Finite-horizon alternating-offer bargaining

Two players alternate proposals for dividing a pie, with a fixed final round. Agreement ends the game; rejection shrinks the next available pie by a factor $0<\alpha <1$. Final rejection gives both players zero. Offers range continuously over the entire available pie. Backward induction in games determines immediate agreement and the proposer's share through the continuous-offer bargaining acceptance lemma and the finite-horizon bargaining parity effect. Discrete offer grids can change both the boundary behavior and the exact shares.

In finite-horizon alternating-offer bargaining, the proposer with $k$ rounds left retains fraction $P_k$ of the currently available pie. Terminal acceptance gives $P_1=1$. After rejection, the responder becomes proposer and can obtain fraction $\alpha P_{k-1}$ of the current pie. The continuous-offer bargaining acceptance lemma therefore gives $P_k=1-\alpha P_{k-1}$. Summing the resulting finite geometric series proves the displayed formula.

Odd horizons give shares above $1/(1+\alpha )$, decreasing toward it; even horizons give shares below it, increasing toward it. Between two available horizons $k_1<k_2$, choose $k_1$ if $k_1$ is odd and $k_2$ if $k_1$ is even. This concerns the initial proposer and immediate agreement, rather than waiting until the selected final round to obtain its share.

Let a proposer divide a pie of size $M>0$ by continuously offering its responder $q\in [0,M]$. Rejection gives fixed continuation payoffs $b$ to the proposer and $c$ to the responder, with $0\le c<M$ and $b<M-c$. Then a subgame perfect equilibrium must have the responder accept $q=c$ with probability one. The proposer offers exactly $c$, retaining $M-c$.

Sequential rationality forces rejection for $q<c$ and acceptance for $q>c$. If acceptance at $c$ had probability $a<1$, that offer would give the proposer $a(M-c)+(1-a)b<M-c$. Every larger offer gives less than $M-c$ and can be improved by a smaller offer still above $c$; every smaller offer gives $b<M-c$. The supremum $M-c$ would therefore be unattained. Randomizing offers cannot attain it either: the nonnegative gap from $M-c$ would be strictly positive almost surely and hence have positive expectation. No best response would exist, contradicting equilibrium. Acceptance at $c$ removes this gap and makes $q=c$ the unique optimal proposal. The assumptions $c<M$ and $b<M-c$ exclude degenerate indifference of the proposer.