Applied decision theory
A breakaway from the causal-ladder curriculum: same engine — expected utility, maximum expected utility, the value of information — but here it is pure practice. A growing set of word-problems from high-stakes fields, each solved step by step, plus a calculator you can drag. The recurring trick: a price or a payoff secretly encodes a probability, and the work is reading it back out.
Several of the problems below are the same move in disguise: read a price backwards to recover the probability it implies. The cleanest version — the one the calculator runs on, and the one the diagram sketches — is the break-even of a risk transfer. It needs only the expected-value reasoning from the causal-ladder page (utility and maximum expected utility); that is the one prerequisite.
You break even between paying a certain cost up front and carrying the risk yourself exactly when $$p^\ast \;=\; \frac{C}{\,L - D\,}.$$ In words: the break-even probability is the up-front cost divided by the slice of the loss that the cost actually removes.
What each symbol means — abstractly, and what it can stand for once you point the equation at a real decision:
| Symbol | Abstractly | Applied, it can be… |
|---|---|---|
| $p^\ast$ | the break-even probability of the bad event — transfer the risk if your true chance is higher, self-bear if it is lower | a screen-crack rate · the claim rate an insurer assumes · how likely you are to actually use the cover |
| $C$ | the certain cost paid up front to hand off the risk, per period | an insurance premium · an extended-warranty price · a maintenance plan · the cost of a hedge |
| $L$ | the full loss you eat if the event happens and you did not pay $C$ | an out-of-warranty repair · a replacement device · the bill for an uncovered failure |
| $D$ | the residual cost you still pay even when covered, so $C$ buys only $L-D$ of real protection | a deductible · a service fee · a copay · $D=0$ when the cover is total |
The other three problems are the same inversion through sibling formulas: in finance the price is the probability; in medicine the "price" is a test and the payoff is the value of the information it buys; in law the break-even is the win-probability that makes a settlement fair. The diagram is that whole family at a glance:
This is a problem set, not an essay, so every entry follows the same five beats — Setup (the scenario and what to find), Principle (the lens, written as a formula), Work (numbered arithmetic), Answer (the boxed result), and Takeaway (the transferable lesson). The tag beside each title names the tool it drills, and new problems slot into the same beats. The theory behind every move lives on the ladder page (Bayes nets → utility → MEU → VPI → counterfactuals); every number is checked by hand.
1 · Insurance — what AppleCare's price believes break-even probability
Apple sells AppleCare+ on the iPhone 17 Pro Max for $C=\$139.99$ a year. Out of warranty, a screen repair is $L=\$379$; with coverage you pay a $D=\$29$ deductible. Below what one-year break rate should you decline the plan?
Self-insuring costs $p\,L$ in expectation; insuring costs $C+pD$. Set them equal and the indifference point is $$p^\ast=\frac{C}{L-D}.$$ Buy coverage iff your true $p$ exceeds $p^\ast$; equivalently, the premium reveals the probability the seller is pricing in.
Decline if you crack a screen less than $\approx\!40\%$ of years; buy if more. The rest of the lineup prices the same way: the base iPhone 17 ($\$119.99$ against a $\$329$ repair) also lands at $119.99/300\approx 40\%$, while the iPhone 17 Pro ($\$139.99$ against $\$329$) needs $139.99/300\approx 46.7\%$ — the worst screen-only value of the three. A risk-averse buyer's threshold sits a hair below these, by a risk premium $\approx\tfrac12 A(W)\,\mathrm{Var}(\text{loss})$ that is negligible when the loss is tiny against your wealth — which is exactly why cheap-item insurance rarely pays.
A posted price is a probability in disguise. Read it back out, remember the deductible shrinks your cover to $L-D$ (not $L$), and buy only when your own break rate beats the rate the premium implies.
2 · Finance — implied probability, and how the vig lies implied probability
Two quick reads. (i) A prediction-market contract pays $\$1$ if a candidate wins and trades at $\$0.62$. What probability does the market imply? (ii) A sportsbook offers decimal odds of $1.90$ on each side of a two-outcome game. What does each price imply, and what is the book keeping for itself?
A $\$1$ binary claim is priced at its risk-neutral probability, so $\text{price}=p$ outright; buy iff your $p>\text{price}$. For decimal odds $d$ the implied probability is $1/d$ — but that is gross of the margin: across outcomes the implied probabilities sum to more than $1$ by the overround, and dividing each by that sum removes the bias.
The market says $62\%$ (buy above that); the two $1.90$ prices look like $52.63\%$ but really encode $50\%/50\%$ plus a $5.26\%$ house edge. The naive implied probability overstates every outcome by the half-vig, so normalize before you trust it — the same price-to-probability inversion as insurance, with the seller's margin folded in.
Inverting a price into a probability is the same move everywhere — options, betting lines, prediction markets. But a quoted price hides the seller's margin, so strip the overround before you trust the number.
3 · Medicine — the value of a diagnostic test value of information
A patient may have disease $D$; your prior is $P(D)=0.40$. Two actions, with payoffs in net-benefit units (read them as thousands of dollars of quality-adjusted value):
| Disease (0.40) | Healthy (0.60) | |
|---|---|---|
| Treat | 90 | 70 |
| Withhold | 30 | 100 |
A perfect test reveals the true state and costs $2$ units. Is it worth ordering before you choose?
With $\mathrm{EU}(a)=\sum_s P(s)\,U(a,s)$ and $\mathrm{MEU}=\max_a \mathrm{EU}(a)$, the VPI of observing the true state is $$\mathrm{VPI}=\Big[\textstyle\sum_v P(v)\,\mathrm{MEU}(v)\Big]-\mathrm{MEU},$$ and you test iff $\mathrm{VPI}$ beats the test's cost.
$\mathrm{VPI}=18$. The test earns its keep only because a "Healthy" result flips the action — blind you would have treated, but the result tells you to withhold. Information is worth exactly what it changes: had both states led to the same choice, $\mathrm{VPI}$ would be $0$, the clinical version of the ladder's $\mathrm{VPI}(F)=0$ result.
Information is worth only what it changes. VPI prices decision-relevance, not how much you learn — a test whose result never flips the action is worth exactly nothing, however informative it feels.
4 · Law — settle or go to trial expected utility
A plaintiff wins an award of $X=\$100{,}000$ with probability $p$, and nothing otherwise. Trial costs $K=\$20{,}000$ in fees either way, and the defendant offers a certain settlement of $S=\$30{,}000$. When should the plaintiff settle — and why might they settle for less than the case is "worth"?
Trial is worth $pX-K$ in expectation, so settle iff $S\ge pX-K$, i.e. below the break-even $$p^\ast=\frac{S+K}{X}.$$ Under risk aversion you compare utilities, not dollars: the gamble's certainty equivalent sits below its expected value by a risk premium, so offers between the two get accepted even when the dollar math "says sue."
The break-even is $p^\ast=50\%$. At $p=0.6$ the EV of trial is $\$40{,}000$, but its certainty equivalent is only $\$35{,}200$ — so a risk-averse plaintiff accepts any offer in $[\$35{,}200,\ \$40{,}000]$, below the expected-value line. (The $\$30{,}000$ on the table is under even the CE, so it is still rejected; nudge it to $\$36{,}000$ and EV says "sue" while caution says "settle.") Expected utility subsumes naive expected value: concavity prices in risk, which is why litigants, insurers, and patients all routinely accept less than a gamble's mathematical expectation.
Expected utility subsumes expected value. Concavity prices in risk, so the rational deal sits below the gamble's dollar expectation by the risk premium — the gap that keeps settlement, insurance, and hedging in business.
More from the anthology new facets
The first four problems all reduce to the same move: read a price into a probability. The next five keep the expected-value and utility engine but push it into genuinely new shapes — a noisy signal updated by Bayes' rule, an optimal quantity rather than a yes/no, growth-optimal sizing over repeated bets, a sequential search with no second chances, and a strategic equilibrium against an opponent who is also optimizing. For the actuary's professional take on pricing the risk itself, see the actuarial track.
5 · Medicine & screening — the base-rate trap of a positive test Bayes + value of information
A disease has prevalence (prior) $P(D)=1\%$ in the screened population. A screening test has sensitivity $P(+\mid D)=95\%$ and specificity $P(-\mid H)=90\%$ — so it falsely flags a healthy person $P(+\mid H)=10\%$ of the time. A patient walks in, gets screened, and tests positive. How likely is it that they actually have the disease?
The post-test probability of disease weights the true-positive mass against all positive mass — true and false: $$P(D\mid +)=\frac{P(+\mid D)\,P(D)}{P(+\mid D)\,P(D)+P(+\mid H)\,P(H)}.$$ A rare disease ($P(D)$ tiny) lets the false-positive term $P(+\mid H)\,P(H)$ dominate, no matter how good the test sounds.
$P(D\mid +)\approx 8.8\%$. Despite a "$95\%$ accurate" test and a scary positive, the patient is still over $91\%$ likely to be healthy — the base-rate fallacy. With disease this rare, a single positive barely moves the needle; the $990$ false positives drown out the $95$ real ones.
A real test is noisy, so its value of information can never exceed the perfect-test ceiling — the $\mathrm{VPI}=18$ units from problem 3: $0\le \mathrm{VEI}\le\mathrm{VPI}$. And like any information, the screen is worth ordering only if its result could flip the treat/withhold choice. At an $8.8\%$ posterior, if you would not treat at that probability, a lone positive buys nothing but anxiety — confirm with a second, independent test before acting.
When the prior is small, even an excellent test produces mostly false positives: sensitivity acts on a tiny slice while the false-positive rate acts on the vast healthy majority. Always ask "out of how many?" before trusting a positive — and remember that a test is worth running only when its answer could change what you do.
6 · Operations & retail — how many to stock critical fractile
A cafe bakes croissants for tomorrow. Each sells for $\$5$ and costs $\$2$ to make; whatever goes unsold by closing is worth $\$0$. So baking one too few forfeits a margin of $C_u=5-2=\$3$, while baking one too many sinks the $C_o=\$2$ it cost to make. Tomorrow's demand is Normal with mean $\mu=100$ and standard deviation $\sigma=20$. How many should the cafe bake this evening?
Bake one more croissant only while its expected gain beats its expected loss. The marginal unit sells with probability $P(\text{Demand}\ge Q)$, earning $C_u$, and sits unsold with probability $P(\text{Demand}<Q)$, costing $C_o$. Equating the two, the optimal stock is the demand quantile whose cumulative probability equals the critical fractile $$F(Q^{\ast})=\frac{C_u}{C_u+C_o}.$$ A ratio of costs again sets a probability threshold, but here it selects an optimal quantity, not a yes/no.
Bake $\approx\!105$ croissants. Because lost margin ($\$3$) outweighs waste ($\$2$), the optimal plan deliberately overshoots the mean of $100$ and stocks out only $40\%$ of days. Slight, intentional waste is cheaper than turning away paying customers.
This is the structural twin of the break-even rule $p^{\ast}=C/(L-D)$: a ratio of costs becomes a probability threshold. The newsvendor's twist is that the threshold is read off the demand distribution as a quantile, so the cost ratio chooses how much to stock rather than whether to act.
7 · Investing & betting: how much to wager on an edge growth-optimal sizing
A repeatable, favourable bet pays $b = 2$ to $1$ on a win, with win probability $p = 0.4$ and loss probability $q = 1 - p = 0.6$. Its expected value is positive: $0.4 \times 2 - 0.6 \times 1 = +0.2$ per unit staked. The question is not whether to bet a clear edge but how much: what fraction of your bankroll should you stake on each repetition to maximize the long-run growth of your wealth?
Because bets compound multiplicatively, the right objective is not expected wealth but expected log-wealth, whose growth rate per bet is $g(f) = p\ln(1 + bf) + q\ln(1 - f)$. Maximizing $g$ over the stake fraction $f$ gives the Kelly fraction $$f^{\ast} = \frac{bp - q}{b} = p - \frac{q}{b}.$$
Stake $f^{\ast} = 10\%$ of bankroll per bet, yielding the maximal log-growth rate of about $+0.97\%$ per wager. Doubling to $20\%$ collapses growth to $+0.07\%$ and a touch more turns it negative; the same edge, recklessly sized, destroys wealth.
Sizing matters as much as the edge itself. The growth curve is steep on the downside of its peak, so overbetting is far more dangerous than underbetting, which is why disciplined practitioners deliberately use fractional Kelly (half or quarter), trading a little growth for a large reduction in volatility and ruin risk.
8 · Search & hiring — when to stop looking optimal stopping
You are filling one role and interview candidates one at a time in random order. After each interview you must accept or reject on the spot, with no recalls, and you only care about hiring the single best of the $n$ applicants. The natural strategy: reject the first $r-1$ as a calibration sample, then hire the first one who beats everyone seen so far. For $n=100$, how many should you let walk, and what are your odds of landing the very best?
Look-then-leap. After rejecting an initial sample of $r-1$, accept the next candidate who is a running record. The best wins exactly when it falls past the sample and the best-so-far at the cutoff lies inside the sample, giving $$P(r)=\frac{r-1}{n}\sum_{i=r}^{n}\frac{1}{i-1}.$$ As $n\to\infty$ the optimal cutoff is $r-1\approx n/e$ and the win probability approaches $1/e\approx 0.368$.
Reject the first $37$ candidates, then hire the first one who beats all of them: you land the very best $\approx 37.1\%$ of the time, against $1\%$ for a random pick. The famous $1/e\approx 36.8\%$ is the large-$n$ limit; at $n=100$ the exact figure is a touch higher.
When choices arrive sequentially and are irreversible, spend the first $\approx 37\%$ of your search purely learning the field, then commit to the next thing that beats everything you have seen. The same $37\%$ rule guides house-hunting, selling an asset into an uncertain market, and dating. This is the only sequential decision on the page; every other example is one-shot.
9 · Game theory & poker — pot odds and the honest bluff implied probability + equilibrium
The pot holds $\$100$. Your opponent bets $\$50$, so to call you must put in $\$50$ to win the $\$150$ that is now out there. Two questions. (i) What is the worst hand (the lowest win-probability) you can still profitably call with? (ii) How often should a balanced bettor bluff so that you cannot exploit them by always calling or always folding?
Pot odds quote you a price: call iff your equity beats the cost of calling divided by what the pot pays you after that call. And a balanced bettor sets the bluff frequency that leaves you exactly indifferent between calling and folding, so neither response gains. $$q^{\ast}=\frac{\text{call}}{\text{pot after call}},\qquad \frac{\text{bluffs}}{\text{value}}=\frac{\text{bet}}{\text{pot}+\text{bet}}.$$
Both answers are $25\%$: you can profitably call with any hand above $25\%$ equity, and a balanced bettor bluffs $25\%$ of the time. The coincidence is not luck: the same $\$50$-into-$\$150$ price drives the caller's threshold and the bettor's mix, so the equilibrium frequency is pinned to the very odds the bet lays.
Pot odds are a price-to-probability inversion, the same move as reading an option or a betting line. The equilibrium bluff frequency is the strategic cousin of the bookmaker's vig: a number chosen not to win each hand but to make the other side indifferent, so they can never adjust their way to a profit against you.