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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.

In the final round, player 1 strictly accepts every $q_2>0$. In any subgame perfect equilibrium it must also accept $q_2=0$ with probability one. Otherwise, if that acceptance probability were $a<1$, offering zero would give player 2 only $a\alpha$. A sufficiently small positive offer would be accepted and yield more. No positive offer maximizes player 2's payoff either, because a smaller positive offer improves it. The same supremum argument rules out a mixed offer distribution attaining the missing maximum. Thus the final-round equilibrium uniquely has $q_2=0$, accepted, with payoffs $(0,\alpha )$.

In round one, player 2 therefore rejects $q_1<\alpha$ and accepts $q_1>\alpha$. At $q_1=\alpha$ it is indifferent, but equilibrium again forces acceptance with probability one: if acceptance were less likely, player 1 could improve on any proposed optimum with an offer just above $\alpha$. Since $\alpha <1$, player 1 strictly prefers the resulting positive payoff to rejection.

Consequently the unique equilibrium outcome is immediate agreement at $q_1=\alpha$, giving payoffs $(1-\alpha ,\alpha )$. The full strategies are: player 1 offers $\alpha$ initially and accepts every nonnegative final offer; player 2 accepts initial offers at least $\alpha$, rejects smaller ones, and offers zero in every reached final-round subgame. These prescriptions cover off-path histories too. Acceptance at indifference follows from the continuous-offer bargaining acceptance lemma, rather than an arbitrary tie-breaking convention.

Let $P_k$ be the initial proposer's share, measured as a fraction of the pie currently available, with $k$ rounds remaining. The last-round argument in part (e) gives $P_1=1$. Rejecting an offer leaves the responder as next proposer with a pie smaller by factor $\alpha$, so its continuation value is $\alpha P_{k-1}$. The current proposer offers exactly that amount, which is accepted by the continuous-offer bargaining acceptance lemma. HenceThe formula follows by summing a finite geometric series, or by induction in the recurrence. Agreement is immediate in every subgame, so this share is player 1's actual initial payoff.

For odd $k$, $P_k=(1+{\alpha }^k)/(1+\alpha )$ lies above $1/(1+\alpha )$ and decreases with $k$. For even $k$, $P_k=(1-{\alpha }^k)/(1+\alpha )$ lies below that value and increases with $k$. Thus prefer odd to even, fewer rounds among odd choices, and more rounds among even choices. For the given $k_1<k_2$, this simplifies to choose $k_1$ if $k_1$ is odd, and choose $k_2$ if $k_1$ is even. All comparisons are strict because $0<\alpha <1$. This is the finite-horizon bargaining parity effect.