Street Fighter: is the game balanced?

Question

Suppose that $A$ is a matrix that describes the matchup information of any pair of Street Fighter characters e.g., considering $3$ characters, assume that the first row/collumn is associated with a character like Ryu, the second row/collumn is associated with Chun-li, the third row/collumn with Dhalsim, etc. Then for a $3\times3$ matrix we have:

$$A=\begin{pmatrix} 5 & 3 & 9 \\ 7 & 5 & 4 \\ 1 & 6 & 5 \end{pmatrix},$$

which tell us e.g., if Ryu fights Chun-li he wins $3$ out of $10$ matches.

Which kind of processing can I do to $A$, so that I know more information about the game?

Answer 1

The most basic evaluation would be that every character should have the same (50%) chance of winning against a random other character. If we rescale $A$ such that $a_{ij}$ is the probability that character $j$ wins against character $i$ ($\frac1{10}$ of every entry in your matrix) then we want $$\|A \cdot e_k\|_1 \stackrel!= \| 0.5 \cdot e \|_1 = N\cdot 0.5$$ Where $N$ is the number of characters and $e_k$ is the $k$-th basis vector and $e$ is the vector of all $1$-s. This means that the column sums of $A$ must all be equal.

In your matrix you can easily see that character $3$ is comparatively stronger than all others.

Answer 2

Let's start by subtracting $5$ from each entry to turn it into a payoff matrix for a zero-sum game: $$A'=\begin{pmatrix} 0 & -2 & 4 \\ 2 & 0 & -1 \\ -4 & 1 & 0 \end{pmatrix}$$

Now we can look for a mixed strategy (a probability with which we select each character) which is a Nash equilibrium. That boils down to $$\begin{pmatrix} 0 & -2 & 4 \\ 2 & 0 & -1 \\ -4 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} p_0 \\ p_1 \\ p_2 \end{pmatrix} = \begin{pmatrix} k \\ k \\ k \\ 1\end{pmatrix}$$ where $k$ is unknown. But we can subtract the first row from each of the other rows of $A'$ to get $$\begin{pmatrix} 0 & -2 & 4 \\ 2 & 2 & -5 \\ -4 & 3 & -4 \\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} p_0 \\ p_1 \\ p_2 \end{pmatrix} = \begin{pmatrix} k \\ 0 \\ 0 \\ 1\end{pmatrix}$$ and then discard the first row to get $$\begin{pmatrix} 2 & 2 & -5 \\ -4 & 3 & -4 \\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} p_0 \\ p_1 \\ p_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}$$ which gives $$\begin{pmatrix} p_0 \\ p_1 \\ p_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{7} \\ \frac{4}{7} \\ \frac{2}{7}\end{pmatrix}$$

So the optimal strategy is to roll a seven-sided die and play Ryu on a 1, Chun-Li on a 2 to 5, and Dhalsim on a 6 or 7. As a tool to assess balance, you could consider using a metric on how far the mixed strategy is from $(\frac13, \frac13, \frac13)$.