Demonstration 1: The second agent's response

In this demonstration, the second agent's response to the risky strategy is analysed, but for now, the clarification alternative is ignored. The piece of the game tree that is examined is described in figure 4.3. For this game tree, the evaluator is expected to produce for the upper branch:

$$\begin{split}&-10 + p.( 0 + 100 ) + (1 - p).80 \\ = &20p + 70 \end{split}$$ (4.1)

For the lower branch, the evaluator is expected to produce:

$$\begin{split}&-10 + p.80 + (1 - p).100 \\ = &-20p + 90 \end{split}$$ (4.2)

These formulas can be verified by looking at the utility function in Section 4.5.1. To find a decision surface between the two alternatives, the points in the belief space at which the alternatives have equal utility are found. There is one solution in this case.

$$20p + 70 = -20p + 90$$ (4.3)

$$\Rightarrow 40p = 20$$ (4.4)

$$\Rightarrow p = 0.5$$ (4.5)

Figure 4.3: Game tree for the second agent's response

Therefore, it can be expected that the second agent will lend the car spanner when $p > 0.5$, and will lend the bike spanner when $p < 0.5$. Figure 4.4 is a plot of the planner's output, which conforms to expectations. This is the plot for $n=8$ which is identical to the plots for the other training levels, 2, 32 and 128.

Figure 4.4: Utility of strategies against P(intend(car-spanner))