<p>Yes. Consider a two player game described by the following matrix</p>

<p>\begin{array}{|c|c|c|}
\hline
&amp;  L &amp; R   \\
\hline
a &amp; 3,0 &amp; 0,0   \\
\hline
b &amp; 0,0 &amp; 3,0    \\
\hline
c &amp; 2,0 &amp; 2,0    \\
\hline
\end{array}</p>

<p>If only pure strategies are allowed, only $a$ and $b$ are rationalizable because they are best responses to $L$ and $R$ respectively while $c$ is not a best response to anything. But if you look at the mixed extension of the game then $c$ strictly dominates 0.5$a$+0.5$b$, hence 0.5$a$+0.5$b$ is not rationalizable. As a result, in the mixed extension all $a$, $b$ and $c$ are rationalizable.</p>

<p>If you insist on an example with three players I can add a third one to this. This third player has exactly one possible strategy, so it does not really matter what his payoffs are and what he does. <br>
If you think that is cheating, the third player can have multiple strategies that do not at all influence the payoff of the first two players.</p>

CC BY-SA 3.0
2015-07-08T08:53:58.643Z
Thank you so much for your answer. I have to admit I'm having a hard time to understand why this example doesn't contradict the well-known result of equivalence of deletion of strictly dominated strategies and rationalizability for two-player games.
Metta World Peace
CC BY-SA 3.0
2015-07-08T10:13:16.870Z
I am not sure what you mean. Perhaps the cause of your problem is that you consider a game and its mixed extension to be the same game, but in face these are different games with different rationalizable strategies.
Giskard
CC BY-SA 3.0
2015-07-05T15:27:20.717Z
2015-07-05T15:47:04.260Z
2015-07-05T15:47:04.260Z
Giskard
<p>It is easy to see that if $\sigma$ is mixed rationalizable strategy, then a pure strategy $a$ such that $\sigma(a)&gt;0$ is also rationalizable, but it seems not vice versa.</p>

<p>For a two-player finite game, let $P_1$ be the set of all pure rationalizable strategies for player 1, then the set of her rationalizable strategies is simply $\Delta(P_1)$, which can be seen as a corollary of equivalence of iterated strict dominance and rationalizability for two-player finite games.</p>

<p>What is an example for a finite game with more than two players such that the set of rationalizable strategies is strictly smaller than the mixed extension of the set of pure rationalizable strategies?</p>

<game-theory><academic-graduate>
If $a$ and $b$ are two pure rationalizable strategies, can $0.5a+0.5b$ fail to be a rationalizable strategy?
Metta World Peace
# If $a$ and $b$ are two pure rationalizable strategies, can $0.5a+0.5b$ fail to be a rationalizable strategy?

(Asked by: username_0 on 2015-07-05)

It is easy to see that if $\sigma$ is mixed rationalizable strategy, then a pure strategy $a$ such that $\sigma(a)>0$ is also rationalizable, but it seems not vice versa.

For a two-player finite game, let $P\_1$ be the set of all pure rationalizable strategies for player 1, then the set of her rationalizable strategies is simply $\Delta(P\_1)$, which can be seen as a corollary of equivalence of iterated strict dominance and rationalizability for two-player finite games.

What is an example for a finite game with more than two players such that the set of rationalizable strategies is strictly smaller than the mixed extension of the set of pure rationalizable strategies?

**Tags:** game-theory, academic-graduate


---

## 1 answers

### Answer by username_1 on 2015-07-05. Score: 6 | Accepted answer

Yes. Consider a two player game described by the following matrix

\begin{array}{|c|c|c|}
\hline
& L & R \\
\hline
a & 3,0 & 0,0 \\
\hline
b & 0,0 & 3,0 \\
\hline
c & 2,0 & 2,0 \\
\hline
\end{array}

If only pure strategies are allowed, only $a$ and $b$ are rationalizable because they are best responses to $L$ and $R$ respectively while $c$ is not a best response to anything. But if you look at the mixed extension of the game then $c$ strictly dominates 0.5$a$+0.5$b$, hence 0.5$a$+0.5$b$ is not rationalizable. As a result, in the mixed extension all $a$, $b$ and $c$ are rationalizable.

If you insist on an example with three players I can add a third one to this. This third player has exactly one possible strategy, so it does not really matter what his payoffs are and what he does.   
If you think that is cheating, the third player can have multiple strategies that do not at all influence the payoff of the first two players.

> **Comments:**
> **username_0**: Thank you so much for your answer. I have to admit I'm having a hard time to understand why this example doesn't contradict the well-known result of equivalence of deletion of strictly dominated strategies and rationalizability for two-player games.  
> **username_1**: I am not sure what you mean. Perhaps the cause of your problem is that you consider a game and its mixed extension to be the same game, but in face these are different games with different rationalizable strategies.  

---

