The Negotiator's Dilemma

Company A offers B exactly 81 units in round 1, keeping 19 units. Company B accepts immediately (81 > what B would get by delaying, and 81 > B's valuation of 60). The shrinking pie makes delay costly for both parties, and backward induction pins down the unique rational outcome.

Apply backward induction. At each round, a company accepts an offer if it is at least as good as what it can get by rejecting and making a counter-offer next round (accounting for pie shrinkage). Step 1: Setup: pie starts at 100. Shrinks by 10% each round. Round 1 pie = 100. Round 2 pie = 90. Round 3 pie = 81. Assume 3 rounds maximum (after that no deal, both get 0). Step 2: Round 3 (B makes offer, pie = 81): A will accept anything ≥ 0 (no more rounds). B offers A exactly 0 and B keeps 81. B's payoff if round 3 is reached: 81. Step 3: Round 2 (A makes offer, pie = 90): B knows that if B rejects, B gets 81 in round 3. So B accepts any offer ≥ 81. A offers B exactly 81, keeping 90-81 = 9. A's payoff if round 2 is reached: 9. Step 4: Round 1 (A makes offer, pie = 100): B knows that if B rejects, A will offer B exactly 81 in round 2. So B's payoff from rejecting round 1 = 81. A offers B exactly 81 in round 1 — but wait, pie is 100 in round 1. A can offer B 81 and keep 19. B is indifferent between 81 now and 81 later — accepts. Step 5: Re-examine: B's outside option (from rejecting round 1) is 81 from round 2. So A must offer B ≥ 81 in round 1 to secure acceptance. A offers B=81, A keeps 19. Total: 19+81=100. Step 6: But does B also compare to its own valuation of 60? B values the deal at 60, so B's surplus from getting 81 > 60 — B definitely accepts. Company A's optimal offer: keep 19, give B 81. Key Insight: Backward induction in bargaining reveals a counterintuitive result: the first mover does not always win. The shrinking pie and the opponent's future leverage determine current offers. The party who makes the last credible offer sets the floor for all prior rounds — their power ripples backward through time.