Chapter 4 of 8

Probability, Expected Value, and Decision Trees

3 sections · ~9 min read · 3 practice questions · part of Business Analytics

Learning objectives

  • Build a decision tree with decision nodes, chance nodes, and payoffs stated in a single currency
  • Update probabilities with Bayes rule after an imperfect test and fold the tree back to an expected value
  • Compute the value of information and the value of perfect information, and know when expected value is the wrong criterion

Trees, folding back, and the discipline of a single currency

A decision tree separates the two things that get conflated in argument: choices you control and events you do not. Squares are decision nodes with one branch per option; circles are chance nodes with one branch per outcome, each carrying a probability, and the probabilities on the branches from a chance node must sum to one. Terminal nodes carry payoffs. Solve by folding back from right to left. At each chance node compute the expected value, which is the probability-weighted sum of the branches. At each decision node take the best available value, because you will be there to choose. The value that rolls back to the root is the expected value of the whole plan, and the branches selected along the way are the plan itself. Two disciplines make this work. First, every payoff must be in the same currency and on the same basis, usually incremental after-tax cash relative to a clearly stated do-nothing baseline. Second, sunk costs never appear. Money already spent is identical down every branch and cannot affect a comparison, so including it wastes effort and invites the escalation error of continuing a losing project because of what has already gone into it.

Imperfect tests and Bayes rule

Most real information is imperfect: a market test, a pilot, a diagnostic, a credit score. The value of such information comes entirely from whether it would change what you do. Bayes rule converts test performance into revised beliefs. The probability that the world is in state H given a positive signal equals the probability of the signal given H, times the prior probability of H, divided by the total probability of the signal across all states. The most reliable way to do this by hand is not to memorise the formula but to build a joint table: multiply each prior by each conditional to get joint probabilities, sum down a signal column to get the marginal probability of that signal, and divide each joint entry by that column total. The rows and columns then have to reconcile to one, which catches arithmetic slips immediately. The lesson that survives from this arithmetic is the base rate effect: when a state is rare, even a fairly accurate test produces mostly false positives, because the many opportunities to be wrong about the common state swamp the few chances to be right about the rare one.

The value of information and the limits of expected value

The expected value of information is the difference between the expected value of the plan when you can act on the signal and the expected value of the best plan without it, before subtracting the cost of obtaining it. That difference is the maximum you should ever pay. A crucial corollary is that information which cannot change your decision is worth exactly zero, no matter how interesting it is; if you would launch on a positive result and launch anyway on a negative one, the test is pure cost. The expected value of perfect information sets the ceiling: it is what a flawless oracle would be worth, and comparing an imperfect test against it shows what fraction of the available value that test captures. Finally, expected value is the right criterion only when the decision repeats often enough for averages to assert themselves, and when no single outcome threatens the firm. A gamble with a positive expected value but a downside that breaches a debt covenant or exhausts the cash runway is not a good bet, because there is no second draw. In those cases decide on a utility that curves with wealth, or add an explicit constraint capping the probability of ruin.

Worked example

Problem

A hypothetical consumer goods firm is deciding whether to launch a new product. Launch requires 4.0 million dollars in fixed cost. If demand turns out high, the launch generates 9.0 million dollars in contribution; if low, 1.5 million dollars. Management judges the probability of high demand to be 0.40. A market test costs 0.25 million dollars. Historically the test returns a positive read for 80 percent of products that turn out high demand and for 30 percent of those that turn out low demand. Should the firm launch without testing, and what is the test worth?

Step by step

  1. Payoffs net of the launch cost. High demand: 9.0 - 4.0 = +5.0 million. Low demand: 1.5 - 4.0 = -2.5 million. Not launching: 0, the baseline.
  2. Expected value of launching with no test = 0.40 x 5.0 + 0.60 x (-2.5) = 2.00 - 1.50 = +0.50 million dollars. This beats 0, so without a test the firm launches, and the tree value at the root is 0.50 million.
  3. Joint probability table for the test. High and positive = 0.40 x 0.80 = 0.32. High and negative = 0.40 x 0.20 = 0.08. Low and positive = 0.60 x 0.30 = 0.18. Low and negative = 0.60 x 0.70 = 0.42. The four sum to 0.32 + 0.08 + 0.18 + 0.42 = 1.00, which checks.
  4. Marginal probability of a positive read = 0.32 + 0.18 = 0.50. Marginal probability of a negative read = 0.08 + 0.42 = 0.50.
  5. Posterior after a positive read: probability of high demand = 0.32 / 0.50 = 0.64, so probability of low demand = 0.36.
  6. Posterior after a negative read: probability of high demand = 0.08 / 0.50 = 0.16, so probability of low demand = 0.84.
  7. Branch value after a positive read: launching gives 0.64 x 5.0 + 0.36 x (-2.5) = 3.20 - 0.90 = +2.30 million, which beats 0, so launch. Branch value = 2.30.
  8. Branch value after a negative read: launching gives 0.16 x 5.0 + 0.84 x (-2.5) = 0.80 - 2.10 = -1.30 million, which loses to 0, so do not launch. Branch value = 0.
  9. Expected value of the test-then-decide plan, before the test fee = 0.50 x 2.30 + 0.50 x 0 = 1.15 million dollars.
  10. Expected value of the test information = 1.15 - 0.50 = 0.65 million dollars, which is the most the firm should pay for this test.
  11. Net of the 0.25 million fee, the test-then-decide plan is worth 1.15 - 0.25 = 0.90 million, against 0.50 million for launching blind, so testing adds 0.40 million dollars of expected value.
  12. Expected value with perfect information = 0.40 x 5.0 + 0.60 x 0 = 2.00 million, so the expected value of perfect information = 2.00 - 0.50 = 1.50 million. The imperfect test captures 0.65 / 1.50 = 43 percent of that ceiling.

Answer. Without testing the firm should launch, with an expected value of +0.50 million dollars. With the test, the firm launches only on a positive read, giving 1.15 million dollars before the fee and 0.90 million after it, so testing is the better plan by 0.40 million dollars. The information itself is worth up to 0.65 million dollars and the fee must stay below that. Perfect information would be worth 1.50 million dollars, so this test captures about 43 percent of the theoretical maximum. All figures are expected incremental contribution in millions of dollars over the product life, excluding sunk development spending, and they assume the stated probabilities and test reliabilities are correct. Note that the unaided launch has a 60 percent chance of losing 2.5 million dollars; if a loss of that size would breach a covenant or exhaust the cash runway, expected value is the wrong criterion and the test is worth more than 0.40 million as insurance.

Practice

Work each question before opening the solution.

  1. Using the same payoffs of +5.0 and -2.5 million dollars, recompute the no-test expected value if the prior probability of high demand is 0.55 instead of 0.40, and state whether the firm would still test.

    Show solution for question 1

    Expected value of launching = 0.55 x 5.0 + 0.45 x (-2.5) = 2.75 - 1.125 = +1.625 million dollars. The firm launches. It is still worth checking the test, but its value falls, because a stronger prior means a negative read is less likely to flip the decision. The general rule is that information is worth most when the prior is near the point of indifference, which here is the probability p satisfying 5p - 2.5(1 - p) = 0, giving p = 2.5 / 7.5 = 0.333. At a prior of 0.55 the decision is more robust and the test buys less.

  2. A fraud model flags 5 percent of transactions. It catches 90 percent of genuine fraud, and true fraud occurs in 1 percent of transactions. Out of 10,000 transactions, how many flagged transactions are actually fraudulent, and what does that imply about the review queue?

    Show solution for question 2

    Of 10,000 transactions, 100 are fraudulent and 9,900 are not. The model catches 0.90 x 100 = 90 of the fraud. It flags 5 percent of all transactions, which is 500, so the remaining 500 - 90 = 410 flags are false. Precision is 90 / 500 = 18 percent, meaning roughly four of every five reviewed transactions are legitimate. This is the base rate effect: fraud is rare, so even a sensitive model fills the queue with honest customers. Whether the model is worth running depends on the cost per review against the loss avoided per catch, not on the 90 percent catch rate.

  3. A proposed pilot would cost 100,000 dollars and give a much better read on demand, but the operations director says the plant is being built either way for capacity reasons. What is the pilot worth?

    Show solution for question 3

    Zero, and the firm should not run it. Information has value only through the decision it changes; if the build proceeds under every possible pilot result, both branches lead to identical payoffs and the expected value with the pilot equals the expected value without it, minus the 100,000 dollars. The correct response is to ask whether the pilot could change any decision, such as capacity sizing, launch timing, or pricing. If it could change one of those, re-scope the pilot around that decision and value it against the payoffs of that choice instead.