# What is a Game Theoretic Best Choice? Part I The aim of this series is to come up with some formalisation of what a game theoretic best choice is without reference to established results before the material is covered in lectures. I will then compare my formalisation to established results.

Notation

In Dr. Elliott’s lectures we defined a Nash Equilibrium to be when $u_i(a)\geq u_i(a_i',a_{-i})$ for all $a_i' \in A_i$ and for all $i$. $a_{-i}$ stands for the actions of everyone else other than person $i$. $u_i$ stands for person $i$‘s utility, i.e. the numerical value or benefit person i gains. $A_i$ is the set (collection) of all possible action profiles. An action profile is a combination of actions taken by everyone. E.g. If we’re in a game with precisely two players, player 1 taking action 1 and player 2 taking action 2 constitutes an action profile ${a_1=1, a_2=2}$.

Finding Expected Utility

We want to find the expected utility of each action. This is the utility you would obtain by doing that action when everyone else does $a_{-i}$ (this action profile could mean Player 2 does A and Player 3 does B.) multiplied by the probability everyone else does $a_{-i}$, summed over all action profiles $a_{-i}$, i.e. all action profiles where you do $a_i$. That is, $E_{a_i}(u_i) = \sum_i u_i(a_i, a_{-i} P(a_{-i}). (1)$

Here, $P(a_{-i}) = P(a_i, a_{-i}|a_i) = P(a | a_i)$. I’m pretty sure they are equivalent. We thus have the condition $\sum P(a|a_i) = 1$.

Let’s apply the equation of expected utility (1) to the Prisoner’s Dilemma to examine the equation further. The game is symmetric, so suppose without loss of generality that we are player 1. The payoffs are as shown below: We denote option Quiet (stay quiet and don’t tell on the other person) as $q$ and option Fink (tell on the other person) as $f$. Then $E_q(u_1) = u_1(q_1, f_2)P(f_2) + u_1(q_1, q_2)P(q_2) = 0\times P(f_2) + 2P(q_2) = 2P(q_2)$. $E_f(u_1)=u_1(f_1,f_2)P(f_2) + u_1(f_1,q_2)P(q_2) = P(f_2) + 3P(q_2)$.

Because it’s symmetric, we can solve for the actions that will be taken.

Naively we can suppose that each player chooses the action that yields the higher expected utility. Then $a_1 = q$ if $E_q(u_1) > E_f(u_1)$,

i.e. if $2P(q_2) > P(f_2) + 3P(q_2) \Rightarrow P(f_2) + P(q_2) < 0 \Rightarrow 1 < 0$, which is never.

Similarly, $a_1 = f$ if $E_q(u_1) < E_f(u_1)$,

i.e. if $2P(q_2) < P(f_2) + 3P(q_2) \Rightarrow P(f_2) + P(q_2) > 0 \Rightarrow 1 > 0$, which is always. So both players always choose to fink.

Generalised

Through this we can write a generalised equation to find out what payoffs would result in different strategies.

As before, we have $a_1 = q$ if $E_q(u_1) > E_f(u_1)$,

i.e. if $u_1(q_1, f_2)P(f_2) + u_1(q_1, q_2)P(q_2) > u_1(f_1,f_2)P(f_2) + u_1(f_1,q_2)P(q_2)$ (eqn 2)

Because the game is symmetric, I claim $P(f_1) = P(f_2)$ and $P(q_1) = P(q_2)$.

Further, I claim that $P(f_1) = P(q_1) = \frac{1}{2}$ if Player 1 is indifferent between $q$ and $f$ (Left hand side of equation 2 = right hand side RHS of equation 2).

Then, when the player is indifferent between $q$ and $f$, we have $u_1(q_1, f_2) + u_1(q_1, q_2) = u_1(q_2, f_2) + u_1(q_1, q_2)$.

That just says that, assuming the other player acts randomly (chooses $q$ or $f$ with probability $0.5$), the expected utilities of Player 1 is the same whether (s)he chooses $q$ or $f$.

Is this a sufficient condition for a player to be indifferent between q and f? No: Each player is still better off if they choose $f$, Fink. And in this case it would be silly to choose $q$, Quiet.

We’ll look into this more in the next post in the series.

For later:

Once we’ve formalised game theoretic ‘best choices’ well enough, we can then define certain axioms like we did in Social Choice Theory and narrow down the space of equilibria we need to explore to only the equilibria that fit those axioms.

Feature Image Credits: Scroll.in, Youtube Video.