# Generating a 3x3 payoff matrix (Game-Theory)

Been working on some economics lately, and was able to figure out how to put together some 2x2 payoff matrix's but not sure how to go about doing a 3x3. I would like to keep the same general idea if possible, i.e. the tikz picture.

Here is what I have used to generate a 2x2.

%third matrix
%learn how to do 3x3
\begin{center}
\begin{tikzpicture}
\matrix[matrix of math nodes,every odd row/.style={align=right},every evenrow/.style={align=left},every node/.style={text width=1.5cm},row sep=0.2cm,column sep=0.2cm] (m) {
2&\bf{3}\\
4&\bf{2}\\
-1&\bf{0}\\
\bf{6}&0\\
};
\draw (m.north east) rectangle (m.south west);
\draw (m.north) -- (m.south);
\draw (m.east) -- (m.west);

\coordinate (a) at ($(m.north west)!0.25!(m.north east)$);
\coordinate (b) at ($(m.north west)!0.75!(m.north east)$);
\node[above=5pt of a,anchor=base] {Left};
\node[above=5pt of b,anchor=base] {Right};

\coordinate (c) at ($(m.north west)!0.25!(m.south west)$);
\coordinate (d) at ($(m.north west)!0.75!(m.south west)$);
\node[left=2pt of c,text width=1cm]  {Up};
\node[left=2pt of d,text width=1cm]  {Down};

\node[above=18pt of m.north] (firm b) {Column};
\node[left=1.6cm of m.west,align=center,anchor=center] {Row};

%\node[above=5pt of firm b]  {Payoff Matrix};
\end{tikzpicture}
\end{center}


• The question is....? – user11232 Feb 15 '15 at 7:56
• @HarishKumar: I think the OP wants to know how to modify his code for a 3 by 3 matrix. – Shahab Feb 15 '15 at 9:33
• @Shahab And how does a 3 x 3 matrix look like? I don't know game theory! – user11232 Feb 15 '15 at 9:35
• @HarishKumar: Basically 9 boxes instead of 4. – Shahab Feb 15 '15 at 9:36
• @Shahab hehe. I think I know that much ;-) – user11232 Feb 15 '15 at 9:44

Please have a look at Will two-letter font style commands (\bf , \it , …) ever be resurrected in LaTeX?

\documentclass[tikz]{standalone}
\usetikzlibrary{matrix,calc,positioning}
\begin{document}
\begin{tikzpicture}

\matrix[matrix of math nodes,draw,
every odd row/.style={align=right},every evenrow/.style={align=left},
nodes={text width=1.5cm},row sep=0.2cm,column sep=0.2cm]
(m) {2&3&6\\4&2&-1\\-1&0&0\\0&0&0\\2&3&6\\4&2&-1\\};
\foreach\x[count=\xi from 2,evaluate={\xx=int(2*\x);\xxi=int(\xx+1)}] in {1,2}{
\draw ({$(m-1-\x)!0.5!(m-1-\xi)$}|-m.north) -- ({$(m-1-\x)!0.5!(m-1-\xi)$}|-m.south);
\draw ({$(m-\xx-1)!0.5!(m-\xxi-1)$}-|m.west) -- ({$(m-\xx-1)!0.5!(m-\xxi-1)$}-|m.east);
}
\foreach\x in{0,1,2}{
\node[text depth=0.25ex,above=2mm] at ($(m.north west)!{(2*\x+1)/6}!(m.north east)$)
\node[left=2mm] at ($(m.north west)!{(2*\x+1)/6}!(m.south west)$)

• Don't you need .append style instead of .style to make keep the node entries math. The OP had some entries \bf, that ought to be \mathbf. – Andrew Swann Feb 15 '15 at 12:41
• @AndrewSwann Oops. I didn't even pay attention. Fixed now via nodes. Thanks. I'll leave the node ornaments to OP though. – percusse Feb 15 '15 at 12:49