Sequential Ultimatum Game

• Content map: SMU H3 Game Theory Map

Setup

Definition:

Sequential Ultimatum Game

• Players: P1 and P2.
• Strategies: P1: choose an offer $x\in\{0,1,\dots,S\}$. Since P1 moves once in the base game, the chosen proposal is the complete pure strategy; P2: a complete contingent response rule $r(x)\in\{\text{Accept},\text{Reject}\}$ for every possible offer $x\in\{0,1,\dots,S\}$.
• Rules:
  • Start with a fixed surplus $S$.
  • Players move sequentially: P1 offers $x$ to P2, then P2 accepts or rejects.
  • The player who chooses optimally by backward induction reaches the subgame-perfect outcome.

Game Tree

Derivation (Backward Induction)

• For any $x\ge 1$, P2 strictly prefers Accept since $x>0$.
• At $x=0$, P2 is indifferent: Accept and Reject both yield $0$.
• Anticipating P2’s rule, P1 chooses the smallest $x$ that will be accepted.

Nash Equilibrium

Result:

• When $x \ge 1$, P2 accepts the split.
• When $x = 0$, P2 is indifferent to accept or reject.

Insights

Insight:

Backward induction selects minimal concession by the proposer: the responder’s ability to punish is limited by what is optimal at the moment of choice.

Extension: Counter-Offers

Setup

Definition:

Two-Round Ultimatum Game with two players

• Start with a first-round proposal and a possible discounted counter-offer round.
• Players alternate proposing and responding across the reached rounds.
• The player who chooses optimally by backward induction reaches the rollback outcome.

Game Tree

Round 2

• P2 proposes $(1-y,y)$.
• If P1 accepts, first-round equivalent payoffs are:

• If P1 rejects, both players obtain $0$.
• Since $1-y\in[0,1]$, P1 weakly prefers accepting any round-2 offer.
• Therefore P2 chooses $y=1$, so the round-2 outcome is:

Rollback to Round 1

• In round 1, P2 compares:
  • accept P1’s offer and receive $1-x$,
  • reject and move to round 2, where P2 receives $\delta$.
• Hence P2 accepts if and only if:

• Equivalently:

• P1 maximises own share $x$, so chooses the largest acceptable value:

• Thus the round-1 offer is:

Nash Equilibrium

Result:

The subgame perfect Nash equilibrium of the two-round counter-offer extension is:

• P1: Offer $(1-\delta,\delta)$ in round 1; Accept all round-2 offers.
• P2: Accept if $x \le 1-\delta$ in $(x,1-x)$; Otherwise reject; in round 2, offer $(0,1)$.

Insights

Insight:

Allowing a counter-offer strengthens P2’s outside option from $0$ to $\delta$. As $\delta$ rises, P2 becomes more patient and captures a larger share of the pie.