1. Construct a payoff matrix for a game of long wait (where m(LongWait) > 0.5 * V) vs. no display. Explain how you worked out each payoff, refering to the war of attrition payoff table when appropriate.
2. Explain whether E(Long, Long) will be a positive or negative number.
Both questions will be answered together.
a. For E(Long, Long): The outcomes of these contests are decided at random and are given by line three of the payoff table: 0.5*V - m(Long). The logic of this payoff was discussed in conjuction with the payoff table.
Now, to answer question 2, if m(LongWait) > 0.5 * V then the payoff to LongWait becomes negative. Thus, they have gained Pyrrhic Victory -- they get the resource but reduce their fitness in the process.
b. For E(Long, No Display): Here Long always wins since No Display forfeits the contest by not playing the game.
c. For E(No Display, Long): Zero, since no display takes off before the contest gets going. It gains nothing and loses nothing.
d. For E(No Display, No Display): We must assume that they divide the goods or remember who won last time and then share (how nice -- this site's mascot Barney would approve!). So the payoff is 0.5 * V but with no costs.
Here's the payoff matrix:
Long |
No Display | |
Long |
0.5*V - m(Long), a NEGATIVE NUMER since m(Long) > V |
V |
No Display |
0 | 0.5*V |
which we can simplify for the discussion we will have back on the previous page to:
Long |
No Display | |
Long |
< 0 (negative) | V |
No Display |
0 | 0.5*V |
What is the game theory term that can be applied to the strategy "play either Rock, Scissors, or Paper at randomwith a probability of 33.3% in each game"?
Mixed ESS. It is mixed since it involves playing three different strategies at a fixed (equilibrial) probability. It is an ESS since it is both of higher fitness than any alternative pure or mixed strategy.