# (Evolutionary Game Theory) Argue informally there can no longer be an ESS in which only b is played

Here is a problem in evolutionary game theory: (So the term I am using should be familiar for people in this field)

The game is called 'Clever Mutants' which is a symmetric two-player game:

For this game, I have managed to find three symmetric Nash Equilibria. They are $(a,a), (b,b)$ and $((1/4,3/4),(1/4,3/4))$, in which the first two are evolutionary stable (ES) and the last one is not.

Now, suppose that mutants have a 'secret handshake'. That is, suppose that mutants can recognize other mutants and play different pure strategies against normal and mutant opponents. For example, a mutant could play $b$ against another mutant but play $a$ against a non-mutant. Argue informally there can no longer be an $ESS$ (Evolutionary Stable Strategy) in which only $b$ is played.

I don't know how to argue for that statement, even informally. Can someone help me, please? Thanks so much.

• Normally the bit "Can someone help me, please? Thanks so much." is not needed, as it is obvious. That is why I removed it. – luchonacho Sep 13 '17 at 17:56

Suppose initially the equilibrium is $(b,b)$. Since the normal population cannot recognize the mutants, they'd always play $b$.
When the mutants enter the population, because they can recognize their own kind through a secret handshake, they can play $a$ against one of their own (when there is a secret handshake), and play $b$ against the normal population (no secret handshake). This way, the mutants are getting $3$ each time they meet another mutant and $1$ each time they meet a normal. Therefore, the mutant's strategy is a best response to the normal's strategy, but a better response to itself than the normal's strategy is. As a result, the normal's strategy fails to be evolutionarilly stable.