Why the Order of Decisions Matters
In a simultaneous game, each player chooses without observing the other player’s current choice. In a sequential game, at least one player acts after observing earlier choices. That difference matters: a decision maker can condition an action on what has already happened, and an earlier player must take the later response into account.
A game tree records the order of decisions. Its branches show available actions, its decision nodes show who chooses, and its terminal points show the payoffs when no further choices remain. This tutorial focuses on finite games of perfect information: every decision is observed by the players who move later, the game ends after finitely many decisions, and no player moves simultaneously with another at a node.
Payoffs are numbers that represent players’ preferences over outcomes; they need not be dollars. A payoff of 8 is preferred to a payoff of 5 for the same player. In the examples, payoff pairs are ordered as \((\text{Player A},\text{Player B})\). If a decision has consequences occurring at different times, treat the listed payoff as the value relevant to the player’s choice; any timing adjustment must already be reflected in that payoff.
The Backward-Induction Method
Backward induction starts at the end of the tree and works toward the beginning. At each decision node, keep the branch that gives the player moving there the highest payoff, assuming choices farther along the tree have already been solved. When the process reaches the first node, the retained choices describe the predicted path.
At each node with no later decisions, compare the terminal payoffs for the player who moves there.
Retain the action that maximizes the moving player’s own payoff. Record the outcome that follows from that action.
At an earlier node, compare the payoffs from the outcomes implied by the already-resolved later choices—not from every branch that was originally available.
The selected action at each node specifies a contingent plan; following that plan from the first node gives the predicted path and terminal payoffs.
The justification is an induction on the number of decisions remaining. At a node with no decisions after it, the moving player can compare the terminal payoffs directly and choose the best one. Now suppose the correct choices have been identified at every node with at most \(k\) decisions remaining. At a node with \(k+1\) decisions remaining, each available action leads to a later node whose choices have already been resolved. The player can therefore compare the resulting payoffs and choose the best action. Repeating this reasoning reaches the first node. The method works because, under the assumptions, players observe the relevant history and optimize their own payoff at each decision.
Here \(h\) is a decision history, \(i\) is the player moving at that history, \(A(h)\) is the set of available actions, and \(z(h,a)\) is the outcome implied by choosing action \(a\) and then following the choices already determined farther along the tree. If two or more actions give the same highest payoff, all are maximizing actions. Backward induction then need not produce a unique path.
Worked Example: A Launch Decision With a Competitive Response
A firm, A, must choose whether to invest in a digital service or wait. If A invests, rival B chooses whether to respond or ignore the launch. After a response, A chooses whether to discount or maintain its price. The terminal payoffs are: discount \((1,2)\), maintain \((0,4)\), and ignore \((5,1)\). If A waits, B chooses to expand or hold its current position; the resulting payoffs are expand \((0,3)\) and hold \((4,1)\). Find the choices implied by backward induction.
Method. Resolve the latest decisions first. After B responds to investment, A compares its own payoff from discounting, 1, with its payoff from maintaining the price, 0. Since \(1>0\), A discounts. That branch therefore implies payoff \((1,2)\).
At the investment node, B compares its payoff from responding, 2, with its payoff from ignoring, 1. Since \(2>1\), B responds. A’s investment decision therefore leads to the response branch and gives A payoff 1.
On the wait branch, B compares its payoff from expanding, 3, with its payoff from holding, 1. Since \(3>1\), B expands, giving A payoff 0. Finally, A compares investing, which yields 1 after the later choices, with waiting, which yields 0. Since \(1>0\), A invests.
Result and check. The path is invest, respond, discount, with terminal payoffs \((1,2)\). At each decision, the selected action gives the moving player a strictly higher payoff than the alternative. The method avoids assuming A can choose B’s action on the wait branch.
Worked Example: An Offer and a Possible Revision
Two organizations are negotiating a division of a project’s benefits, measured in payoff units. Organization A first chooses an offer: either \((6,4)\) or \((4,6)\). Organization B may accept or reject. If B rejects, A chooses whether to revise the offer to \((5,5)\) or withdraw, which gives \((0,0)\). After a revision, B chooses whether to accept it or reject it; rejection then gives \((0,0)\). Determine A’s initial offer and the resulting path.
Method. Begin with B’s response to a revised offer. B receives 5 by accepting and 0 by rejecting, so B accepts. Knowing that, A compares revising, which gives A 5, with withdrawing, which gives A 0. A revises.
Now consider B’s response to the initial offer \((6,4)\). Accepting gives B 4. Rejecting leads to A’s revision and then B’s acceptance, giving B 5. Because \(5>4\), B rejects the initial offer. For the initial offer \((4,6)\), accepting gives B 6, while rejecting still leads to a payoff of 5. Since \(6>5\), B accepts that offer.
At the first decision, A compares the consequences of the two offers. Choosing \((6,4)\) leads to a rejection, a revision, and a final payoff of 5 for A. Choosing \((4,6)\) is accepted immediately and gives A 4. Since \(5>4\), A chooses \((6,4)\).
Result and check. A makes the \((6,4)\) offer; B rejects; A revises to \((5,5)\); B accepts. The final payoffs are \((5,5)\). The initially proposed allocation is not itself the outcome because B’s later choice is part of the tree.
Worked Example: A Policy Choice With Several Decision Nodes
A university chooses whether to run a pilot program or launch a full program. A faculty group then chooses to support or resist the proposal. After support, the university chooses whether to expand or maintain the program; after resistance, it chooses whether to consult or cancel. Payoffs are \((\text{university},\text{faculty})\). For a pilot, the outcomes are: support then expand \((8,5)\), support then maintain \((5,6)\), resist then consult \((4,4)\), and resist then cancel \((1,2)\). For a full launch, they are: support then expand \((9,3)\), support then maintain \((6,5)\), resist then consult \((3,2)\), and resist then cancel \((2,4)\). Find the implied choices.
Method. At each of the university’s final decision nodes, compare the university’s own payoffs. Under a pilot with support, expanding gives 8 rather than 5, so the university expands. Under a pilot with resistance, consulting gives 4 rather than 1, so it consults. Under a full launch with support, expanding gives 9 rather than 6, so it expands. Under a full launch with resistance, consulting gives 3 rather than 2, so it consults.
The faculty group now compares its payoffs. After a pilot, support leads to payoff 5 for the faculty group, while resistance leads to payoff 4; it supports. After a full launch, support gives the faculty group 3, while resistance gives it 2; it supports again.
At the first decision, the university compares the outcome of a pilot, which gives it 8, with the outcome of a full launch, which gives it 9. It chooses the full launch.
Result and check. The path is full launch, support, expand, giving payoffs \((9,3)\). The faculty group’s payoff of 3 on this path is lower than its payoff of 5 after a pilot, but the faculty group chooses support at the full-launch node because its alternative there gives only 2. Each decision is judged using the options available at that node.
Common Mistakes and Limits
A frequent error is to compare an earlier player’s choices using a branch that a later player would not select. In the launch example, a mistaken calculation gives A payoff 4 from waiting by assuming B holds its position, then compares that 4 with the investment payoff of 1 and concludes that A waits. But B would not hold: B gets 3 from expanding and only 1 from holding. The correct wait payoff for A is therefore 0, so A invests. The correction is to solve each later decision using the payoff of the player who actually moves there.
- Do not maximize total payoffs. A player chooses to maximize that player’s own payoff, not the sum of everyone’s payoffs.
- Do not treat an available branch as certain. An earlier player’s comparison must use the choices implied by later decision makers.
- Do not confuse a plan with the realized path. A contingent plan specifies what a player would do at each of that player’s decision nodes; the realized path visits only some of them.
- Check for ties. If a player has equally good actions, more than one continuation may be consistent with payoff maximization, and the first decision may consequently have more than one predicted outcome.
Backward induction is useful whenever choices occur in a known order and later players observe earlier actions. It translates a complicated tree into a sequence of payoff comparisons, while preserving the fact that a player’s best action depends on the choices that follow. The method is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.
Check Your Understanding
For each game, apply backward induction and give the predicted path and terminal payoffs. Assume each player maximizes that player’s own payoff.
- Firm A chooses launch or defer. After launch, B chooses imitate or specialize. If B imitates, A chooses cut or hold, with payoffs \((4,3)\) and \((2,5)\), respectively. If B specializes, payoffs are \((6,1)\). After defer, B chooses expand, giving \((1,4)\), or wait, giving \((3,2)\). (Answer: Launch, imitate, cut; payoffs \((4,3)\).)
- A proposes either \((7,3)\) or \((3,7)\). B accepts or rejects. Following rejection, A chooses to revise to \((4,6)\) or exit for \((0,0)\); B then accepts or rejects the revision, with rejection yielding \((0,0)\). (Answer: A proposes \((7,3)\), B rejects, A revises to \((4,6)\), and B accepts; final payoffs \((4,6)\).)
- A park authority chooses redesign or closure, then a business group chooses support or oppose. The authority then chooses permit or deny. For redesign, the payoffs after support are permit \((7,4)\), deny \((3,2)\), and after opposition are permit \((5,1)\), deny \((2,3)\). For closure, after support they are permit \((6,2)\), deny \((4,1)\), and after opposition permit \((3,4)\), deny \((1,5)\). (Answer: Redesign, support, permit; payoffs \((7,4)\).)
- A chooses P or Q, then B chooses x or y. The payoffs are P-x \((5,4)\), P-y \((2,4)\), Q-x \((3,1)\), and Q-y \((4,2)\). Account for the tie at B’s decision after P. (Answer: There are two backward-induction outcomes: P then x, with \((5,4)\); or Q then y, with \((4,2)\).)
- A chooses North or South, then B chooses Bid or Pass. After Bid, A chooses Accept or Reject. For North, these payoffs are Accept \((6,5)\), Reject \((2,1)\), and Pass \((4,2)\). For South, they are Accept \((5,3)\), Reject \((1,4)\), and Pass \((3,2)\). (Answer: North, Bid, Accept; payoffs \((6,5)\).)