MBA Decision Trees Concept Card Template

Decision Trees Concept Card: [Your decision or course]

A concept card for decision trees, the core tool of the decision analysis session in an MBA data and decisions course. It covers expected value and rollback, the value of perfect and sample information (EVPI, EVSI), risk attitude with utility functions and certainty equivalents, and sensitivity analysis. It is for MBA students preparing for a class or exam, and for managers who need to structure a launch, investment or testing decision under uncertainty. Source basis: von Neumann and Morgenstern (1944), Raiffa and Schlaifer (1961), Raiffa (1968), Howard (1966) and Pratt (1964), written up in original words with original fictional examples (Northfield Outdoor Co., Corvane Robotics, Tallow Creek Medical Center). Every number was computed in Python and is shown with its Excel formula. Replace every [bracketed] value and every row marked Example.

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When to use it

One-sentence rule: use a decision tree when you must choose now, the result depends on uncertain events you cannot control, and later choices may depend on what you learn in between.

Typical MBA uses: product launch versus regional pilot, build versus license a technology, bid or no bid, settle or go to court, invest in a capacity expansion before demand is known, and whether a market test is worth its cost.

The idea in plain language

A decision tree is a map of a decision that unfolds over time. It uses three kinds of node:

1. Decision nodes (drawn as squares): points where you choose. Each branch is an option you control.
2. Chance nodes (drawn as circles): points where something uncertain happens. Each branch is an outcome, and the probabilities on the branches leaving one chance node add up to 1.
3. End nodes (drawn as triangles or simply as the payoff): the value you end up with on that path, usually a net present value or a profit, net of all costs incurred along the path.

You build the tree from left to right, in the order things happen in time. You then solve it from right to left, a procedure called rollback (or folding back): at each chance node, replace the branches by their probability-weighted average (the expected value); at each decision node, keep the best branch and prune the others. When you reach the root, you have the best first decision and a full contingent plan ("launch nationally if the test is favourable, regionally if not").

Two extensions turn the tree from arithmetic into management:

• Value of information. Before you act, you can often buy information. The most you should ever pay for any information is the expected value of perfect information (EVPI): the gain if a clairvoyant told you the outcome before you chose. Real tests are imperfect, so their worth is the expected value of sample information (EVSI), which you compute with Bayes' rule. Pay for a test only when EVSI exceeds its cost.
• Risk attitude. Expected value treats a 50 percent chance of USD 10m and a sure USD 5m as equal. Most firms, and most people, are not indifferent when the stakes are large relative to their wealth. A utility function, built on the axioms of von Neumann and Morgenstern (1944), converts payoffs into utilities; you then roll back expected utility and translate the result into a certainty equivalent, the sure amount you would accept in place of the gamble.

Formulas and notation

Expected value of an alternative $a$ with states $s_1, \dots, s_n$:

$EV(a) = \sum_{i=1}^{n} p_i \, v(a, s_i)$

Rollback. At a chance node take $EV$; at a decision node take $\max_a EV(a)$. The value of the tree without information is $EV_{\text{woi}} = \max_a EV(a)$.

Expected value of perfect information:

$EVPI = \sum_{i} p_i \max_a v(a, s_i) \; - \; \max_a \sum_i p_i \, v(a, s_i) = EV_{\text{wpi}} - EV_{\text{woi}}$

Bayes' rule for a test result $r$ (for example favourable or unfavourable):

$P(s_i \mid r) = \frac{P(r \mid s_i)\, p_i}{\sum_j P(r \mid s_j)\, p_j}, \qquad P(r) = \sum_j P(r \mid s_j)\, p_j$

Expected value of sample information:

$EVSI = \sum_{r} P(r) \max_a \sum_i P(s_i \mid r)\, v(a, s_i) \; - \; EV_{\text{woi}}, \qquad \text{efficiency} = \frac{EVSI}{EVPI}$

Exponential utility and certainty equivalent (constant absolute risk aversion, risk tolerance $R$):

$U(x) = 1 - e^{-x/R}, \qquad CE = -R \ln\!\Big(\sum_i p_i \, e^{-x_i/R}\Big), \qquad \text{risk premium} = EV - CE$

Two properties to remember. First, $EVSI \le EVPI$ always, and $EVPI \ge 0$ always: information can never have negative expected value to a decision maker who uses it rationally (it can, of course, be overpriced). Second, information has value only if some possible result would change your decision. If every test result leads to the same action, EVSI is zero, however accurate the test.

Assumptions and limits

A deeper limit: a tree describes one decision maker. In organisations, people who estimate probabilities may have incentives to inflate them (the champion of a project) or deflate them (a sandbagging sales team). Separating the people who estimate inputs from the people who benefit from the decision is a governance choice, not a mathematical one. Behavioural research also shows that people do not naturally follow expected utility (Kahneman and Tversky 1979); the tree is a prescriptive tool for how to decide, not a description of how people do decide.

How managers use it in practice

• Structure before numbers. In practice the biggest value of a tree is that it forces a team to agree on the options, the uncertainties and the order in which things happen. Many arguments disappear once the tree is drawn.
• Contingent plans. The output is a policy, not a single action: "test first; launch nationally only if the test is favourable". That policy is what goes into the board paper.
• Pricing information. EVPI is a quick ceiling for market research budgets, pilots, due diligence and expert advice. If a research agency quotes more than EVPI, decline without further analysis.
• Thresholds instead of point estimates. Managers rarely agree on a probability, but they often agree on which side of a threshold it lies. A break-even probability ("national wins if the chance of high demand exceeds 16.7 percent") is a more useful output than an EV.
• Risk attitude as policy. Large firms often set a corporate risk tolerance (for example a fraction of equity or of annual net income) so that business units value risky projects consistently. Use the value in your own firm's policy and treat any rule of thumb as a starting point only.
• Link to valuation. End values are usually NPVs from a cash flow model (mba/mba-dcf-valuation-card); a tree with an option to expand or abandon later is a real option (mba/mba-options-real-options-card).

Worked example A (fully solved): Northfield heated-jacket launch

Situation (fictional). Northfield Outdoor Co. has developed a heated jacket. The team can launch nationally, launch in one region only, or not launch. Demand next season is uncertain: high, medium or low. A two-month test market in three stores costs USD 0.40m. All payoffs are NPVs in USD m, net of launch costs, before the test cost.

From past tests, the market team estimates the test's likelihoods: P(favourable | high) = 0.80, P(favourable | medium) = 0.50, P(favourable | low) = 0.15.

Step 1: expected values without information

Put probabilities in B2:D2 and the payoff rows in B3:D5.

$EV_{\text{woi}} = 3.90$ (national launch), Excel =MAX(E3:E5).

Step 2: value of perfect information

With a clairvoyant, Northfield would launch nationally if demand is high (12.0), nationally if medium (3.0) and not at all if low (0.0).

$EV_{\text{wpi}} = 0.30(12) + 0.50(3) + 0.20(0) = 5.10, \qquad EVPI = 5.10 - 3.90 = 1.20$

Excel: =SUMPRODUCT(B2:D2,B7:D7)-E7, where row 7 holds =MAX(B3:B5) copied across. No test, survey or consultant can be worth more than USD 1.20m to this decision.

Step 3: posterior probabilities with Bayes' rule

Excel: joint =B2*B3, P(fav) =SUMPRODUCT(B2:D2,B3:D3), posterior =B4/$E$4.

Step 4: roll back the "test first" branch

$EV_{\text{wsi}} = 0.52(6.6346) + 0.48(1.3125) = 4.08, \qquad EVSI = 4.08 - 3.90 = 0.18$

Efficiency = 0.18 / 1.20 = 15 percent. Excel: posterior EVs =SUMPRODUCT(posterior_row,B3:D3), then =P_fav*MAX(fav_EVs)+P_unfav*MAX(unfav_EVs)-E_woi.

Step 5: decision

Net of its cost, testing is worth 4.08 - 0.40 = 3.68 < 3.90. The test destroys USD 0.22m of expected value: its information is worth only USD 0.18m because a favourable result is common even in medium demand and an unfavourable result only shifts the choice from national to regional, which saves 0.38m in that branch. Decision: launch nationally without the test, unless a test can be bought for less than USD 0.18m or made more accurate.

Step 6: sensitivity to the chance of high demand

Hold P(medium) at 0.50 and let P(high) = $x$, P(low) = $0.5 - x$. Then $EV_{\text{national}} = 18x - 1.5$ and $EV_{\text{regional}} = 6x + 0.5$. They cross at $x = 2/12 = 0.1667$. National beats no launch when $x > 0.0833$, and regional beats no launch for every $x$ in the range. Excel: build a one-variable data table (What-If Analysis, Data Table) on B2 with =0.5-B2 in D2, or solve the break-even with Goal Seek.

Step 7: risk attitude

Suppose Northfield's board sets a risk tolerance of $R$ = USD 10m for this business unit. With $U(x) = 1 - e^{-x/10}$:

Excel: expected utility =SUMPRODUCT($B$2:$D$2,1-EXP(-B3:D3/$R$1)), certainty equivalent =-$R$1*LN(1-E3).

With $R$ = 10 the ranking flips: the regional launch has the higher certainty equivalent, because the national launch carries a 20 percent chance of losing USD 6m. The two certainty equivalents are equal at $R$ = 11.02 (found with Goal Seek or a root finder). With $R$ = 5 the national launch is worth only 0.17 for sure against 1.86 for the regional launch. Recommendation to the board: the answer depends on how much downside the firm can absorb; state the threshold risk tolerance (about USD 11m) rather than only one answer.

The same rollback in Python

The script below reproduces Steps 1 to 4 and prints $EV_{\text{woi}}$, EVPI and EVSI. Change the arrays to reuse it for your own decision.

import numpy as np
p = np.array([0.30, 0.50, 0.20])
pay = {"National": np.array([12.0, 3.0, -6.0]),
       "Regional": np.array([5.0, 2.0, -1.0]),
       "No launch": np.zeros(3)}
lik_fav = np.array([0.80, 0.50, 0.15])

ev_woi = max(p @ v for v in pay.values())
ev_wpi = p @ np.max(np.vstack(list(pay.values())), axis=0)
p_fav = p @ lik_fav
post_f, post_u = p * lik_fav / p_fav, p * (1 - lik_fav) / (1 - p_fav)
ev_wsi = p_fav * max(post_f @ v for v in pay.values()) \
    + (1 - p_fav) * max(post_u @ v for v in pay.values())
print(round(ev_woi, 2), round(ev_wpi - ev_woi, 2), round(ev_wsi - ev_woi, 2))
# 3.9 1.2 0.18

Worked example B (half solved): Corvane Robotics gripper

Situation (fictional). Corvane Robotics needs a new gripper for its warehouse robot. It can license an existing design now (NPV USD 8.0m) or build in-house at an R&D cost of USD 4.0m. Engineering puts the chance of success at 0.60. If the build succeeds, the product line is worth USD 15.0m (before R&D). If it fails, Corvane must license later on worse terms, worth USD 6.0m (before R&D).

1. Draw the tree: one decision node (build or license), one chance node after "build". Payoffs net of R&D: success $15.0 - 4.0 = 11.0$, failure $6.0 - 4.0 = [\ ]$, license 8.0.
2. $EV(\text{build}) = 0.60 \times 11.0 + 0.40 \times [\ ] = [\ ]$. Excel =0.6*11+0.4*[cell].
3. Best decision without information: [ ], $EV_{\text{woi}} = [\ ]$.
4. With perfect information: if success is foreseen, build (11.0); if failure is foreseen, [ ] ([ ]). $EV_{\text{wpi}} = 0.60 \times 11.0 + 0.40 \times [\ ] = 9.80$, so $EVPI = [\ ]$.
5. Break-even probability of success at which build and license have equal EV: solve $p(11.0) + (1 - p)(2.0) = 8.0$, so $p = [\ ]$. Excel: Goal Seek on the EV cell.
6. A three-month prototype sprint that would reveal success or failure perfectly costs USD 1.5m. Is it worth doing? [ ] Explain in one sentence.

Exercise C (unsolved): Tallow Creek imaging capacity

Situation (fictional). Tallow Creek Medical Center must expand imaging capacity. Demand over five years is high (probability 0.45) or low (0.55). Net NPVs in USD m, after all costs:

1. Draw the tree and compute the EV of each alternative. Which is best?
2. Compute EVPI. A regional demand study costs USD 0.25m. Should the board consider it before computing EVSI?
3. The study is right 80 percent of the time in either state (P(says high given high) = P(says low given low) = 0.80). Compute the posteriors, roll back, and find EVSI. Is the study worth USD 0.25m?
4. Find the probability of high demand at which buying and leasing have equal EV.
5. The hospital's risk tolerance is USD 3m. Compute the certainty equivalents with exponential utility. Does the choice change?

Connections to other cards

• Probability and Bayes: posteriors are Bayes' rule in a table; revise base rates with academic/stats-probability-basics and academic/stats-bayesian-inference.
• NPV: end values are NPVs of cash flows, built in mba/mba-npv-irr-card and mba/mba-dcf-valuation-card.
• Real options: a tree with later decisions (expand, abandon, wait) is the discrete form of mba/mba-options-real-options-card.
• Game theory: when the "chance" node is really a rival's choice, switch to mba/mba-game-theory-card.
• Simulation: when a few branches cannot represent the uncertainty, use the Monte Carlo session in mba/mba-decision-analytics-guide.
• Regression: probabilities and likelihoods can be estimated from data, for example a test's hit rate from past launches, with mba/mba-regression-for-managers-card or academic/stats-logistic-regression.
• AI-assisted decisions: model scores feed priors and likelihoods; see mba/mba-ai-decision-making-card for when to trust them.

Common mistakes and misconceptions

1. Drawing the tree in the wrong time order. A chance event that resolves after a decision must sit to its right. Putting the test result after the launch decision makes the information useless by construction.
2. Double counting costs. The test cost is charged once on its branch; payoffs at the end nodes are net of costs incurred on that path, not of costs on other paths.
3. Probabilities that do not add to 1 at a chance node, often because a state was forgotten.
4. Using priors instead of posteriors after a test. The whole point of the test branch is that probabilities change.
5. Confusing likelihoods and posteriors. P(favourable given high) = 0.80 does not mean P(high given favourable) = 0.80; in Worked example A it is 0.46.
6. Thinking EVPI is what a test is worth. EVPI is a ceiling; a real test is worth EVSI, which is often much less.
7. Valuing information that cannot change the decision. If every result leads to the same action, EVSI = 0.
8. Reading EV as the likely outcome. In Worked example A, the national launch never pays USD 3.90m: it pays 12, 3 or -6. EV is a long-run average and a ranking device.
9. Ignoring risk attitude for bet-the-company decisions. Use certainty equivalents when a loss would threaten the firm.
10. Point answers without sensitivity. Always report the break-even probability or risk tolerance where the decision flips.

Self-test questions

1. Name the three node types and the rollback rule at each.
2. Why is EVPI never negative, and why is EVSI never larger than EVPI?
3. In Worked example A, what is P(favourable)? Show the formula.
4. A test has P(positive given good) = 0.9 and P(positive given bad) = 0.9. What is its EVSI, and why?
5. In Worked example A, how cheap would the test have to be for testing to be worth it?
6. State the certainty equivalent formula for exponential utility. What does a larger $R$ mean?
7. Two alternatives have the same EV. One has a wider spread of payoffs. Which does a risk-averse decision maker prefer?
8. What do the von Neumann and Morgenstern axioms guarantee, in one sentence?
9. In Worked example A, P(high) falls from 0.30 to 0.15 with P(medium) unchanged. Which launch is best now?
10. Give one reason a manager might prefer to report a break-even probability rather than an EV.

My notes and next steps

Sources and further reading

• von Neumann, J., and Morgenstern, O. (1944). Theory of Games and Economic Behavior. Princeton University Press.
• Raiffa, H., and Schlaifer, R. (1961). Applied Statistical Decision Theory. Division of Research, Graduate School of Business Administration, Harvard University.
• Raiffa, H. (1968). Decision Analysis: Introductory Lectures on Choices under Uncertainty. Addison-Wesley.
• Howard, R. A. (1966). Information value theory. IEEE Transactions on Systems Science and Cybernetics, 2(1), 22-26.
• Pratt, J. W. (1964). Risk aversion in the small and in the large. Econometrica, 32(1/2), 122-136.
• Bayes, T. (1763). An essay towards solving a problem in the doctrine of chances. Philosophical Transactions of the Royal Society of London, 53, 370-418.
• Kahneman, D., and Tversky, A. (1979). Prospect theory: an analysis of decision under risk. Econometrica, 47(2), 263-291.
• Textbooks (further reading, by name only): Clemen and Reilly, Making Hard Decisions; Hammond, Keeney and Raiffa, Smart Choices; Keeney and Raiffa, Decisions with Multiple Objectives.
• Open resource: OpenStax, Introductory Statistics 2e (CC BY-NC-SA 4.0), chapters on probability and Bayes' rule, at openstax.org.

Answer key

Worked example B. (1) Failure payoff $6.0 - 4.0 = 2.0$. (2) $EV(\text{build}) = 0.60(11.0) + 0.40(2.0) = 7.40$. (3) License, $EV_{\text{woi}} = 8.00$. (4) If failure is foreseen, license (8.0); $EV_{\text{wpi}} = 0.60(11.0) + 0.40(8.0) = 9.80$, $EVPI = 9.80 - 8.00 = 1.80$. (5) $11p + 2(1 - p) = 8 \Rightarrow 9p = 6 \Rightarrow p = 0.6667$. (6) Yes on expected value: a perfect sprint is worth up to EVPI = 1.80 and costs 1.50, a net gain of 0.30. The margin is thin, so check the sprint's real accuracy: an imperfect sprint is worth EVSI, which may be below 1.50.

Exercise C. (1) EV buy $= 0.45(2.8) + 0.55(-1.2) = 0.60$; EV lease $= 0.45(1.6) + 0.55(0.4) = 0.94$; EV outsource $= 0.50$. Lease is best. (2) $EV_{\text{wpi}} = 0.45(2.8) + 0.55(0.5) = 1.535$; $EVPI = 0.595$. The study costs less than EVPI, so it is worth computing EVSI. (3) P(says high) $= 0.45(0.80) + 0.55(0.20) = 0.47$; posteriors given "high" 0.7660 and 0.2340, given "low" 0.1698 and 0.8302. After "high", buy (EV 1.864 against lease 1.319); after "low", lease (0.604 against outsource 0.500 and buy -0.521). $EV_{\text{wsi}} = 0.47(1.864) + 0.53(0.604) = 1.196$, so $EVSI = 1.196 - 0.94 = 0.256$, efficiency 43 percent. The study is worth only USD 0.006m more than its cost of 0.25m: break-even in practice, so it pays only if its accuracy is really 80 percent. (4) $2.8q - 1.2(1 - q) = 1.6q + 0.4(1 - q)$ gives $q = 1.6/2.8 = 0.5714$. (5) With $R = 3$: CE buy 0.01, lease 0.88, outsource 0.50. Leasing stays best, and buying loses almost all its value to the risk premium. Check these with the Python rollback in Worked example A by changing the inputs.

Self-test.

1. Decision node (square): take the maximum. Chance node (circle): take the probability-weighted average. End node: the payoff of that path.
2. With perfect information you can always do at least as well as without it, by choosing the same action, so $EV_{\text{wpi}} \ge EV_{\text{woi}}$. Sample information is a noisy version of perfect information, so it cannot beat it on expected value.
3. $P(\text{fav}) = 0.30(0.80) + 0.50(0.50) + 0.20(0.15) = 0.52$; Excel =SUMPRODUCT(B2:D2,B3:D3).
4. Zero. The result is equally likely in every state, so posteriors equal priors and the decision cannot change.
5. Below USD 0.18m (EVSI).
6. $CE = -R\ln(\sum_i p_i e^{-x_i/R})$. A larger $R$ means more tolerance for risk; as $R \to \infty$, $CE \to EV$.
7. The one with the narrower spread: its certainty equivalent is higher because its risk premium is smaller.
8. If preferences satisfy completeness, transitivity, continuity and independence, there is a utility function such that the decision maker prefers the gamble with the higher expected utility.
9. At $x = 0.15$: national $18(0.15) - 1.5 = 1.20$, regional $6(0.15) + 0.5 = 1.40$. Regional is best, since 0.15 is below the break-even 0.1667.
10. Managers often disagree on the exact probability but agree on which side of a threshold it lies; the threshold also shows how robust the choice is.

Revision history

Standard and licence note

Based on the decision analysis tradition of von Neumann and Morgenstern (1944), Raiffa and Schlaifer (1961), Raiffa (1968), Howard (1966) and Pratt (1964). The text, examples, data and exercises are original and fictional; no textbook, published case or course material was copied or paraphrased. This template is not affiliated with or endorsed by any business school, publisher or author. Check the cited works for exact definitions before you rely on them in graded work. Template licence: MIT.