Which (pure strategies) in each game are dominated? For each dominated strategy specify the (mixed) strategy that dominates it.
The solution manual says that
in Game 1, R is dominated by $\sigma_2 = (1/2, 1/2, 0)$
and B is dominated by $\sigma_1 = (1/2, 1/2, 0)$
in Game 2, B is dominated by $\sigma_1 = (3/5, 2/5, 0)$
and M is weakly dominated by $\sigma_2 = (1/4, 0, 3/4)$
I don't understand this solution completely. And also, how these sigma values are written?
We will see these in order:
In Game 1, R is dominated by $\sigma_2 = (1/2, 1/2, 0)$
Notation: $\sigma_2$ is a mixed strategy for player-2 to play the choices $L,M$ with probability $\frac{1}{2}$ each.
Explanation for dominance: $\sigma_2$ dominates R as in R, player-2 wins 1 irrespective of what player-1 plays (see the right side of the column 3 values on table 1). Thus the expected earnings for player-2 when player-1 plays strategies $\{T,C,B\}$ is $\{1,1,1\}$. On the other hand, if Player-2 plays $\sigma_2$, then the expected earning is $\{\frac{4+2}{2},\frac{0+4}{2},\frac{3+1}{2}\} = \{3,2,2\}$ when player-1 plays $\{T,C,B\}$.
Note that $\{3,2,2\}$ dominates $\{1,1,1\}$. Therefore, irrespective of what player-1 plays, player-2 should play $\sigma_2$. Thus, $\sigma_2$ dominates R
In Game 1, B is dominated by $\sigma_1 = (1/2, 1/2, 0)$ The expected payoff to player-1 under B when player-2 plays $\{L,M,R\}$ is $\{1,1,1\}$ The expected payoff to player-1 under $\sigma_1$ when player-2 plays $\{L,M,R\}$ is $\{\frac{4+2}{2},\frac{0+4}{2},\frac{3+1}{2}\} = \{3,2,2\}$, which dominates $\{1,1,1\}$. Thus $\sigma_1$ dominates B.
In Game 2, B is dominated by $\sigma_1 = (3/5, 2/5, 0)$
The expected payoff to player-1 under B when player-2 plays $\{L,M,R\}$ is $\{2,2,2\}$ The expected payoff to player-1 under $\sigma_1= (3/5, 2/5, 0)$ when player-2 plays $\{L,M,R\}$ is $\{\frac{3}{5}\times 0 + \frac{2}{5}\times 6,\frac{3}{5}\times 3 + \frac{2}{5}\times 1,\frac{3}{5}\times 5 + \frac{2}{5}\times 6\} = \{\frac{12}{5},\frac{11}{5},\frac{27}{5}\}$, which dominates $\{2,2,2\}$. Thus $\sigma_1$ dominates B.
In Game 2, M is weakly dominated by $\sigma_2 = (1/4, 0, 3/4)$
The expected payoff to player-2 under M when player-1 plays $\{T,C,B\}$ is $\{3,3,6\}$ The expected payoff to player-2 under $\sigma_2= (\frac{1}{4}, 0, \frac{3}{4})$ when player-1 plays $\{T,C,B\}$ is $\{\frac{1}{4}\times 2 + \frac{3}{4}\times 6,\frac{1}{4}\times 6 + \frac{3}{4}\times 2,\frac{1}{4}\times 0 + \frac{3}{4}\times 8\} = \{5,3,6\}$, which weakly dominates $\{3,3,6\}$.
We say $\sigma_2$ weakly dominates M because (i) Payoff($\sigma_2$,Player-1 move) $\geq$ Payoff(M,Player-1 move) $\forall$ Player-1 moves, but (ii) $\exists$ Player-1 move such that Payoff($\sigma_2$,Player-1 move) $>$ Payoff(M,Player-1 move)