Use probability to make fair decisions and to analyze strategies in product testing, medical screening and games

Learning goal: Use probabilities to make fair decisions, for example by drawing lots or using a random number generator, and analyze decisions and strategies using probability concepts, including product testing, medical testing and pulling a hockey goalie at the end of a game.

Scope note. S.MD.6 and S.MD.7 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.

What a fair decision is. When a choice among people or options should not favor anyone, the choice can be handed to a chance process. A procedure is fair in the probability sense when each eligible person or option has the same probability of being chosen, or, when the intent is proportional rather than equal, when each has the intended probability. Fairness is a property of the probabilities of the procedure, not of how the outcome turned out; a fair draw can still select the same person twice in a row.

Drawing lots. Writing each of n names on identical slips, mixing thoroughly, and drawing one blind gives each name probability 1/n. The conditions matter: identical slips, thorough mixing and a blind draw are what make the outcomes equally likely. To choose an ordered list, such as who presents first, second and third, keep drawing without replacement; every ordering of the n names then has probability 1/n!, so every person is equally likely to be in any position.

Random number generators. A random number generator that produces each integer from 1 to n with equal probability does the same job. Number the people 1 to n and generate one value. When the generator's range does not match, adapt it without distorting the probabilities. To choose one of 5 people with a fair six-sided die, assign faces 1 to 5 to the people and reroll on a 6: every eventual outcome is then one of the five faces, each equally likely, so each person has probability 1/5. The tempting shortcut "a 6 goes to person 1" is unfair: person 1 would have probability 2/6 and the others 1/6 each. Similarly, to choose among three people with a coin, toss twice: HH for the first, HT the second, TH the third, and toss again on TT; each person ends with probability 1/3.

Intended unequal probabilities. Sometimes fair means proportional. If four clubs contributed 10, 20, 30 and 40 hours to an event and one club is to be chosen to keep the equipment, a draw with 1, 2, 3 and 4 slips respectively gives probabilities 0.1, 0.2, 0.3 and 0.4, matching the shares of effort. The decision is fair relative to the stated rule. What matters is that the probabilities are chosen and announced in advance and that the procedure actually produces them.

Analyzing decisions with probability: product testing. A factory inspector can test only a sample from a batch. Suppose a hypothetical batch of 20 items contains 2 defective ones, and the inspector tests 3 items chosen at random without replacement, rejecting the batch if any tested item is defective. What is the probability that the batch passes? The batch passes if all 3 tested items come from the 18 good ones: P(pass) = 18C3/20C3 = 816/1,140 = 68/95 ≈ 0.716. By the Multiplication Rule the same value is (18/20)(17/19)(16/18) = 4,896/6,840 = 68/95. So this plan lets a batch with 10% defectives through about 72% of the time. That is the number that decides whether the plan is adequate. Testing 5 items instead gives P(pass) = 18C5/20C5 = 8,568/15,504 = 21/38 ≈ 0.553; the inspector can compare the cost of testing more items against the cost of accepting bad batches, and the probabilities make the trade-off explicit.

Medical testing and conditional probability. A screening test for a condition is described by two conditional probabilities: the probability of a positive result given the condition is present (sensitivity) and the probability of a positive result given it is absent (the false-positive rate). The question a patient cares about is the reverse conditional: the probability of having the condition given a positive result. These are different, as the everyday-language lesson showed, and the difference can be enormous when the condition is rare.

Use a hypothetical population of 10,000 people (invented numbers, not data about any real test or disease) in which 1% have the condition, the test detects 95% of those who have it, and it gives a positive result to 5% of those who do not. Then 100 people have the condition and 95 of them test positive; 9,900 do not have it and 5% of them, 495, test positive anyway. Total positives: 95 + 495 = 590. P(condition | positive) = 95/590 = 19/118 ≈ 0.161. A positive result on this test means only about a 16% chance of the condition, because the 495 false positives from the large healthy group outnumber the 95 true positives from the small affected group. In this model most positive results are false positives, which is why screening programs often follow a positive screen with a more specific confirming test; a real decision also weighs the harms and benefits of each next step, which this model does not include. If the condition were more common, say 20% of the population, the same test would give P(condition | positive) = 1,900/(1,900 + 400) ≈ 0.826, so the base rate drives the result.

Independence in the screening model. The test result and the condition are dependent, and that is what makes the test useful. P(positive) = 590/10,000 = 0.059, but P(positive | condition) = 95/100 = 0.95; P(condition) = 0.01, but P(condition | positive) = 95/590 ≈ 0.161. If the two events were independent, both conditional probabilities would equal the unconditional ones: a positive result would leave the chance of the condition at 1%, and the test would carry no information. The product rule gives the same verdict: P(condition and positive) = 95/10,000 = 0.0095, while P(condition) × P(positive) = 0.01 × 0.059 = 0.00059.

Pulling the goalie. In ice hockey, a team trailing late in the game may replace its goalie with an extra attacker, leaving its net empty. The decision looks reckless, since the opponent scores far more easily into an empty net, but probability shows the logic. Take hypothetical rates for the final 90 seconds when trailing by one goal: with the goalie in, the trailing team scores with probability 0.10 and concedes with probability 0.05; with the goalie pulled, it scores with probability 0.25 and concedes with probability 0.40. As a simplification, assume at most one goal is scored in the time left, so scoring and conceding are mutually exclusive outcomes (no goal has probability 1 − 0.10 − 0.05 = 0.85 with the goalie in and 1 − 0.25 − 0.40 = 0.35 with it pulled). If the trailing team does not score, it loses, whether the score ends one goal or two goals apart. Conceding therefore costs nothing extra; the loss was already the outcome. What matters is the probability of tying the game: 0.25 with the goalie pulled versus 0.10 with the goalie in. Pulling the goalie more than doubles the chance of tying the game and so avoiding a loss in regulation time, and the higher chance of conceding is irrelevant to the result. The strategy is justified because the payoff structure makes one bad outcome no worse than another. With more time left, when a conceded goal could still be answered, the analysis changes and the numbers must be redone.

The general method. Each analysis followed the same steps: list the possible outcomes of each option, assign probabilities from the model or the data, attach a payoff or a yes-or-no consequence, and compare. The probabilities were exact for the fair-decision procedures and the sampling plan, and hypothetical for the medical and hockey examples; a real analysis would replace them with measured rates and would state where they came from.

Misconceptions. In fair decisions, students accept "a 6 means person 1" because every person can still be chosen; fairness needs equal probabilities, not merely possibility. In medical testing, students report the sensitivity, 95%, as the chance a positive person is ill; that reverses the conditional. In the goalie problem, students compare the conceding probabilities and stop; the comparison must be of the outcomes that change the result.

Worked fair procedure: Choose 2 of 7 students for a task with a ten-sided die numbered 0 to 9. Assign 1 to 7 to the students and reroll on 0, 8 or 9; the first accepted roll picks one student with probability 1/7 each. For the second, reroll until a different accepted number appears, so each remaining student has probability 1/6. Every pair of students then has probability 2/42 = 1/21, and 7C2 = 21 pairs are possible, so all pairs are equally likely.

Worked sampling plan: Batch of 20 with 2 defective; test 3 without replacement. P(at least one defective found) = 1 − 68/95 = 27/95 ≈ 0.284. If the plan instead tests 3 items with replacement (returning each after testing, which is pointless in practice but useful as a comparison), P(none defective) = (18/20)³ = 0.729, slightly higher than 0.716, because replacing a good item keeps the pool of good items full.

Worked reverse conditional: In the hypothetical screening population, P(no condition | negative) is the fraction of negatives who are healthy. Negatives: 5 with the condition (100 − 95) and 9,405 without (9,900 − 495), total 9,410. P(no condition | negative) = 9,405/9,410 ≈ 0.9995. A negative result is very reassuring in this population, even though a positive result leaves the condition unlikely (about 16%); both facts follow from the same table.

Represent and record: Write three fair procedures for choosing one of 5 people (slips, a six-sided die with a reroll rule, a three-toss coin scheme with a repeat rule: assign five of the eight equally likely toss sequences to the five people and toss again on the other three) and beside each show that every person has probability 1/5. Then build the 10,000-person table for the hypothetical screening test with rows condition and no condition, columns positive and negative, fill all four cells (95, 5, 495, 9,405), and read off P(condition | positive) = 95/590 and P(no condition | negative) = 9,405/9,410.

Independent performance: Part 1: design a fair procedure for a decision in your school or class that currently is not made by chance (assigning seats, choosing an order of presentations), state the intended probabilities, and prove the procedure gives them. Part 2: analyze one strategy decision with probability: a sampling plan for a batch (state batch size, number defective, sample size; compute the probability of passing), or a two-option end-of-game decision with hypothetical scoring rates, or a screening test with hypothetical rates. Success criteria: probabilities are computed, not asserted; hypothetical numbers are labeled hypothetical; conditional probabilities name the given event; the recommendation follows from the comparison and states what would change it.