In a book on game theory I saw the following example of a game, a modified version of Roshambo (or Rock-paper-scissors), which is described by the following payoff-matrix:
$$
\begin{array}{c|c|c}
0 & -1 & +1 \\ \hline
+1 & 0 & -2 \\ \hline
-1 & +2 & 0
\end{array}
$$
To solve this game, the book proceeds as follow. Say the strategy of the first player will be $\{ a,b,c \}$. Then he wants to make the EV of the second player from the varios strategy options equal. For the different selections (columns) of the seconds player we got the following EVs for the first player:
First column: $0\cdot a + 1\cdot b + (-1)\cdot c = b – c$
Second column: $(-1)\cdot a + 0\cdot b + 2\cdot c = 2c – a$
Third column: $1\cdot a + (-2)\cdot b + 0 \cdot c = a – 2b$
Equating the above equations yields:
$$
b – c = 2c – a = a – 2b
$$
and with the restriction $a+b+c=1$, the system of equations
\begin{align*}
b – c & = 2c – a \\
2c – a & = a – 2b \\
a + b + c & = 1
\end{align*}
results, which has the solution $a = 1/2, b = 1/4, c = 1/4$.
Now I wanted to generalise the above solution technique and wanted to apply it to the following modified version of the above game.
$$
\begin{array}{c|c|c|c}
0 & 1 & -1 & -1 \\ \hline
-1 & 0 & 1 & -1 \\ \hline
1 & -1 & 0 & 1 \\ \hline
1 & 1 & -1 & 0
\end{array}
$$
Let ${a,b,c,d}$ a mixed strategy for the first player, and regardless of which strategy player two chooses, the EV should be equal. So I calculated the EVs for the four different pure strategies of player two and equated them.
Player two selects first column: $0\cdot a + (-1)\cdot b + 1\cdot c + 1\cdot d = -b + c + d$
Second column: $a – c + d$
Third column: $-a + b – d$
Fourth column: $-a -b + c$
Equating them and introducing the additional restriction $a+b+c+d = 1$, I get the following system of equations:
\begin{align*}
a + b – 2c &= 0 \\
2a + b – 2c + d &= 0 \\
2b – c – d &= 0 \\
a + b + c + d &= 1
\end{align*}
which has no solution???? But I read also (called the fundamental theorem) that every matrix game has a solution??? So I am confused, but I can't spot if I made an error?
EDIT: Changed "Third column: $-a + b – d$" to "Third column: $-a + b + d$".
EDIT: The Game Matrix was wrong, in the third column the last entry should be "-1", not "1". Changed this, now the third column again reads: $-a + b – d$.
Best Answer
As has been pointed out in a comment, you're making the unwarranted assumption that the expected values of all of the opponent's moves must be equal. Only the expected values of the moves to which the opponent's mixed strategy assigns positive probability must be equal. That's because if they weren't, the opponent could shift probability from a worse move to a better move; but she can't shift probability away from a move that already has zero probability.