<p>The situation is not described by <span class="math-container">$K_A(K_B(3))$</span>. <span class="math-container">$A$</span> does not know that the draw was 3, and hence does not know that <span class="math-container">$B$</span> thinks it is in the set <span class="math-container">$K_B(3)$</span>. What <span class="math-container">$A$</span> does know is that the draw
is in set <span class="math-container">$K_A(3)$</span>, hence <span class="math-container">$A$</span> thinks <span class="math-container">$B$</span> should have narrowed it down to <span class="math-container">$K_B(K_A(3))$</span>.</p>

CC BY-SA 4.0
2020-09-23T09:35:50.797Z
2020-09-23T09:35:50.797Z
Giskard
<p>Person <span class="math-container">$A$</span> has information partition <span class="math-container">$\{\{1,2\},\{3,4\},\{5,6\}\}$</span></p>
<p>Person <span class="math-container">$B$</span> has information partition <span class="math-container">$\{\{1\},\{2,3\},\{4,5\},\{6\}\}$</span></p>
<p><span class="math-container">$w=3$</span> has been realized.</p>
<p>I'm confused about what <span class="math-container">$A$</span> knows that <span class="math-container">$B$</span> knows.</p>
<p>At first I thought of it as <span class="math-container">$K_aK_b(3)=K_a(\{2,3\})=\{1,2,3,4\}$</span>, but the example I found online says <span class="math-container">$K_aK_b(3)=\{2,3,4,5\}$</span>. why?</p>

<game-theory>
Information Partition and Common Knowledge
ph8ndstne
# Information Partition and Common Knowledge

(Asked by: username_0 on 2020-09-23)

Person $A$ has information partition $\{\{1,2\},\{3,4\},\{5,6\}\}$

Person $B$ has information partition $\{\{1\},\{2,3\},\{4,5\},\{6\}\}$

$w=3$ has been realized.

I'm confused about what $A$ knows that $B$ knows.

At first I thought of it as $K\_aK\_b(3)=K\_a(\{2,3\})=\{1,2,3,4\}$, but the example I found online says $K\_aK\_b(3)=\{2,3,4,5\}$. why?

**Tags:** game-theory

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## 1 answers

### Answer by username_1 on 2020-09-23. Score: 3 | Accepted answer

The situation is not described by $K\_A(K\_B(3))$. $A$ does not know that the draw was 3, and hence does not know that $B$ thinks it is in the set $K\_B(3)$. What $A$ does know is that the draw
is in set $K\_A(3)$, hence $A$ thinks $B$ should have narrowed it down to $K\_B(K\_A(3))$.


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