Recently, Tyler Cowen pointed to a 'Rubinstein bargaining game where players fail to reach an agreement, thereby eating up more and more of the pie. Each individual plays "chicken" and hopes the other will give in.'

Here is a nice and simple example of the Rubinstein bargaining game (taken from a game theory midterm exam (University of Texas at Austin); thanks to co-blogger Michael Sigmund for the promt delivery) :

Consider the following three-period, alternating game:

In period one, player 1 makes an offer (m₁) to player 2. Player 2 has the option of accepting the offer or rejecting the offer. If player 2 accepts, the players receive the payoffs (1 - m₁, m₁) and the game ends. If player 2 rejects, the game moves on to period two. {m_. ∈ [0,1], δ ∈ (0,1)}

In period two, player 2 makes an offer (m₂) to player 1. Player 1 has the option of accepting the offer or rejecting the offer. If player 1 accepts, the players receive the payoffs (δm₂, δ(1-m₂)) and the game ends. If player 1 rejects, the game moves on to period three.

In period three, player 1 makes an offer (m₃) to player 2. Player 2 has the option of accepting the offer or rejecting the offer. If player 2 accepts, the players receive the payoffs (δ²(1-m₃), δ²m₃) and the game ends. If player 2 rejects, the game ends and the players receive (0.0).

Finding the subgame perfect equilibrium:

Start with the 3^rd period. Player 1 offers m₃. If player 2 accepts, u₁ = δ²(1-m₃), u₂ = δ²m₃. If player 2 rejects, u₁=u₂=0. The subgame perfect equilibrium for this stage is m₃=0, accept m₃ ≥ 0. u₁ = δ², u₂ = 0.

Now in the 2^nd period, player 2 offers m₂. If player 1 accepts, u₁ = δm₂, u₂ = δ(1-m₂). If player 1 rejects, they go on to the 3^rd period where u₁ = δ² and u₂ = 0. The subgame perfect equilibrium for this stage is accept m₂ ≥ δ, m₂ = δ. u₁ = δ², u₂ = δ(1 - δ).

Now in the 1^st period, player 1 offers m₁. If player 2 accepts, u₁ = (1-m₁, u₂ = m₁). If player 2 rejects, they go on to the 2^nd period where u₁ = δ² and u₂ = δ(1-δ). The subgame perfect equilibrium for this stage is m₁ = δ(1- δ), accept m₁ ≥ δ(1 - δ). u₁ = 1 - δ(1 - δ), u₂ = δ(1 - δ).

The subgame perfect equilibrium is [m₁ = δ(1 - δ), accept m₂ ≥ δ, m₃ = 0], [accept m₁ ≥ δ(1 - δ), m₂ = δ accept m₃ ≥ 0].

Question of the day: In the subgame perfect equilibrium one player has a higher payoff regardless of δ. Which player is it?

Answer: Player 1 has a higher payoff than player 2 if 1 - δ(1 - δ) > δ(1- δ). This occurs if 2δ² - 2δ + 1 > 0, which is true for any δ ≥ 0. Player 1 always has a higher payoff.