<p>If you look at the definition for Nash equilibrium, you can see that if both $A$ and $B$ are in $\texttt{State 4}$, then if $A$ changes their mind that would mean that they would be defecting from a situation where they get $1$ point to one where they get $0$, a net decrease since at $\texttt{State 4}$ they are both defecting. Similarly for player $B$.</p>

<p>The reverse is true for the case where they both cooperate ($\texttt{State 1}$). The reason this isn't a Nash equilibrium is that, if any one of the two chooses to change strategy (i.e. go from cooperation to defection) they would see a net increase in their reward. </p>

CC BY-SA 3.0
2016-01-17T21:41:20.990Z
2016-01-17T21:41:20.990Z
A.K.
<p>Imagine two prisoners held in separate cells, interrogated simultaneously, and offered deals (lighter jail sentences) for betraying their fellow criminal. They can "cooperate" (with the other prisoner) by not snitching, or "defect" by betraying the other. However, there is a catch; if both players defect, then they both serve a longer sentence than if neither said anything. Lower jail sentences are interpreted as higher payoffs (shown in the table).</p>

<p><a href="https://i.sstatic.net/4GmAt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4GmAt.png" alt="enter image description here"></a></p>

<p>Wiki says</p>

<blockquote>
  <p>The prisoner's dilemma thus has a single Nash equilibrium: both players choosing to defect i.e. State (4)</p>
</blockquote>

<p>But how can <code>State (4)</code> be in <code>Nash Equilibrium</code>. </p>

<ul>
<li>A can take gain by changing its decision i.e. <code>State (4)</code> -> <code>State (2)</code>
and.</li>
<li>B can take gain by changing its decision i.e. <code>State (4)</code> -> <code>State (3)</code>.</li>
</ul>

<p>According to me State (1) should be in <code>Nash Equilibrium</code>. Am I getting it correct ...</p>

<p><a href="https://en.wikipedia.org/wiki/Nash_equilibrium" rel="nofollow noreferrer">Referrence</a></p>

<gt.game-theory><nash-equilibrium>
Understanding Prisoner's dilemma
Atinesh
# Understanding Prisoner's dilemma

(Asked by: username_0 on 2016-01-17)

Imagine two prisoners held in separate cells, interrogated simultaneously, and offered deals (lighter jail sentences) for betraying their fellow criminal. They can "cooperate" (with the other prisoner) by not snitching, or "defect" by betraying the other. However, there is a catch; if both players defect, then they both serve a longer sentence than if neither said anything. Lower jail sentences are interpreted as higher payoffs (shown in the table).

[![enter image description here](https://i.sstatic.net/4GmAt.png)](https://i.sstatic.net/4GmAt.png)

Wiki says

> The prisoner's dilemma thus has a single Nash equilibrium: both players choosing to defect i.e. State (4)

But how can `State (4)` be in `Nash Equilibrium`.

* A can take gain by changing its decision i.e. `State (4)` -> `State (2)`
  and.
* B can take gain by changing its decision i.e. `State (4)` -> `State (3)`.

According to me State (1) should be in `Nash Equilibrium`. Am I getting it correct ...

[Referrence](https://en.wikipedia.org/wiki/Nash_equilibrium)

**Tags:** gt.game-theory, nash-equilibrium

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## 1 answers

### Answer by username_1 on 2016-01-17. Score: 1 | Accepted answer

If you look at the definition for Nash equilibrium, you can see that if both $A$ and $B$ are in $\texttt{State 4}$, then if $A$ changes their mind that would mean that they would be defecting from a situation where they get $1$ point to one where they get $0$, a net decrease since at $\texttt{State 4}$ they are both defecting. Similarly for player $B$.

The reverse is true for the case where they both cooperate ($\texttt{State 1}$). The reason this isn't a Nash equilibrium is that, if any one of the two chooses to change strategy (i.e. go from cooperation to defection) they would see a net increase in their reward.


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