Class 3: Game Theory II: Sequential Moves and Commitment
Textbook reading: Kreps, Chapter 2, Section 2.1 (extensive-form games, information sets, moves by nature, strategies), Section 2.3 (backward induction), the end of Section 2.4 (Nash equilibrium in extensive-form games), and Sections 2.5 and 2.6 (the case-let).
Key Ideas
From tables to trees
- Last class every player moved at once, BUT most business situations are sequential:
- One firm enters and the other responds
- One side makes an offer and the other accepts or refuses
- One party invests and then finds out whether it paid off
- When the order of moves matters, draw the game as a tree instead of a table: extensive-form game
Nature as a player
- Sometimes, outcomes can’t be predicted exactly because nature is the one in control. Ex:
- Demand uncertainty
- Success of research and development
- Oil and mineral exploration
- Regulatory and legal outcomes
- Nature: a “player” who makes the moves no one controls, with probabilities on the branches
- A player facing Nature evaluates a gamble
Solving extensive form games: backward induction
- Start at the last move, where a player is choosing between payoffs. Mark their best response
- Step back one node: that player can anticipate what comes next, so they are also choosing between known outcomes
- Keep going to the start of the game
- The prediction assumes every player looks ahead and trusts everyone else to do the same
- A strategy is a full plan: what you would do at every node where you might have to move, including nodes you may never reach
The weak point is the trust
- The payoffs may leave out something players care about (fairness, resentment)
- A long chain of “I know that you know” is fragile
- Commitment, credibility, and reputation (Classes 4 and 5) are about what turns an empty threat into a real one
- When a prediction fails, look at the model first; the usual culprit is the assumption that people care only about money
Problems of the Day
Q7: Jan Moves First
Same Sam and Jan as last class, same payoffs (Sam’s first in each cell):
| Jan: Old Pros | Jan: Art Museum | Jan: Cafeen | |
|---|---|---|---|
| Sam: Old Pros | 6, 4 | 4, 3 | 4, 2 |
| Sam: Art Museum | 2, 1 | 5, 5 | 2, 2 |
| Sam: Cafeen | 1, 1 | 1, 3 | 3, 6 |
But now the rules change: Jan picks a place, goes there, and texts Sam “I’m at X and I’m not moving.” Sam reads the text and then chooses where to go.
Step 1: Draw this as a tree, left to right. Draw the letter J with a circle around it to represent Jan’s decision node. Then draw 3 lines coming out of it toward the right, and label each line with an O (Old Pros), A (Art museum), or C (Cafeen). At the end of each line, draw a S with a circle around it to represent Sam’s decision nodes, given Jan’s choice. Draw 3 lines out from each of Sam’s decision nodes, labeling each of those O, A, or C. Finish the tree with payoffs at the end of each of Sam’s lines, listing Sam’s payoffs first.
Step 2: Backward induction. Start at the last move. Cross off any lines that are not Sam’s best responses. Then move back a step to analyze what Jan should do, given what she expects Sam to do.
Step 3: Find the Nash Equilibrium. Just like last class, a Nash Equilibrium is a combination of strategies, one per player, such that no player can do better by changing their strategy alone, given what everyone else is doing.
What payoff does Jan end up with? She is the one to move first; why doesn’t she choose to go to Cafeen to get the highest payoff she possibly can?
Extension. Before Jan leaves the house, Sam texts her: “If you go to the museum, I’m going to Old Pros without you. If you go to Old Pros, I’ll meet you there.” Should Jan believe him? Is this a credible threat? Explain.
Q8: Entry Under Uncertainty
Part 1: Known costs. Two firms, A and B, are considering a new market. Firm A decides first whether to enter. Firm B sees whether A entered and then decides. If A is alone in the market it earns 30; if B is alone it earns 25. If both enter, A earns −5 and B earns −10. A firm that stays out earns 0.
Draw the tree, then solve for the Nash Equilibrium using backward induction.
Part 2: Uncertain costs. Same game as before, but now Nature makes the last move. Production costs are HIGH with a probability of 0.7 and LOW with a probability of 0.3.
Profits are:
| Costs high | Costs low | |
|---|---|---|
| A alone in the market | 30 | 60 |
| A sharing the market with B | -5 | 10 |
| B alone in the market | 25 | 50 |
| B sharing the market with A | -10 | 5 |
A firm that stays out earns 0.
Draw the tree same as before, but adding nature’s move at the end, labeling lines with the probabilities they occur. Calculate expected profits as weighted averages. Then use backward induction. Does the Nash Equilibrium change?
Q9: Two Bakeries
Two bakeries sell the same bread in the same town. Each morning each bakery decides how many hundred loaves to bake: 3, 4, or 6. The price depends on the total baked: with Q hundred loaves on the market the price is 12 − Q dollars per loaf. Ingredients cost nothing (bear with us), so a bakery baking q hundred loaves earns q × (12 − Q) hundred dollars. The table gives profits, Bakery A’s first:
| B bakes 3 | B bakes 4 | B bakes 6 | |
|---|---|---|---|
| A bakes 3 | 18, 18 | 15, 20 | 9, 18 |
| A bakes 4 | 20, 15 | 16, 16 | 8, 12 |
| A bakes 6 | 18, 9 | 12, 8 | 0, 0 |
Suppose both bakeries decide at the same time each morning. Star best responses for A: how many loaves will they never bake? Star best responses for B: how many loaves will they never bake? Use iterated dominance to explain how the bakeries will settle on the Nash Equilibrium. Why don’t both bakeries make 3 loaves, raising the price and both earning more?
Now suppose Bakery A bakes first, and Bakery B sees Bakery A’s loaves on the shelf before deciding. Draw the extensive form and use backward induction to solve for the Nash Equilibrium. Why is it so different from the part A equilibrium?
Moving first is an advantage when your move is observed, irreversible, and shapes the other player’s response in your favor. But moving first is not always an advantage: think about rock paper scissors!
Answers
- Q7:
- Step 1: Jan’s node on the left with branches O, A, C; a Sam node at the end of each with branches O, A, C; nine endpoints with the payoffs from the table, Sam’s first.
- Step 2: Sam’s best responses: if Jan is at Old Pros, Old Pros (6). If Jan is at the museum, the museum (5). If Jan is at Cafeen, Old Pros (4 beats 3 and 1). Cross off the other six branches. Jan now compares three outcomes: Old Pros gives her 4, the museum gives her 5, Cafeen gives her 2. She goes to the museum.
- Step 3: Jan’s strategy is “museum.” Sam’s strategy is a full plan: Old Pros if she goes to Old Pros, museum if she goes to the museum, Old Pros if she goes to Cafeen. Outcome (5, 5). Jan gets 5. Cafeen only pays 6 if Sam follows her there, and he will not: once she is at Cafeen, Sam’s best response is Old Pros, leaving her with 2. Moving first lets Jan pick which of last class’s two equilibria happens, but it does not let her pick a cell that is not an equilibrium.
- Extension: the threat is not credible. If Jan is at the museum, Sam gets 5 by joining her and 4 by going to Old Pros; carrying out the threat costs him. Jan should go to the museum anyway. Notice why Sam sends the text: if she believed it, she would go to Old Pros (4 beats 3), which is Sam’s preferred equilibrium. Nothing in the payoffs makes him follow through, so backward induction ignores it. Class 5 is about what he would have to do to make it real.
- Q8:
- Part 1: if A enters, B compares −10 with 0 and stays out; A earns 30. If A stays out, B compares 25 with 0 and enters; A earns 0. A enters. Nash equilibrium: A enters; B stays out if A entered, enters if A did not. Outcome (30, 0).
- Part 2: expected profits, using 0.7 and 0.3 as weights. A alone: 0.7(30) + 0.3(60) = 39. A sharing: 0.7(−5) + 0.3(10) = −0.5. B alone: 0.7(25) + 0.3(50) = 32.5. B sharing: 0.7(−10) + 0.3(5) = −5.5. Replace each endpoint with its expected value and solve as before: if A enters, B stays out (−5.5 < 0); if A stays out, B enters (32.5 > 0); A enters. Same equilibrium, outcome (39, 0) in expectation. The uncertainty changes the numbers but not their order, so nobody’s decision changes. It would matter if the low cost state were likely enough to make sharing profitable for B.
- Q9:
- A’s best responses: 4 against B’s 3 (20), 4 against B’s 4 (16), 3 against B’s 6 (9). A never bakes 6: 4 beats 6 against every column (20 > 18, 16 > 12, 8 > 0). By symmetry B never bakes 6. Delete both 6s and 4 now beats 3 for each bakery (20 > 18, 16 > 15), so each bakes 4. Nash equilibrium (4, 4) with payoffs (16, 16). Both baking 3 would pay 18 each, but whichever bakery is trusted to bake 3, the other earns 20 by baking 4 instead; the same prisoner’s dilemma as Q5, with bread instead of confessions.
- B’s best responses: 4 if A baked 3 (20), 4 if A baked 4 (16), 3 if A baked 6 (9 beats 8 and 0). A anticipates: baking 3 gives 15, baking 4 gives 16, baking 6 gives 18. A bakes 6. Nash equilibrium: A bakes 6; B bakes 4 after 3, 4 after 4, 3 after 6. Outcome (18, 9). In part (a), 6 was A’s worst choice because B would be baking 3 or 4 alongside. Now A’s 6 loaves are on the shelf before B decides, and B’s best reply to a full market is to cut back. A’s move is observed and irreversible, so it changes what B does rather than simply guessing at it. Total profit (27) is lower than in part (a) (32); the first mover gains by taking a bigger share of a smaller pie.