So far we have considered games where the two players are competing. To conclude this Chapter we will briefly consider what might happen if they agree to cooperate.

We will see that even here the correct strategy is not entirely straightfoward.

We will illustrate our definitions and results with the following two player game, where for convenience we will label the different pure payoff vectors with letters:

Example 4.11

Consider the game

Strategy 1 Strategy 2
Strategy 1 \(W=(4,2)\) \(X=(1,4)\)
Strategy 2 \(Y=(0,1)\) \(Z=(3,0)\)

Definition 4.12

We define the payoff polygon for a bimatrix game \(A\) to be the convex hull of the payoff vectors appearing as entries of \(A\), i.e. the smallest convex set which contains all those points.

Example 4.11a:

Here is the payoff polygon for our example.

Example completed by hand in the lecture

By cooperating, the two players can achieve any payoff in the payoff polygon (and no others).

For example, to achieve the payoff halfway between \(W\) and \(Z\), player 1 can play Strategy 1 half the time, and Strategy 2 half the time, with player 2 playing Strategy 1 whenever player 1 does, and Strategy 2 otherwise.

Note that this is **not* the same as player 1 choosing \(\underline{x}=(0.5,0.5)\) and player 2 choosing \(\underline{y}=(0.5,0.5)\) as these are not coordinated.

We see that by cooperating, the two players can choose any point in the payoff polygon. But which is the fairest point to choose?

Recall that each player can always guarantee that they receive their safety value, regardless of how the other player plays. So neither player should accept less than their safety value.

We will denote the corresponding pair of payoffs by \((u_0,v_0)\), the status quo point.

If \((u,v)\) lies in the payoff polygon and there is another point \((u',v')\) also in the polygon with \(u'\geq u\) and \(v'\geq v\) then the players would be foolish to agree on \((u,v)\) as the new point is no worse for either of then (and better for at least one).

Based on these considerations we define

Definition 4.13

The negotiation set for a matrix game is the set of points \((u,v)\) in the payoff polygon such that \(u\geq u_0\) and \(v\geq v_0\) and there is no point \((u',v')\) in the polygon with \(u'\geq u\) and \(v'\geq v\) with \((u',v')\neq (u,v)\).

It is easy to determine the negotiation set for a matrix game from the payoff polygon and safety values, as it will consist of the points on the boundary that are above and to the right of the status quo \((u_0,v_0)\) and which do not have a point that is above and to the right of them.

Example 4.11b

Here is the negotiation set for our example.

Example completed by hand in the lecture

If the negotiation set consists of a single point then that will be the best strategy to cooperate on. But what if there is more than one point in that set?

Once again, Nash came up with an answer to this question.

Definition 4.14

The Nash bargaining solution for a matrix game is the unique element \((u,v)\) of the negotiation set which maximises the value of \[f(u,v)=(u-u_0)(v-v_0).\]

To find this solution we use calculus. The negotiation set will be made up of one or more straight lines. For each line we can write \(u\) as a function of \(v\), and so \(f\) will be a function of one variable on this line.

The maximum of \(f\) on such a line occurs either at the stationary point or one of the end points. By calculating these for each line and choosing the largest, the Nash solution can be found.

Example 4.11c

Here is the Nash bargaining solution for our example.

Example completed by hand in the lecture

What is it about the Nash bargaining solution that makes it the right choice?

Nash suggested that any reasonable solution should satisfy the following four axioms:

A1: Rationality: The solution is in the negotiation set.

A2: Linear invariance: If \(u\) (or \(v\)) is transformed by a function of the form \(au+b\) with \(a>0\) then the solution should be transformed by the same function.

A3: Symmetry: If the payoff polygon is symmetric about the line of slope \(1\) through the status quo point, then the solution lies on this line.

A4: Independence of Irrelevant Alternatives: If \((u,v)\) is the Nash solution for a polygon \(P\) with status quo \((u_0,v_0)\) and \(Q\) is another polygon containing \((u,v)\) and with the same status quo, and \(Q\subset P\), then \((u,v)\) is the Nash solution for this polygon.

Then Nash was able to show

Theorem 4.15

The Nash bargaining solution is the unique solution satisfying A1-A4.

On the face of it, all 4 axioms seem quite reasonable. However, some game theorists do not agree with axiom A4, and other solution schemes have also been proposed which satisfy alternative axioms.

Example 4.16

An example where the Nash bargaining solution does not seem fair.

Example completed by hand in the lecture

Further details can be found in the discussion on page 204 of Game Theory, Alive, by Karlin and Peres.