An at-the-money call with a zero interest rate is worth one over the root of two pi, which is 0.39894, times the absolute swing of the terminal price, so a 10 dollar standard deviation prices it at 3.9894 and the closest of the offered 1, 5 and 10 is 5. The two-step estimate is exact under a symmetric terminal law, where the option finishes above the strike exactly half the time and the average gain when it does is 7.979. Calibrate a lognormal to the same 10 dollar swing and those two factors become 0.4801 and 8.285, whose product is still 3.98, which is why the qualifier about half cannot be cut.
Sine is differentiable everywhere and never leaves the band from minus one to one on the real axis, which makes it the counterexample everyone reaches for, and the modulus of sine at 10i is already 11013.23. The Cauchy estimate caps every Taylor coefficient by M over r to the n on a circle of radius r, so growing r kills every coefficient above the constant one and nothing but a constant survives. The same estimate with a polynomial growth bound gives more: an entire function bounded by C times one plus the modulus of z, all to the k, is a polynomial of degree at most k, and one corollary of that is the fundamental theorem of algebra.
A plain t sitting next to the t squared in a Gaussian exponent looks like a new function and is only a shift. Completing the square turns the integral of e to the minus a t squared over two plus b t, from x to infinity, into e to the b squared over 2a times the root of 2 pi over a times the standard normal at a rescaled and shifted argument, never at x itself. The worked case comes out as exactly half a bell, e root pi over two or 2.40901455, but only because its lower limit happens to land on the centre b over a.
The characteristic roots of u'' + u' + u are the primitive cube roots of unity, so the homogeneous part decays with envelope e to the minus x over two and oscillates with period 4 pi over root 3, which is 7.2552. Substituting that homogeneous solution back into the equation leaves a residual of exactly minus one, and that residual is the whole distance between the common wrong answer and the right one. The constant u = 1 solves the equation by itself, so every solution settles on 1, and the constant trial works only because the coefficient on u is not zero.
The ratio and root tests both return 1 on the sum of e to the minus root n, which settles nothing, and the usual write-up of the problem then quotes 4 over e as the answer. That number is the floor rather than the cap: the sum is 1.6704068, which is 13.52 percent above it, and the usable bound comes from integrating from 0 instead of from 1, giving exactly 2. Where each bar of width one sits relative to its index is the single step that decides which way the inequality points.
An estimated line of expected return against market sensitivity that sits entirely above the theoretical one is not a market on sale, because pricing errors scatter above and below instead of lifting everything by the same amount. In a two-factor world where every asset carries the same 0.75 exposure to the second risk, the fitted single-factor line comes out exactly parallel to the theoretical one and exactly 3 percentage points above it, with residuals of zero, while a mispricing world engineered to have the same average lift leaves errors of both signs as large as 6.5 points. Let the second exposure grow with sensitivity and the slope moves too, at which point the two lines can cross inside an ordinary sample.
Two stocks each swinging 20 percent a year with a correlation of one half give their product a volatility of 20 root 3, or 34.641 percent, rather than 40. Adding the two numbers is correct at exactly one correlation, namely 1, because the composite volatility is the law of cosines with the correlation as the cosine of the angle between two arrows. Pricing the call at 40 percent overstates it by 14.2 percent and ignoring the correlation understates it by 16.9 percent, and the other diagonal of the same parallelogram prices the ratio of the two stocks.
Buying two-year notes because you expect the curve to steepen is a position on the level of rates: the same correct view loses two dollars if the steepening arrives with everything rising. Matching the two legs on dollar duration removes the parallel part of the move as an algebraic identity, so half a point of widening pays four dollars whatever the level does. Matching market value as well is impossible with only two bonds and needs a third leg carrying no duration, and on a full cash-flow reprice the level survives at second order, worth 0.07 against a four-dollar profit at fifty basis points.
A riskless zero-coupon bond at 100 has a six-month forward of 102.531512, a premium. Give the same bond an 8% coupon and the forward drops to 98.511194, a discount, because the sign of the premium is the sign of the rate minus the coupon and nothing else. Quoting the forward at spot when the coupon is rich hands the other side a riskless 1.518802 per hundred, and the discrete-coupon version shows the answer also turns on whether a payment date falls before delivery.
A futures settles up every day, so its fair price is a plain expectation, while a forward settles once, so its fair price is a discounted expectation renormalised. The difference between the two is exactly the covariance of the discount factor with the contract price divided by the expected discount factor, and for a deposit contract quoted as 100 minus the rate both fall together, so the fair forward price is 95.019999 against the futures' 95.000000. Long the forward and short the futures is worth 0.0195 points at inception, the gap reaches 39.48 basis points at ten years, and on an asset whose price rises with rates the whole answer reverses.
In a rally you want positive convexity, and a mortgage pool has negative convexity, because the borrowers hold the right to prepay and you are short that option. Modelled as a ten-year 6 percent bond minus a three-year call struck at 105, the straight bond has convexity plus 68.8 and the pool minus 177.4, and doubling a rally from 100 to 200 basis points takes the bond from 7.79 percent to 16.35 while the pool goes only from 4.21 to 6.13. The pool still gains, so the reflex is right about the sign and wrong about the size, and the single parameter set in 108 with positive curvature is one whose prepayment option is far out of the money.
A top-rated issuer picks the coupon that prices its ten-year bond at exactly 100 off its own flat 5 percent curve, and the same cash flows discounted off a swap curve 25 basis points lower come to 101.954087. Two facts do the work: a present value is strictly decreasing in every rate it is discounted at, and for a top-rated name the swap curve sits below its own bond curve because a swap risks no principal and is margined daily. A modified duration of 7.7217 times the spread accounts for 1.9304 of the lift, and a convexity of 74.9977 supplies the last two cents.
A share at 50 goes to 65 or to 40, and the right to buy it at 50 is worth exactly 6, because three fifths of a share against 24 borrowed pays the option in both states and costs 6 today. That bill contains no probability at all, which symbolic differentiation shows and a sweep never could, so the price may be computed under whichever beliefs are convenient and the artificial 2/5 returns the same 6. Discounting the mean payoff at the share's required 15 percent gives 9.13, and the rate that does work is the option's own 75 percent, which cannot be known before the price is.
Give a pulled-back log price the same 20 percent instantaneous swing as a free-wandering one and its horizon variance stops being sigma squared times T: at one reversion time only 0.432332 of it survives, the volatility that prices a one-year call is 13.1504 percent, and the call falls from 7.9656 to 5.2425. The same pull makes consecutive returns fight each other, with a first-order autocorrelation of exactly minus half of one minus phi, and that is the independence the pricing model rests on. The formula still returns the right European price and has lost the hedging argument that justified it.
A six-month at-the-money call on a 50 dollar share sheds a cent and a half a night to time decay, and its expected price tomorrow is higher anyway. The deterministic total differential gives minus 0.66 cents and predicts the opposite of the truth, while Ito's third term, half the gamma times the squared move, adds plus 1.13 and runs on variance rather than direction. Substituting the pricing equation for theta cancels that term exactly and leaves an expected return of the riskless rate plus elasticity times the premium, which is 35.54 percent a year here and turns negative below a real drift of 4.18.
A one-year at-the-money call on a 100 share with a 22 percent swing costs 8.7591 and carries 54.3795 dollars of share exposure, an elasticity of 6.2084. Multiply the share's market sensitivity of 1.10 by that and the call follows the market at 6.8292, so a share expected to earn 7.7 percent a year sits under a call expected to earn 47.8. The reflex answer, that an option price is a fair game with no drift, is true under the pricing measure and false under the one you live in, and both halves are measured here rather than asserted.
Daily, weekly and monthly returns give per-day variance estimates of 1.0000, 1.3225 and 1.4000, and the reflex is to average them into 1.2408, a figure no horizon produced. The variance ratio is a weighted sum of autocorrelations, so a forty percent overshoot at twenty periods measures dependence rather than noise, and the coefficient that reproduces it is 0.17554. With twenty years of daily data that ratio sits 4.6 standard errors above one and with five years only 2.3, which is why the number means nothing without the sample size attached.
If any order is equally likely to come from any of twenty traders, a buy order puts the posterior at 21/40 and forces an honest ask five cents above a mid of zero, so the spread is 2/N whatever the crowd's size. Each of the nineteen uninformed traders then loses exactly five cents a trade, which sums to the insider's 95 cents because (N-1)/N and 1 - 1/N are the same number. Volume falling is a comparative static on top of that spread rather than a theorem of the model, and naming the insider would have repaired the market instead of breaking it, since a known informed trader can simply be refused.
A call is a forward with a put stapled to it, because (s-X)+ minus (X-s)+ equals s-X for every terminal price, so with rates at zero and the strike at today's price the call and the put cost exactly the same 11.9235. Every penny of that premium buys protection against a fall the question has ruled out, which is why the forward pays 20 on a certain rise to 120 against the call's 8.0765, a factor of 2.476. The volatility fixes the size of the mistake and never its direction: at 60 percent the call actually loses 3.58 on a certainty.
Short a call struck at 100 with the share at 113.40 and two months left, the hedge holds 0.90028 shares; a month later at 112 it wants 0.94591, so you buy 4.56 per hundred. The reflex to sell is not a blunder, because freezing the clock and letting the same fall happen alone really does take the hedge to 0.87726, but the time effect is 2.7 times larger and points the other way. Holding the hedge constant traces the curve S(T) = X exp(d1 sigma root T minus half sigma squared T), which puts the break-even fall at 3.51 percent and, at expiry, at the strike itself.
Sketch a one-year call struck at 100 with a five percent rate. Deep in the money the curve straightens into a line of slope one, and that line crosses at 95.122942 rather than at 100, so drawing it through the strike is out by 4.877058 for ever. That gap is the interest saved on the strike, and it is also why an American call on a share paying no dividends is never exercised early. Plot the same option against the futures price and the crossing returns to 100 while the slope drops to 0.951229.
You own one-month calls struck at 110 with the share at 100, and you short 0.1452 shares against each one. If the share rallies to exactly 110 and stops, the calls expire worthless while the short has lost ten dollars a share, so the hedged position is down 2.074208 where the unhedged one would have lost only its 0.622212 premium. The worst case sits at the strike because the profit is piecewise linear with slopes of -0.1452 and +0.8548, and a rebalanced hedge on the same path loses 3.058738.
A bond paying 100 in ten years costs 67.5564 at a four percent yield. The first two points of yield cost 11.7169 and the next two only 9.5201, so the curve bends. The usual explanation blames duration falling as yields rise, and this bond refutes it: with a single cash flow its Macaulay duration is exactly ten at every yield. The slope is minus duration times price over one plus the yield, and the general statement needs no duration at all, only that every discount factor is convex.
A share at 100, a one-year call struck at 100, a zero rate and a twenty percent swing return 0.539828 and 0.460172. The first is the number of shares in the replicating portfolio, and reading it as the chance of finishing in the money hands you the complement of the right answer, since at the money with a zero rate the two numbers sum to one. The article carries the density identity that makes the share count exact, the measure under which the first number is a probability after all, and the four cents the straight line misses over a two-dollar move.
Two properties are worth a million each, one an empty field and the other a beach collecting admission. The six-month forward is 1,020,000 for the field and 990,000 for the beach, and the gap is exactly the income the forward buyer never collects. Today's spot already capitalises every future admission, which is why income enters the forward as a subtraction, and why a carrying cost on the field would only widen the gap.
A 20-year 7 percent bond on a flat 10 percent curve prices at 744.5931, and a one-point rise costs exactly 63.1262. The tangent alone says 67.7028, and adding the second-order term of 4.8416 lands at 62.8612, inside 27 cents of the truth. Note that the correction and the error it corrects are two different numbers, which is why the estimate ends up on the wrong side of the answer.
Let a fair coin decide at the start of the year whether the share runs at 15 or 35 percent, price calls in that world, then read the volatilities back out with the constant-volatility formula: 28.43, 25.91, 24.97, 25.68 and 27.16 percent across five strikes. The floor sits at the money and below the 25 percent average of the two regimes, which one second derivative settles without any numerics. The usual explanation for the wings is refuted here, because the coin-flip world is less likely to clear 130 than a flat 25 percent and its option is still worth 27 percent more.
Two weights that add to 0.98 give the variance forecast a half-life of 34.31 days; delete the second one and the half-life is 0.2744 days, gone before the next open. The same recursion turns strictly normal daily draws into a year with kurtosis exactly 297/67, and one shuffle of those same numbers separates the fat tail from the clustering. It also has a condition nobody quotes: stationarity is not enough for that kurtosis to be finite.
Whether an up move followed by a down move lands where a down move followed by an up move lands decides between a quadratic node count and an exponential one, and at twenty steps the gap is a factor of 9,078.6. Both sums carry N+1 terms rather than N, because a twenty-step tree has twenty-one dates on it, and the off-by-one costs the entire final row. The article also states the recombination hypothesis exactly, which is weaker than the usual ud = 1.
Held to expiry the position needs the full five dollars and a two dollar move loses three, but the same move sold the next morning is worth 5.4405. Nothing is assumed to get there: the five dollar price pins the volatility at 35.5424 percent and the position's slope at exactly a tenth. That slope is why the gain is lopsided, and why one dollar down loses money while two dollars down gains six cents.
At the money the shorter maturity always wins, and a single negative derivative settles it for every volatility and every maturity: curvature runs 0.06907 against 0.02814 for one month against six. Ten percent out of the money the order reverses, 0.01854 against 0.02352, and the two curves cross 8.845 percent above the strike. What forces a crossover to exist is a conservation law, since every option in the family carries exactly the same total curvature and can only choose how to spread it.
A pays the floating rate L and receives 24 per cent minus 2L, which nets to 24 minus 3L and factors as three times 8 minus L: three vanilla swaps at eight per cent, so the fixed rate was never the twenty-four printed on the deal. Reading it as twenty-four is a 48-point error at a floating rate of twenty-four. The article also carries the version that does need a model, where a floor on the inverse leg adds two caplets struck at twelve and the factorisation fails.
A ticket paying a hundred dollars if a share finishes above its strike is squeezed between two ordinary call spreads at every width, so its value is pinned by prices already quoted with no distribution assumed anywhere. The limit is minus the derivative of the call price in the strike, which equals e to the minus rT times N(d2) because two density terms cancel exactly at every strike. Here that is 53.2325, against the 62.35 a real-world drift would give.
Cutting a hundred million of a thirty-year bond down to fifty takes five hundred futures, and the usual arithmetic of fifty million over a hundred thousand lands there only because the contract's duration per dollar of face happens to match the bond's. What a hedge matches is dollars per basis point: 56,288.92 against 112.5778. Hold a thirty-year zero instead and the same job needs 1,234 contracts, while five hundred three-month contracts would cover 22.2 per cent of it.
An eight per cent thirty-year bond at par loses 27.49 dollars when its yield rises 25 basis points, and its yield moves because the principal is collateralised in United States Treasuries. The answer that circulates, about thirty-five dollars, needs a duration of fifteen, and a par bond at an eight per cent yield cannot have one: its modified duration is its own annuity factor, capped at 12.5 at any maturity whatsoever. The pass-through, the only soft number in the chain, is swept from an eighth to a half.
The map x to 4x(1-x) contains no randomness and is still useless for prediction, because substituting x = sin squared of pi t turns it into angle doubling: one binary digit of your measurement is spent per step, so fifty steps eat fifteen decimal digits. The resulting series has autocorrelation exactly zero at every lag, proved by orthogonality of distinct cosine frequencies rather than measured. What that does not establish is anything about real return series, and the article says so.
A share at 150, a call struck at 100, a year to run and no dividend: cashing out pays 50 while the option is worth 54.97. The floor S minus X e to the minus rT assumes no distribution at all, and it beats immediate exercise by exactly one year of interest on the strike, 4.8771. The article carries the cusp where the gap peaks at 10.4506, the shelf it settles onto far in the money, and the dividend condition that makes early exercise optimal after all.
Heads pays $7 in eighteen months, tails costs $2 today, and the curve gives 12% for one year and 18% for two. Averaging the amounts gives $2.50, which is 38.68% too high, because expectation and discounting only commute when every cash flow lands on the same date. The answer is about $1.80, and four defensible compounding conventions spread it from 1.7862 to 1.8381, so one decimal is honest and two are not.
Moving the average inside the exponential returns 1, and 1 happens to be the exact median and the exact geometric mean of e^X, which is why the mistake survives every re-check of the arithmetic. Completing the square in the exponent gives the true value e^(sigma squared over two), or 1.6487 at unit spread, because multiplying a Gaussian density by e^x slides its centre and scales its mass. Convexity settles the direction before any integral is set up, and on a heavy-tailed variable the quantity stops being finite at all.
Shade the region between a diffusing particle and the time axis over one second. The box is one wide and about one tall, so the eye guesses a variance of one, and the answer is one third because each increment counts only for the time remaining after it. No stochastic integration is needed to define the object, only continuity of the path, and the constant is pinned twice over: once by the weight (T minus t) and once by integrating the covariance min(s,t) across the square.
A flat belief about a coin's bias is an input to the calculation, not a conclusion of it, and a single head does not leave it standing. The density tilts to 2p, the cumulative law becomes p squared, the average bias moves to 2/3, and the old answer of one half is demoted to the lower quartile. The general update is the Beta conjugate family, which sends 750 heads in 1000 to Beta(751, 251) with mean 0.749501.
Traders treat opposite signs for theta and gamma as a law of the desk, but it is the pricing equation rearranged, and the equation names its own exceptions. At a zero rate the identity is exact and unbreakable; with a positive rate the interest on the bond leg buys the exception, and a deep in-the-money put has theta +7.0053 and gamma +0.0040317 together. The change of variables to the heat equation shows where the interest was hiding.
Both seats in the marble game average a dollar a play, and that arithmetic stays true to the last line. Seat A carries variance 3/2 against seat B's 1, and seat A's law turns out to be seat B's law with one prize smeared outward, so every concave utility prefers B without variance ever being mentioned. Once both players stop flipping coins, seat B is ahead on the average too, at 1 against 3/4.
More volatility is worth more is a theorem about convex payoffs, and a step function is not convex. An at-the-money cash-or-nothing digital falls from 46.02 to 42.07 cents when the swing doubles, and the sensitivity is positive only below X exp(-(r + sigma squared over two)(T-t)). The cap on the payoff is only half the explanation; the falling median is the other half.
A share swinging twenty dollars a year gives an at-the-money call that looks like it should cost ten, half the swing collected half the time. It costs 7.98, because the upper half of a bell curve averages 0.798 of a standard deviation rather than a whole one. The article derives the general arithmetic-Brownian price, checks both limits, and quantifies the negative-price defect that got the model retired and then rehabilitated.
A holding pays $200 if a team wins four games first, you must take a symmetric position on every game, and committing the whole hundred to game one produces the right payoffs a week too early. Backward induction on the lattice fixes the amount at half the gap between the two successor values, $31.25. The same number is 5/16 of the holding, which is the chance the other six games split three each, and no win probability appears anywhere in the derivation.
Two stocks with equal expected returns, variances 0.10 and 0.40, and correlation 0.5: the reflex differentiates the portfolio variance and reports an interior weight. The minimum sits at 100% in the calmer stock, and the usual explanation for that, which blames the no-shorting rule, is wrong. The vertex of the variance parabola lands exactly on w = 1, so the constraint does no work at all and the answer survives dropping it.
Acceptance restricts the value to below your bid, where a uniform variable averages half of it, and doubling half your bid returns exactly your bid. The expected profit is therefore identically zero at every bid up to 100 and 100 minus b above it, so there is no optimal bid to find. With a general multiplier the profit is b squared times k minus 2, over 200, making doubling the exact break-even multiple, and the article shows a value distribution starting at 50 where the same bidder profits.
Requiring the fifty-fifty at every amount you could open forces the weights to satisfy f(x) = f(x/2)/2, whose only solutions are proportional to 1/x, and that integrates to infinity at both ends. Conditional on the pair, the swap gains the smaller amount or loses it with equal chance, which is zero and needs no assumption at all. The article carries a proper spread where the conditional answer is genuinely x/2, and the infinite-mean spread where swapping really is right at every observable amount.
The put reaches its strike more often, 0.3348 against 0.2821, and the call is still worth more, 4.2920 against 3.5891. The mechanism is not the unbounded-upside story, which would predict a gap at the money where put-call parity provably gives none; at a zero rate the 110 call equals 1.1 times a put struck at 90.909, and the put on offer is struck lower than that. The article also records two circulating claims that fail at these strikes, since the in-the-money chances at r = sigma^2/2 are 0.3168 and 0.2992 rather than equal, and the price ratio is 1.63 rather than 2.
Six sevenths in the quieter share beats holding it alone, because what the jumpy share adds at the margin is its correlation times its own swing, fifteen against twenty. The derivative of the variance at a full allocation is exactly +1/50, so the informal test and the first-order condition are one statement. The article carries the exact optimum sqrt(27/700), the convexity making it a minimum, and the correlation threshold of two thirds above which the dip disappears entirely.
A shift leaves every deviation from the mean untouched, and a stretch multiplies the covariance and one standard deviation by the same factor, so both cancel out of the ratio. The tempting answer of five times rho is worse than wrong: at rho = 0.40 it names 2.0, which Cauchy-Schwarz forbids any correlation from reaching. The article carries the general affine rule, the sign flip a negative factor produces, and the curved maps the invariance does not survive.
At the money the log term in d1 vanishes and what remains is strictly positive for every non-negative rate and every volatility, so the delta always beats 0.5 and is 0.6554 at twenty percent. A square rather than a derivative gives the sharp floor: at six percent over a year the delta can never fall below 0.6355. The article also kills the sentence that sounds like a restatement of the answer, since the chance of finishing in the money falls to 0.4801 at forty percent volatility.
Turning all fifty-two cards lands on exactly zero, which makes zero the floor rather than the value. Backward induction over the grid of remaining cards gives the exact rational 41984711742427/15997372030584, and a two-line argument shows the optimal policy can never finish below zero in any deal. The article carries the small-deck ladder, the stopping boundary the table actually produces, and two plausible rules that lose money against it.
The standard deviation of a sum is not the sum of the standard deviations, so quadrupling the horizon only doubles the risk. The article carries the general square-root law, the ratio that diagnoses the mistake, and the controls showing a bell curve does none of the work: a two-point yearly return lands on 0.20026 and a uniform one on 0.20008. It also carries what actually breaks the rule, which is dependence rather than fat tails.
Both games pay 3.5 million dollars on average, so "they match" is true and the inference that it is a wash is not. Shrinking the ticket by a million divides the spread by a million while adding a million independent rolls multiplies it back by only a thousand, so the ratio of standard deviations is exactly root of a million: 1,707,825 against 1,707.83. The whole argument rests on independence, and at a correlation of 1 the diversified game becomes the single roll exactly.
Because the horizons are ten and five, both sides of the no-arbitrage equation are fifth powers and the root disappears, leaving 1 + f = 1.15 squared over 1.10 = 529/440, so f = 89/440 exactly. Reflecting 10 percent around 15 to get 20 is low by exactly (b - a) squared over (1 + a), a square over a positive number, which is why the reflection can never overshoot for any pair of rates. Under continuous compounding the same problem is linear and 20 percent is exactly right, so the instinct is correct machinery pointed at the wrong convention.
Logarithmic differentiation turns the exponent into a factor and gives x^x times (1 + ln x), which is exactly the sum of the power-rule answer x^x and the exponential-rule answer x^x ln x. That is a theorem rather than a coincidence: the two rules are the partial derivatives of u^v, and walking the diagonal u = v = x adds both partial effects. The power rule accidentally returns the correct slope at x = 1, which is the one point nobody should use to test a rule.
There is no elementary antiderivative to evaluate, and the checkable slice of that is one line: if p is a polynomial then p' - 2xp has degree deg p + 1, which can never equal the degree of 1. Squaring the integral turns it into a rotationally symmetric integral over the plane, where the polar area element supplies the factor r that makes the radial integral elementary, so I squared equals 2 pi times one half. The same idea survives without polar coordinates via the substitution y = xt, and it fails for e to the minus x to the fourth because x^4 + y^4 is not a function of the radius.
The fraction that maximises long-run growth is exactly the edge, 2p-1, which is 0.2 on this coin, and one derivative gets you there. Double it and the growth rate is -0.0024469 a flip, negative on a game that leans your way three hundred times in a row, and the crossing happens at 0.3894 rather than at 0.4. The article carries the exact median over 300 flips, 25 dollars to 10504.19 at the optimum and to 12.00 at double, the reason about 48 percent of overbettors still finish ahead anyway, and the place where the textbook approximation mean minus half the variance returns the opposite sign.
A contract struck at the current share price is worth roughly 0.3989 S sigma root T, so doubling the time to expiry multiplies the price by root two and takes 100 dollars to about 141 rather than 200. The exact ratio erf(s/2) over erf(s/(2 root 2)) is always strictly below root two because erf is concave, so 141.42 is a ceiling never reached. Strip out the strike condition and the rule collapses: the same doubling multiplies a strike 30 percent above spot by 4.19 and one 30 percent below by 1.01.
Differentiating y = L tan(wt) gives a spot speed of w R squared over L, so the footprint accelerates with the square of its distance from the lamp: a tenth of pi directly opposite, and exactly pi miles per second nine miles along. The 9 is the along-shore leg, which makes 90 the squared hypotenuse rather than the square of nine, and that misreading is the usual failure. Nothing physical moves at that speed, and a straight coast running 2310 miles would carry a nominally faster-than-light spot carrying no information at all.
Six months of a sixty dollar year carries 60 over root two, which is 42.43 rather than 30, because variances add over disjoint intervals and standard deviations do not, so the digital is worth exactly $239,750. The figure of $250,000 in circulation comes from rounding the z score 0.7071 up to 0.75 and then reading the tail at 0.75 as 0.25, but Phi(0.75) = 0.773373, so even the rounded chain gives 0.2266. Rounding z upward has to make the tail smaller, and 0.25 is larger, which is the tell that a symbol changed meaning mid-calculation.
The 95 percent margin on a proportion is almost exactly 1 over the square root of the sample size, because p(1-p) is flat enough near its peak to call a quarter and 1.96 is close enough to 2, and those two roundings are reciprocal so they annihilate. At N = 1000 the shortcut gives 3.16 percent against an exact 3.04, and it always errs on the conservative side. Reporting one standard error instead, 1.55 percent, describes a 68 percent interval rather than a 95 percent one.
The standard deviation of 1, 2, 3, 4, 5 is either 1.4142 or 1.5811, and offering one of them without asking which question you are answering is the only wrong move. The sum of squared deviations is 10 either way, so everything turns on whether you divide it by 5 or by 4. Bessel's correction makes the variance unbiased and leaves the standard deviation biased low by about six percent at this sample size, and a third divisor beats both of them if you optimise for mean squared error instead.
The two correlations you are handed do not pin the third one down, but they fence it into exactly [−1/50, 1], and that interval dips below zero. The fence falls out of a 3×3 determinant read as a quadratic in the unknown, and out of a picture: 0.7 is an angle of 45.573°, both stocks live on a cone of that half-angle around the index, and putting them on opposite sides opens 91.146° between them. Also here: why the real tipping point is ab ≥ 0 together with a² + b² ≥ 1 rather than "both above 0.707", why 0.9 and 0.5 force a positive answer while 0.9 and −0.9 allow −1, why standing on the floor costs a rank, and why three Bernoulli(0.5) indicators with the same two correlations are confined to [0.40, 1] instead.
A fair coin cuts probabilities into halves and quarters, and a short argument about the prime factorisation of two shows it can never reach one third in a bounded number of flips. Dropping the bound fixes it: flip twice, bin the tail-tail, and each child holds exactly a third for 8/3 flips on average. That naive scheme turns out to be the best any coin-flipping procedure can do for three outcomes, which stops being true at five.
A three-month at-the-money call on a stock at 100 with 40% volatility is worth about eight dollars, and you can get there in two multiplications. The constant four tenths turns out to be the height of the normal bell at its peak, and the whole error of the mental rule is one rounding plus one cubic term. Scaling volatility linearly with time instead of with its square root gives ten dollars, which is 25.5% too high.
Take the centre, then mirror every move through it, and you place the last coin. The proof has three requirements and only one of them needs that opening move, which is the step a one-line answer skips. Central symmetry alone is not the condition: an annulus is centrally symmetric and the first player loses on it.
A stranger says out loud what every islander can already see, and ten days later ten people leave. The fact was mutual knowledge all along; what the announcement supplied was the nine levels of nested knowledge above it. An explicit count over 4096 possible worlds settles the induction without trusting it.
An at-the-money call has no ceiling on its payoff and an at-the-money put is capped at the strike, yet at a zero interest rate the two cost exactly the same. The reason is put-call parity and it uses no model at all: the difference of the two payoffs is a straight line, so pricing it needs only the risk-neutral mean. The equality was checked on five terminal distributions with mean at the strike, and on a sixth whose mean is 120, where the gap is exactly 20.
A die is rolled up to three times and you are paid the face you stop on. The reflex answer of 3.5 is the value of the same game with the right to stop deleted, and the real value is 14/3, reached by computing the game from its last roll backwards. The thresholds move as rolls run out, which is why a four is worth keeping late and worth rejecting early.
Raise your right hand at a mirror and the reflected hand stays on the same side of the room, which means the usual question has a false premise. A plane mirror is the matrix diag(1, 1, -1): it fixes both axes lying in the glass and reverses only the direction you look along. Its determinant is -1, so no rotation reproduces it, and the sideways flip everyone reports belongs to the half turn you perform in your head.
Every family averages exactly one girl and contains exactly one boy, so the ratio of expected counts is exactly one half and a large town splits evenly. The expected share inside a single family is not one half but ln 2, and it is still 0.5249 across ten families, with the excess falling off like one over four m.
The pile really does average exactly one dollar, which is why almost everyone answers one dollar and why the trap is a correct calculation of the wrong quantity. The play is worth two thirds, because the roll that ends the game pays on two of its three faces and that roll is independent of how big the pile grew.
A pebble climbing four boxes on coin flips needs 18/5 flips on average, and the two-line renewal argument that gives 4 is wrong. Its premise is true, since half of all games really do end on flip two, but the non-finishing half is two different states: tails-tails sends the pebble home while heads-heads leaves it on box 3, one flip from the exit and worth only 14/5.
Coins have no memory, which is true, and nobody said this coin is fair, which is the whole problem. A fair coin explains the run with probability two to the minus one hundred while a two-headed coin explains it every time. The article locates the threshold exactly and reconciles the answer with the companion piece on ten heads, which asks a different question about a different setup.
Unfolding two faces into a 2 by 1 rectangle turns the walk into a straight segment of length root five, crossing the shared edge at half height. The reflex answer of one plus root two is the same one-parameter family evaluated at the end of that edge instead of its middle, so the trap and the answer are two points on one curve.
The hedge on a long call is the slope of its value, and a convex curve flattens as you slide left, so a falling share forces a smaller short and a smaller short is a purchase. No volatility, maturity or distribution enters that argument. The rebalance buys twenty shares, and the position gains 0.9824 per share on the fall.
Halving the length and timing the flame is not a biased estimator of half the time, it is unrelated to it: across four thousand random cords the midpoint method scattered from under fifteen seconds to over forty-five. Lighting both ends gives exactly thirty on every cord, by an argument that never evaluates the burn rate.
The North Pole really is a solution, so the trap is only the words "and nowhere else". A mile north of the parallel whose whole lap measures 1/n of a mile, the eastward mile is n exact revolutions, which puts a starting circle 1.159 miles from the South Pole for one lap, 1.080 for two, 1.053 for three. Every point of every circle works, so the honest count is uncountable rather than infinite.
You hit one time in ten, your two opponents three and six, and you shoot first. Firing into the air is worth 965/4736 = 20.376%, which beats removing the strongest player by 0.195 percentage points, because a landed hit drops you into the duel you must enter second at 7/37 rather than first at 10/37. The article carries all three option values, the fixed points they solve, and the single Nash equilibrium that turns the usual assumption into a conclusion.
Two people arrive at random inside the same hour and each waits fifteen minutes, so the reflex answer is a quarter. Drawing both arrival times as one point in a 60 by 60 square turns the question into an area, and the two corner triangles it leaves out have legs of 45, giving 7/16 rather than 1/4. The general formula n(2T-n)/T squared shows why the first minutes of patience buy the most.
A stock at 100 goes to 130 with probability 0.8 or 70 with probability 0.2, rates are zero, and the right to buy at 110 is worth 10 rather than 16. A third of a share funded by borrowing 70/3 reproduces both payoffs and costs 10 today, which prices the option without using a probability anywhere. The general risk-neutral probability (S-d)/(u-d) is a half here only because rates are zero and 100 sits midway between the two outcomes.
Two random breaks, three pieces, and three inequalities that collapse into one. The quarter falls out of a square with no integral at all, and the average longest piece, 11/18, explains why the answer feels too low but isn't.
Draw X and Y uniformly from the unit interval and their product beats a half with probability (1 - ln 2)/2, about 15.3 percent. The reflex answer of a quarter counts a condition that is genuinely necessary and treats it as sufficient, which is why 0.8 times 0.6 sits inside the quarter square and still loses. The hyperbola y = 1/(2x) cuts the winners down to a sliver, and one integral measures it.
The reflex answer is around 180, half the calendar. The real threshold is 23, because a match needs a pair and 23 people carry 253 of them. The same reasoning puts the answer to "does anyone share MY birthday" at 253 people, eleven times the crowd, and the square-root threshold behind both is why a 64-bit random id collides after five billion draws rather than eighteen quintillion.
You toss five fair coins, I toss four, and you win on strictly more heads: the answer is exactly 256 of the 512 outcomes. Because you hold one coin more, "not strictly more heads" and "strictly more tails" are the same event, and turning every coin over is a bijection between them. The fifth coin is worth nearly fourteen percentage points over the 93/256 you would have without it, and none of that is an edge.
Independent guessing gives the prisoners 7.9 x 10^-31. Following the slip you just found gives them 0.311828, and the gap is thirty orders of magnitude from a rule you can state in one sentence. The strategy never raises anyone's individual chance above one half; it only makes the failures coincide, which is the whole lesson.
Told that one of two children is a girl, the chance both are girls is 1/3. Watch a girl open the door instead and it is 1/2, from the same four families and the same prior. One likelihood separates them: a mixed family always satisfies the statement, but sends the girl to the door only half the time. Push the identifying detail to a girl born on a Tuesday and the answer slides to 13/27.
Draw one coin from a thousand, flip ten heads, and the chance it is the two-headed one is 0.5062. Both reflex answers miss, in opposite directions: ninety-nine percent ignores the bag, one in a thousand ignores the flips. Counting patterns gets the exact figure with no Bayes notation at all, and the reason it lands on a coin flip is that 2^10 happens to sit next to the size of the bag.
Drop chips at random into dough, cut it into a hundred cookies, and ask how many chips guarantee no bare cookie nine times out of ten. Five hundred chips, five per cookie on average, works about half the time. Inclusion-exclusion pins the answer at 683, a closed form you can solve on a whiteboard agrees, and the coupon collector's mean of 518.7 is the sophisticated wrong answer.
On the unit interval the indicator of the rationals has upper sum 1 and lower sum 0 for every partition ever written, so the two never meet and Riemann returns nothing at all. Give the k-th rational an interval of width ε/2^k and the whole countable set sits inside a total length of ε, which puts its measure at 0 and its Lebesgue integral at 0. Lebesgue's criterion turns that into the general law — a bounded function on a compact interval is Riemann integrable exactly when its discontinuities have measure zero — which is why Thomae's function, discontinuous on the same dense set, is integrable and this one is not. The trade is not free: sin x over x on the half line has an improper Riemann value of π/2 and no Lebesgue integral at all.
A counterfeit coin that might be heavy or light, a balance that only reports which side falls, and a hundred dollars a weighing. Counting rules out four; only a construction gets you five.
A disease one person in two hundred carries, and a test with no false negatives at all. The reflex answer to a positive result is above ninety percent, and it is wrong by more than a factor of ten. A crowd of a thousand people shows why before the algebra does, and Bayes puts the exact figure at 100/1493.
A stock with zero volatility, a call struck at the money, and the reflex answer — zero — that fails real trading interviews. One arbitrage argument prices it three different ways, and Black–Scholes agrees on the way out.
Drop three random points on a circle and the triangle contains the center exactly a quarter of the time. Two proofs: an honest average, then two coins wearing a geometry costume.