Weigh decisions by expected value: payoffs of games of chance and comparison of strategies

Learning goal: Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values; find the expected payoff for a game of chance; evaluate and compare strategies, such as a high-deductible versus a low-deductible insurance policy, on the basis of expected values under various but reasonable chances of the relevant events.

Scope note. S.MD.2 and S.MD.5, with its parts a and b, are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

Decisions as random variables. A decision with uncertain consequences can be modeled by listing the possible outcomes, assigning each a payoff (a number: money won, cost paid, points scored) and a probability, and computing the expected value of the payoff. The expected value is the mean payoff per decision over a long run of similar decisions. When one option has a higher expected payoff than another, it does better on average; whether "on average" is the right criterion for a single decision is a separate question addressed at the end.

Expected payoff of a game of chance. A hypothetical raffle, invented for this lesson, sells 500 tickets at 2 dollars each. One ticket wins 300 dollars, two tickets win 50 dollars each, five tickets win 10 dollars each, and the remaining 492 win nothing. Let V be the prize on one ticket. The expected prize is

E(V) = 300(1/500) + 50(2/500) + 10(5/500) + 0(492/500) = (300 + 100 + 50)/500 = 450/500 = 0.90 dollars.

The expected net payoff, prize minus the 2-dollar price, is 0.90 − 2 = −1.10 dollars per ticket. Interpretation: over many tickets the buyer loses an average of 1.10 dollars each; the organizer collects 1,000 dollars and pays out 450, keeping 550. A negative expected net payoff is the normal condition of a fund-raising raffle or a commercial game; the buyer pays for the chance, and possibly for the cause. Knowing E(V) = 0.90 tells a buyer what a ticket is worth on average, which is different from what it might pay on one occasion.

A second hypothetical game, of the kind printed on a food wrapper: a card gives a 1-in-20 chance of a 5-dollar item and a 1-in-5 chance of a 1-dollar item, otherwise nothing, with the card free. E(payoff) = 5(1/20) + 1(1/5) + 0(3/4) = 0.25 + 0.20 = 0.45 dollars. Since the card costs nothing, its expected net payoff is +0.45; a free chance whose outcomes are all prizes or nothing always has a non-negative expected net payoff. If a real promotion is examined, its probabilities must be read from its published rules, not assumed.

Comparing strategies by expected cost. Insurance decisions are the classic case. The following policy terms and probabilities are hypothetical, chosen to make the arithmetic clear; real premiums and accident rates vary. A driver chooses between two one-year policies. Policy H has a high deductible of 1,000 dollars and an annual premium of 600 dollars. Policy L has a low deductible of 250 dollars and a premium of 900 dollars. The deductible is the amount the driver pays on a claim before the insurer pays the rest. Suppose a minor accident would cost 800 dollars to repair and a major accident 6,000 dollars.

The driver's out-of-pocket cost for the year is premium plus, for each accident, the smaller of the repair cost and the deductible. Under H a minor accident costs the driver 800 (below the 1,000 deductible, so the driver pays it all) and a major accident costs 1,000. Under L a minor accident costs 250 and a major accident costs 250.

Take a first set of reasonable chances for the year: P(minor) = 0.10, P(major) = 0.02, P(neither) = 0.88, treating at most one accident as the model. Expected annual cost:

H: 600 + 0.10 × 800 + 0.02 × 1,000 = 600 + 80 + 20 = 700 dollars.
L: 900 + 0.10 × 250 + 0.02 × 250 = 900 + 25 + 5 = 930 dollars.

Under these chances the high-deductible policy is cheaper on average by 230 dollars. The 300-dollar premium saving outweighs the expected extra deductible payments of 70 dollars.

Varying the chances. A comparison is only as good as its probabilities, so the standard asks for various, but reasonable, chances. Suppose a driver with a worse record faces P(minor) = 0.30 and P(major) = 0.10. Then H: 600 + 240 + 100 = 940; L: 900 + 75 + 25 = 1,000. H still wins, by 60. Push further to P(minor) = 0.40 and P(major) = 0.15: H: 600 + 320 + 150 = 1,070; L: 900 + 100 + 37.50 = 1,037.50. Now L is cheaper on average. The break-even condition is that the expected saving from the low deductible equals the 300-dollar premium difference: 550 × P(minor) + 750 × P(major) = 300. When that expected saving is less than 300, H has the lower expected cost; when it exceeds 300, L does. Check: 550(0.30) + 750(0.10) = 240 < 300, and 550(0.40) + 750(0.15) = 332.50 > 300. The decision depends on the driver's own chances, and computing the expected values under several plausible values shows how sensitive the choice is.

Beyond the expected value. Expected value is the average, and averages hide spread. Under policy H a major accident costs the driver 1,000 dollars at once; under L it costs 250. A driver who could not pay 1,000 dollars without serious hardship might reasonably choose L even when H has the lower expected cost, because avoiding a ruinous outcome is worth more to them than the average saving. Expected value is the right criterion for decisions that repeat many times and whose individual outcomes are affordable; for a one-time decision with an unaffordable worst case, the distribution of outcomes, not only its mean, belongs in the analysis. A complete comparison reports the expected values, the range of outcomes, and the assumptions.

A fully worked model: choosing a project. A student can enter one of two hypothetical science-fair categories. In category A, the probability of a 200-dollar prize is 0.15 and of a 50-dollar prize is 0.25; otherwise nothing. In category B, the probability of a 100-dollar prize is 0.40; otherwise nothing. E(A) = 200(0.15) + 50(0.25) = 30 + 12.50 = 42.50 dollars. E(B) = 100(0.40) = 40 dollars. A has the higher expected prize by 2.50 dollars. Both categories give the same 40% chance of winning something (0.15 + 0.25 = 0.40 for A), so they differ only in what a win is worth: in A a win is 200 dollars or 50 dollars, in B it is always 100 dollars. The expected values are close enough that the student's preference between A's wider spread and B's single prize level could reasonably decide.

Misconceptions. The first is judging a game by its largest prize rather than its expected value: a 300-dollar prize does not make a 2-dollar ticket a good buy when the expected prize is 90 cents. The second is fixing one probability and treating the conclusion as universal: the insurance comparison flipped when the chances changed. The third is confusing expected net payoff with the expected prize: the ticket price must be subtracted.

Worked game payoff: A hypothetical carnival game costs 3 dollars. Roll two dice: a sum of 7 wins 10 dollars, doubles win 6 dollars, anything else wins nothing (7 and doubles cannot both occur). P(sum 7) = 6/36, P(doubles) = 6/36. E(prize) = 10(6/36) + 6(6/36) = 96/36 ≈ 2.67 dollars. Expected net payoff = 2.67 − 3 = −0.33 dollars per play. The player loses about 33 cents per play on average.

Worked strategy comparison: Two hypothetical delivery options for a shop: Option 1 costs 20 dollars per delivery with a 0.02 chance of a lost item costing 400 dollars; Option 2 costs 26 dollars with a 0.005 chance of the same loss. Expected cost per delivery: Option 1: 20 + 0.02 × 400 = 28 dollars; Option 2: 26 + 0.005 × 400 = 28 dollars. The expected costs are equal, so the choice turns on other factors: the shop that cannot absorb a 400-dollar loss may prefer Option 2. If the lost-item probability for Option 1 were 0.03, Option 1 would cost 32 on average and Option 2 would be better.

Worked sensitivity check: For the insurance policies (H: premium 600, deductible 1,000; L: premium 900, deductible 250; minor repair 800, major 6,000), find the chance of a minor accident at which the two policies have equal expected cost when P(major) = 0.05. Set 600 + 800p + 50 = 900 + 250p + 12.50, so 550p = 262.50 and p ≈ 0.477. Below a 47.7% chance of a minor accident, H is cheaper on average; above it, L is. A 47.7% chance of a minor accident in a single year is very high compared with the other chances used in this lesson, so in this hypothetical comparison H is the expected-value choice unless the 1,000-dollar worst case is unaffordable.

Represent and record: Make a payoff table for the raffle (prize, probability, product) and total the products to get E(V) = 0.90; then subtract the ticket price. Make a second table for the two insurance policies with rows "no accident", "minor", "major", columns for probability, cost under H, cost under L, and the two products; total each column for the first probability set (700 versus 930), then redo it for the second and third sets and record which policy wins each time.

Independent performance: Design a hypothetical game of chance with at least three prize levels and a ticket price; compute the expected prize and the expected net payoff and state what the organizer keeps per ticket on average. Then compare two strategies of your own devising (two insurance policies, two warranty options, two routes with different chances of delay) under three different but reasonable sets of probabilities, and find the break-even probability at which the strategies tie. Success criteria: every payoff has a probability and the probabilities sum to 1; expected values are shown as sums of products; the net payoff subtracts the cost; the strategy comparison states the winner under each probability set and includes one sentence on what the expected value does not capture (the worst case and its affordability); all games and policies are labeled hypothetical.