Week 2: Math/Stats Review II: Probability & Expected Value

This week’s big question: when the outcome is uncertain, how do you combine the chances of things happening, and then compare your options on average? Probability gives you the chances; expected value turns a whole spread of “what might happen” into one comparable number that powers every decision model in Weeks 4–6.

Before the week

Read: Taylor, Introduction to Management Science (13e), Chapter 11, pp. 539–553. Fifteen pages, but two stretches inside them are not yours this week, so the real load is about ten. The chapter PDF is in this week’s Canvas module.

Stop where “The Normal Distribution” begins on p. 553. That section is background for Week 12’s PERT, not for this week.

One thing to know before you open it. Taylor writes the multiplication and addition rules without ever calling them rules, and never uses the word “complement.” Those three names are ours, so the shapes will be familiar from the chapter even though the labels are new on Monday.

Then the warm-up self-check. It is ungraded, retryable, and takes five minutes; it tests the arithmetic Monday leans on rather than the rules themselves.

Session 1: Mon, Aug 31 · Probability (TBD)

🧩 Puzzle

The clinic’s two screens. A walk-in clinic runs two independent screening tests on each patient. Test A flags a problem with probability \(0.15\); test B flags one with probability \(0.10\). What’s the chance both tests flag the same patient? And what’s the chance at least one of them does? Those two questions sound similar but use opposite rules. Getting them straight is the whole skill.

Build the skill 1: “and” for independent events (multiply)

In probability, an experiment is a repeatable process with an uncertain result (run a screen, draw a record, roll a die) and an event is the outcome or set of outcomes whose chance you want. Here the experiment is screening one patient; “Test A flags” and “both tests flag” are two different events.

Two events are independent when one happening tells you nothing about the other, exactly the assumption for two separate tests. For independent events the multiplication rule gives the chance they both happen:

\(P(A \cap B) = P(A)\,P(B)\).

So both tests flagging is \(P(A \cap B) = 0.15 \times 0.10 = 0.015\), about a 1.5% chance. Multiplying shrinks the number fast: requiring several independent things at once is hard, which is why “and” probabilities are small.

Multiplying the two rates does something worth seeing directly. Screening one patient has four possible results, not two, and the multiplication rule prices every one of them:

Those four cover everything that can happen, so they sum to 1. Drop patients through the widget below and watch the running shares settle onto exactly those numbers. Both flag is the rarest by a long way, and how long it takes to appear at all is the point: 1.5% is about one patient in every 67.

Build the skill 2: “or” (addition rule and the complement trick)

“At least one” is an or question. The general addition rule avoids double-counting the overlap:

Here \(P(A \cup B) = 0.15 + 0.10 - 0.015 = 0.235\). Often the fastest route to “at least one” is the complement: the opposite of “at least one flags” is “neither flags,” and for independent events neither-flags is \(0.85 \times 0.90 = 0.765\), so at-least-one is \(1 - 0.765 = 0.235\), the same answer.

The overlap term only vanishes when events are mutually exclusive, and that is not the same as independent. Two events are mutually exclusive when they cannot both happen, so \(P(A \cap B) = 0\) and the addition rule collapses to \(P(A \cup B) = P(A) + P(B)\). Two events are independent when one tells you nothing about the other. These are opposite ideas, and confusing them is the most common error in the chapter. Pull one warehouse item: “damaged” (\(0.08\)) and “expired” (\(0.05\)) are not mutually exclusive. A unit can be both, with \(P(\text{both}) = 0.02\), so you must subtract the overlap: \(P(\text{damaged or expired}) = 0.08 + 0.05 - 0.02 = 0.11\). But a single shipment is graded exactly one of on-time / late / lost, so “late” (\(0.12\)) and “lost” (\(0.03\)) are mutually exclusive and \(P(\text{late or lost}) = 0.12 + 0.03 = 0.15\) with no overlap to remove. Note the trap: mutually exclusive events are the opposite of independent. If one happening forces the other not to happen, then learning the first changes the second’s probability to \(0\).

The picture below is why you subtract. Take 100 warehouse items: 8 are damaged, 5 are expired, and 2 are both. Mark the damaged ones, then mark the expired ones, and then add the two counts: \(8 + 5 = 13\). Now count the squares that actually carry a mark. There are only 11. The 2 items that are both got marked twice, once in each pass, and subtracting them once is the entire addition rule:

Dividing by 100 turns that count straight back into the probability: \(11/100 = 0.11\), and the 89 unmarked squares are the items that are neither.

Build the skill 3: joint probabilities, marginals, and the two-way table

The cleanest way to organize “and” probabilities for two events is a joint (contingency) table: rows for one event, columns for the other, with each inside cell a joint probability \(P(A \cap B)\) and the row/column totals the marginal probabilities: the plain probability of one event on its own, like \(P(A)\). Suppose 400 surveyed students are cross-classified by how they get to campus (rides the bus, written \(R\), vs. drives, written \(D\)) and whether they hold a parking pass \(K\) or no pass \(N\). The letters are deliberately not \(A\) and \(B\): those stay reserved for the general rule, so that nothing below has to be read twice.

The survey does not arrive as probabilities. It arrives as counts, which is how data always arrives, so start there and divide at the end:

Now divide every cell by the 400 students surveyed. Nothing else changes: the same table, on a 0-to-1 scale.

Every inside cell is a joint probability, and every one of them is just a count over 400. \(P(D \cap K) = 140/400 = 0.35\) means 35% of students both drive and hold a pass. Sum a row or column and you get a marginal: \(P(D) = 0.35 + 0.10 = 0.45\), which is the \(180\) drivers over \(400\). A conditional probability rescales a joint by a marginal. Written for two generic events, it asks “given the second one happened, how likely is the first?”:

Put the campus letters in and it asks, among drivers, how likely is a parking pass: \(P(K \mid D) = \dfrac{P(K \cap D)}{P(D)} = \dfrac{0.35}{0.45} \approx 0.78\), far above the overall \(P(K) = 0.45\). In counts that is \(140/180\), and reading it that way tells you what conditioning actually does: it throws away the 220 bus riders and re-bases the question on the 180 drivers who are left. The denominator changed from 400 to 180, and that is the whole idea. Because \(P(K \mid D) \ne P(K)\), driving and holding a pass are dependent: knowing someone drives changes the chance they have a pass. (When events are independent, the conditional changes nothing: \(P(A \mid B) = P(A)\), the trivial case from the two screens above.) Conditional probability is where the information lives. It is the engine behind the decision trees of Weeks 4–5 and Week 6’s Bayes updating, and reading it off a table like this one is the single most testable skill in the chapter.

In practice: independence is an assumption, not a fact

• “Independent” is often assumed because it’s convenient, not because it’s true. Two medical tests can share a failure mode (both fooled by the same confounder); two loans can default together in the same recession. Assuming independence then understates the chance of a joint bad outcome.
• Where do the probabilities come from? A probability is one of three kinds: classical (counted from equally likely outcomes, like a die), relative-frequency (the share observed in past data, and the most widely used kind in business), or subjective (an informed expert’s degree of belief). The clinic’s \(0.15\) and \(0.10\) are relative frequencies estimated from past patients, so they carry sampling error. A rule applied to a shaky input gives a shaky answer, and naming the source type tells you how much to trust it.
• Watch the wording. “And,” “or,” “at least one,” and “given that” point to different rules. Mis-reading the question, not the arithmetic, is the usual source of wrong probabilities.

Back to the puzzle

Both screens flag the same patient with probability \(0.15 \times 0.10 = \mathbf{0.015}\) (about 1.5%), and at least one flags with probability \(0.15 + 0.10 - 0.015 = 1 - (0.85)(0.90) = \mathbf{0.235}\) (about 23.5%). The clinic should expect roughly 1 in 4 patients to trip at least one screen, far more than the rare double-flag. Multiply for “and,” add (minus the overlap) or use the complement for “or.”

In-class group practice (wrap-up)

To close Session 1, work an original probability scenario by hand in your group: a two-way table of visit counts, the three kinds of probability that come off it, an independence check, and one “at least one” question you answer twice. Write your group’s names on the sheet and hand it to your TA before you leave. It counts toward participation.

• Download: Week 2 group practice (Word)

Session 2: Wed, Sep 2 · Expected value (TBD)

🧩 Puzzle

The \$2 raffle ticket. A community raffle sells tickets for \$2 each. Most tickets win nothing, but some pay \$5, a few pay \$50, and one lucky ticket pays \$500. The published odds are: \$0 with probability \(0.939\), \$5 with probability \(0.05\), \$50 with probability \(0.01\), and \$500 with probability \(0.001\). Is a ticket worth buying? “It could pay \$500” is true but beside the point. You need the average.

Build the skill 1: expected value sums payoff × probability

The expected value of an uncertain payoff weights each outcome by its probability and adds them up:

\(E[X] = \sum_i x_i\, p_i\).

It is the long-run average outcome, not the most likely one (here that’s \$0), and not the best case (\$500). For the raffle’s gross winnings:

So an average ticket returns \$1.25 in prizes. Expected value is the single most important number in decision analysis: it collapses a whole distribution of “what might happen” into one comparable figure.

Build the skill 2: decisions compare expected values (net it out)

A payoff you have to pay for is judged on its net expected value. The ticket costs \$2, so the expected net result per ticket is

A negative expected value means that, played over and over, you lose about 75 cents per ticket on average. Comparing options is just comparing their expected values: “buy” has \(E = -\$0.75\), “don’t buy” has \(E = \$0\), so don’t buy wins.

The picture below puts the two sides of that comparison on one scale. The top bar is what a ticket returns, built up one coloured piece at a time, each piece a prize’s payoff times its probability. The bottom bar is what a ticket costs. Step through it, and the thing to watch for at the end is the gap between the two bar ends: that distance is the net expected value.

Two pieces of that top bar are worth pausing on. The \$0 outcome has no width at all, just a tick mark, even though it is what happens to 93.9% of tickets: an outcome worth nothing contributes nothing, however likely it is. And the \$500 jackpot piece is exactly the same width as the \$50 one, 50 cents each, because it pays ten times as much one tenth as often. That is what “it could pay \$500” is worth once you price it properly. The whole top bar comes to \$1.25, the bottom bar is \$2, and the ticket is a bad deal by precisely the distance between their two ends.

Build the skill 3: expected value has a partner: variance (the spread)

Expected value tells you the center of a distribution but nothing about how spread out it is, and spread is exactly what “risk” means. Pair every expected value with the variance, the probability-weighted average of the squared distance from the mean:

\(\operatorname{Var}(X) = \sum_i (x_i - E[X])^2\, p_i, \qquad \text{SD}(X) = \sqrt{\operatorname{Var}(X)}\).

Work it for a small random variable. A campus help-desk logs \(x = 0, 1, 2, 3\) server outages per week with probabilities \(0.50, 0.30, 0.15, 0.05\). First the center: \(E[x] = 0(0.50) + 1(0.30) + 2(0.15) + 3(0.05) = 0.75\) outages per week. Now the spread, row by row:

Summing the last column gives \(\operatorname{Var}(x) = 0.7875\) and \(\text{SD}(x) = \sqrt{0.7875} \approx 0.89\) outages. The standard deviation is in the same units as \(x\), which is what makes it readable: the help-desk averages \(0.75\) outages a week, and a typical week misses that average by about \(0.89\) outages. “About \(0.89\)” is not a promise that every week lands within \(0.75 \pm 0.89\); it is the square root of the average squared miss, so it is the size of a normal-sized error, not a boundary. Some weeks will be further out, and a week can never be below zero outages, so the misses are not even symmetric here. Two choices can share an expected value while one has a tiny SD (steady) and the other a large SD (a wild swing), and a decision-maker who hates surprises will not treat those as equal. Next week that gap is all you have: Week 3 takes the probabilities away entirely, so a criterion that states how much downside you can absorb is the only thing left to choose by.

In practice: expected value ignores risk and “the long run” may never arrive

• EV says nothing about spread (risk). Two choices can share an expected value while one is a sure thing and the other swings wildly. The variance and standard deviation you just computed (\(\text{SD} \approx 0.89\) for the outages) are how we measure that spread, and most people, and most businesses, care about it, not just the mean. A small certain loss and a rare-but-huge loss can carry the same EV yet feel nothing alike. Week 3 strips the probabilities away and makes you choose by a criterion that names your tolerance for the downside; Week 6 brings the probabilities back and puts a curve on that tolerance, called utility.
• A single play is not the long run. “Lose \$0.75 on average” describes thousands of tickets; buy one and you almost certainly get \$0 (or, very rarely, \$500). EV is the right lens for a business making the same bet repeatedly, and a weaker lens for a one-shot, high-stakes choice.
• Garbage in, garbage out. The whole calculation rests on the published probabilities being honest and complete. If the odds are wrong or a prize tier is missing, the tidy \$1.25 is fiction.

Where this is heading: flipping a conditional with Bayes

Conditional probability runs both directions, and the reverse direction is where decisions with information get made. A food truck’s espresso machine is well-calibrated with probability \(0.70\) or drifting with probability \(0.30\). A well-calibrated machine pulls a bad shot only \(5\%\) of the time; a drifting one \(25\%\) of the time. The multiplication rule for dependent events, \(P(A \cap B) = P(A)\,P(B \mid A)\), gives the joint “drifting and bad” \(= 0.30 \times 0.25 = 0.075\), and the chance of a bad shot overall is

Now flip the question: you just pulled a bad shot. How likely is the machine actually drifting?

A single bad shot revised your belief from \(0.30\) up to about \(0.68\). That update from a prior to a posterior using evidence is Bayes’ rule, and it is the named payoff of conditional probability we cash in during Week 6 (decision analysis with sample information).

Back to the puzzle

The gross expected winnings are \(0(0.939) + 5(0.05) + 50(0.01) + 500(0.001) = \mathbf{\$1.25}\) per ticket. Subtract the \$2 price and the expected net result is \(1.25 - 2.00 = \mathbf{-\$0.75}\), a losing bet on average. That’s why the running mean in the widget settles near \(-\$0.75\) rather than at \$0 or \$500. Play for fun or to support the cause if you like, but as an investment the math is clear: on expected value, skip it. (If the cause matters to you, the \$0.75 “loss” is really a donation, which is a perfectly good reason to play that EV alone can’t capture.)

After-class check

Questions spanning the addition and multiplication rules, independence vs. mutual exclusivity, joint and conditional probability from a two-way table, and expected value with its spread (variance). Retry freely; nothing is submitted.

Weekly take-home (graded: submit on Canvas)

Everything for the take-home lives in one starter workbook, and nothing in it is pre-computed. The model sheet opens with a case brief: a term of shuttle trip logs, dispatch’s own record of call-outs by month, and a service-contract quote priced by the year for a decision taken by the month. In the Your model block you name the random quantity and its time period and say what has to be converted before the two options can be compared at all; then you turn the log into a two-way table and test whether the two routes really behave independently. Neither of the figures you need arrives on the basis the manager decides on, so the contract price and the failure risk each get converted before they meet. You also measure the spread, because expected value alone points one way here and risk points the other.

Then the workbook turns the question round. Rather than asking what the quoted contract costs, it asks what the cover is actually worth: the monthly price at which the manager would not care either way. That ceiling is the number she can negotiate with, and it is yours to derive, not to look up. One short response plus the required AI-use disclosure go on the Free responses sheet. Upload the completed .xlsx on Canvas. That one file is your entire submission.