<p>The "secret handshake" idea was first formalized in economics by <a href="http://www.dklevine.com/archive/refs4540.pdf" rel="noreferrer">Robson (1990)</a>.</p>

<p>Suppose initially the equilibrium is $(b,b)$. Since the normal population cannot recognize the mutants, they'd always play $b$. </p>

<p>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.</p>

CC BY-SA 3.0
2017-09-13T19:43:47.227Z
2017-09-13T19:43:47.227Z
Herr K.
<p>Here is a problem in <strong>evolutionary game theory</strong>: (So the term I am using should be familiar for people in this field)</p>

<p>The game is called 'Clever Mutants' which is a symmetric two-player game:</p>

<p><a href="https://i.sstatic.net/Ej8hV.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Ej8hV.png" alt="enter image description here"></a></p>

<p>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. </p>

<blockquote>
  <p>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. <strong>Argue informally there can no longer be an $ESS$ (Evolutionary Stable Strategy) in which
  only $b$ is played.</strong></p>
</blockquote>

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

CC BY-SA 3.0
2017-09-13T17:56:57.300Z
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
<game-theory>
(Evolutionary Game Theory) Argue informally there can no longer be an ESS in which only b is played
Y.X.
# (Evolutionary Game Theory) Argue informally there can no longer be an ESS in which only b is played

(Asked by: username_0 on 2017-09-13)

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:

[![enter image description here](https://i.sstatic.net/Ej8hV.png)](https://i.sstatic.net/Ej8hV.png)

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.

**Tags:** game-theory

> **Comments:**
> **username_1**: Normally the bit "Can someone help me, please? Thanks so much." is not needed, as it is obvious. That is why I removed it.  

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## 1 answers

### Answer by username_2 on 2017-09-13. Score: 6

The "secret handshake" idea was first formalized in economics by [Robson (1990)](http://www.dklevine.com/archive/refs4540.pdf).

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.


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