multiply. Convert "unknowable" into a chain of individually guessable factors; think in powers of ten; state the band and refuse the false point. Confidently within threefold beats precisely wrong by a million.

Shape

The eye is a confident witness and a poor one. Two closers: a puzzle where the picture lies, and one where a two-line argument outsees every picture that could ever be drawn.

★★★ 22 · The Missing Square

Four flat pieces: a right triangle 8 units long and 3 high; a right triangle 5 long and 2 high; and two L-shaped pieces of 7 and 8 unit cells. Arranged one way, they fill what appears to be a right triangle 13 units long and 5 high. Rearranged, the same four pieces form the same 13-by-5 "triangle" — with one interior unit cell empty. Cut them from graph paper: both assemblies look impeccable. Where did the square go?

Where can a whole square hide?

Solution. In the hypotenuse. Neither figure is a triangle. The 8×3 piece's slanted edge rises at slope 3/8 = 0.375; the 5×2 piece's at 2/5 = 0.4. Laid end to end they form a line that bends — by barely more than a degree, well under what the eye resolves at pencil width. In one arrangement the false hypotenuse sags concave, and the figure holds area 32; in the other it bulges convex, area 33. A true 13×5 triangle would hold 13 × 5 ÷ 2 = 32.5: the two impostors sit half a unit on either side of it. The missing square never left. It is smeared along the seam as a sliver-thin parallelogram of area exactly 1. Paul Curry, a New York amateur magician, built the effect in 1953; Gardner gave it the classic exposition in Mathematics, Magic and Mystery (1956); its grandfather is Sam Loyd's vanishing-warrior card Get Off the Earth (1896).

The Move — Verify the diagram. A picture is a claim, not a proof. Sum the areas; check the slopes; measure. The magician's version plays fair — it announces itself as a trick. The chart in the slide deck that shows exactly what everyone hoped to see makes no such announcement.

★★ 20 · The Mutilated Chessboard

Take a chessboard and cut away two diagonally opposite corner squares, leaving 62. You have 31 dominoes, each covering exactly two edge-adjacent squares. Can the 31 dominoes tile the mutilated board?

Try a few tilings in your head before answering.

Solution. No — and the proof never tries a single tiling. The two removed corners share a color; a chessboard's opposite corners always do. What remains is 32 squares of one color and 30 of the other. Every domino, wherever it lies, covers exactly one dark and one light square — adjacency guarantees it. So 31 dominoes cover 31 dark and 31 light. But the board offers 32 and 30. Done.

Feel what just happened. By hand, attempt after attempt would fail without ever revealing whether the failure was the board's or the attempt's. The coloring argument refutes all arrangements at once — it examines zero of them — by exhibiting a quantity that no legal placement can change. This is why John McCarthy proposed the puzzle as a benchmark for automated theorem-provers: a machine that only searches arrangements cannot see a coloring. The puzzle was posed by the philosopher Max Black (Critical Thinking, 1946) and spread by Gamow & Stern and by Gardner's 1957 column, and it remains the cleanest two-sentence aha in mathematics.

The Move — Find the invariant. To prove a thing impossible, find a quantity every legal move conserves and show that start and goal disagree on it. Chapter 9 forges this single flash into an arsenal — colorings, parities, and conserved sums that execute whole families of impossible tasks in a paragraph.

The moves you now own

• Borrow a unit, return it (★ 26, the camels) — extend the system, read its structure, retract the loan.
• Squeeze the tightest constraint (★★★ 41, SEND+MORE) — start where the structure is rigid and let the forcing cascade.
• Seek the fixed point (★★★ 48, the self-describing number) — when the description describes itself, demand consistency: x = f(x).
• Count the complement, and count pairs (★★ 32, birthdays) — "at least one" means compute "none"; interactions scale as n².
• Condition on the procedure, not the picture (★★★ 33, Monty Hall) — evidence is how likely the sight was under each hypothesis.
• Stratify before comparing (★★★ 35, Simpson's) — find the variable steering cases into groups; distrust the flattering aggregate.
• The sampling defines the odds (★★★★ 34, the Tuesday boy) — no conditional probability without the generating procedure.
• Decompose, estimate, multiply (★★ 46, piano tuners) — powers of ten; bands, not points.
• Verify the diagram (★★★ 22, the missing square) — a picture is a claim.
• Find the invariant (★★ 20, the chessboard) — impossibility by conservation.

One meta-move rode along with all ten. Every failure of intuition in this chapter announced itself as certainty — the answer that felt most obvious was the trap, every time. The stronger the click of obviousness around a number, the more it has earned the two-line check.