The reason this doesn't work is the same as the one I gave in this answer to this question of yours. Indifference must hold only between options that are assigned non-zero probabilities, and in the present case the strategies in the only Nash equilibrium assign zero probability to $S$, so there's no indifference between $S$ and $G$ at equilibrium.
That Wikipedia article is rather misleading in that, though it doesn't say so explicitly, it gives the impression that that algorithm can always be used to find a Nash equilbrium, whereas, as you've now twice confirmed, a mixed strategy supported on all pure strategies is in fact rather a special case. You can perhaps most easily see this by noting that no mixed strategy could ever cause your opponent to be indifferent between two options of which one is strongly dominated by the other.
Firstly, as you noted that game has no equilibrium in pure strategies.
It is then true that because player II has only two pure strategies that there must be an equilibrium where both players have support size 2 (if the game was degenerate there could be other types of equilibria too, but it turns out not to be degenerate, i.e. for every mixed strategy with support size $k$ there are at most $k$ best responses; thus, because the game is non-degenerate and zero-sum, there is a unique equilibrium where both players play exactly two strategies).
Which supports of player I are candidate supports? Well we only need to consider $2 \times 2$ sub-games with "cyclic preferences", since only these can allow to make the other player indifferent.
The only two possibilities for player I's supports are $A_1,A_3$ and $A_2,A_3$ (for $A_1,A_2$ we see that player II prefers $B_2$ against both rows and so cannot be made indifferent).
Now we try both. Since the game is zero-sum and non-degenerate we know that in one case, even though we can find probabilities for player II to make player I indifferent between the two strategies, the third will give a better payoff (which happens for $A_1,A_3$).
So, it turns out that the unique equilibrium is for player I to play $A_1,A_2,A_3$ with probabilities $0,0.8,0.2$ and player II to play $0.2,0.8$. Against player II's strategy, player I gets the same, best payoff for $A_2$ and $A_3$ as required.
For $2 \times n$ or $n \times 2$ games (even non-zero sum games) it is easy to analyse them with a diagram of best responses plotted against the mixed strategies of the player with two strategies. See my answer here
Mixed strategy nash equilibria in from $2\times N$ bimatrix form
Also check out our online game solvers:
http://banach.lse.ac.uk/
and
http://gte.csc.liv.ac.uk/gte/builder/
Best Answer
Here's one sensible sequence of steps:
Step 1: Notice that T strictly dominates $B$, since $(3,1,4)$ is componentwise strictly greater than $(1,0,3)$. Remove $B$ and we are left with a $2 \times 3$ game.
Step 2: In this new game, with $B$ removed, $R$ dominates $C$, since $(2,3)$ is componentwise strictly greater than $(1,2)$. After removing $C$ we are left with a $2 \times 2$ game:
$$ \left( \begin{array}{c|cc} & L & R\\ \hline T &3,0& 4,2\\ M &3,4& 2,3\\ \end{array} \right) $$
Step 3: Having found two pure equilibria already, look for non-pure equilibria. Player 2 can be made indifferent between $L$ and $R$ as we see below. But, player 1 cannot be made indifferent between $T$ and $M$ because $T$ weakly dominates $M$: as soon as there is any positive probability on $R$, player 1 strictly prefers $T$. Thus player 2 cannot mix in equilibrium, and actually the pure equilibrium $(M, L)$ is actually only the endpoint of a range of equilibria:
$$ ((1-p, p), L)\ \text{where } p \in [2/3, 1] $$
The threshold of $p=2/3$ is the point at which player II is indifferent between $L$ and $R$ against $(1-p,p)$. When $p=2/3$ both L and R give expected payoff $1/3 \cdot 0 + 2/3 \cdot 4 = 1/3 \cdot 2 + 2/3 \cdot 3 = 8/3$.
A range of equilibria like this is only possible is a degenerate game. A $2 \times 2$ game is necessarily degenerate. More generally, a game is degenerate if there exists a mixed strategy $x$ with support size $k$ (i.e. , $|\{x_i \ | \ x_i >0\}| = k$) and more than $k$ best responses against $x$. In this example, the strategy $x$ is the pure strategy $L$, which has support size 1 but two best responses against it, $TM$ and $M$.
You can check the equilibria at:
The output from the former for this game is:
2 x 2 Payoff matrix A: 3 4 3 2 2 x 2 Payoff matrix B: 0 2 4 3 EE = Extreme Equilibrium, EP = Expected Payoff Decimal Output EE 1 P1: (1) 0.333333333333 0.666666666667 EP= 3.0 P2: (1) 1.0 0.0 EP= 2.66666666667 EE 2 P1: (2) 1.0 0.0 EP= 4.0 P2: (2) 0.0 1.0 EP= 2.0 EE 3 P1: (3) 0.0 1.0 EP= 3.0 P2: (1) 1.0 0.0 EP= 4.0 Rational Output EE 1 P1: (1) 1/3 2/3 EP= 3 P2: (1) 1 0 EP= 8/3 EE 2 P1: (2) 1 0 EP= 4 P2: (2) 0 1 EP= 2 EE 3 P1: (3) 0 1 EP= 3 P2: (1) 1 0 EP= 4 Connected component 1: {1, 3} x {1} Connected component 2: {2} x {2}Here one of the two equilibrium components (the range of equilibria described above in terms of $p$) is denoted by
Connected component 1: {1, 3} x {1}The other equilibrium component is the pure strategy equilibrium $(T,R)$.