An extreme equilibrium $(x,y)$, where $x$ and $y$ are the mixed strategies of players 1 and 2, is a Nash equilibrium where for both players mixed strategies cannot be described as convex combinations of other mixed strategies that form equilibria.
Fact: There are always finitely many extreme equilibria.**
For a mixed strategy $x$, let the support of $x$ be defined as the set of pure strategies that $x$ uses with positive probability, i.e.,
$$\text{support}(x) := \{ i : x_i > 0 \}.$$
A game is called non-degenerate game if for all mixed strategies $x$, if $|\text{support}(x)| = k$ then the number of best responses against $x$ is at most $k$.
Fact: Every non-degenerate game has only extreme equilibria.
For a simple example of a game with non-extreme equilibria consider a $2\times 2$ game where both players get $0$ for all strategy profiles. In this case, all four pure profiles are extreme equilibria and all strict mixtures are non-extreme Nash equilibria. For a less trivial example, consider any $2 \times 2$ bimatrix game where there is no strict dominance for either player but exactly one of the players has a weakly dominant strategy. Then, the game is degenerate and the game will have infinitely many equilibria. As an example, consider the following game.
$$A = \left( \begin{array}{ccc}
1 & 0 \\
0 & 1 \\
\end{array} \right)
\quad
B = \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \\
\end{array} \right)
$$
The game has exactly two extreme equilibria. Here is the output from the online game solver http://banach.lse.ac.uk, which shows these two extreme equilibria:
2 x 2 Payoff matrix A:
1 0
0 1
2 x 2 Payoff matrix B:
1 1
1 0
EE = Extreme Equilibrium, EP = Expected Payoff
Decimal Output
EE 1 P1: (1) 1.0 0.0 EP= 1.0 P2: (1) 1.0 0.0 EP= 1.0
EE 2 P1: (1) 1.0 0.0 EP= 0.5 P2: (2) 0.5 0.5 EP= 1.0
Rational Output
EE 1 P1: (1) 1 0 EP= 1 P2: (1) 1 0 EP= 1
EE 2 P1: (1) 1 0 EP= 1/2 P2: (2) 1/2 1/2 EP= 1
Connected component 1:
{1} x {1, 2}
The final line of this output shows that against the unique equilibrium strategy of player 1 (top row), player 2 can play any convex combination of $(1,0)$ and $(1/2,1/2)$.
For a thorough technical exposition, which covers both finding extreme equilibria and then from these all equilibria of bimatrix games, see, e.g.,
David Avis, Gabriel D. Rosenberg, Rahul Savani, and Bernhard von Stengel (2010).
Enumeration of Nash equilibria for two-player games.
Economic Theory, 42:9–37.
On the relationship between correlated equilibria and Nash equilibria:
Fact:: The set of correlated equilibrium
distributions of an $n$-player noncooperative game is a convex
polytope that includes all the Nash equilibrium distributions. In fact, the Nash equilibria all lie on the boundary of the polytope.
See:
Robert Nau, Sabrina Gomez Canovas, and Pierre Hansen (2004).
On the geometry of Nash equilibria and correlated equilibria.
International Journal of Game Theory 32: 443-453.
Yes, there are infinitely many, with player 2 mixing.
Assume that player 2 mixes: Here it may be possible because the payoff for player 2 in $(s,s,n)$ and in $(s,n,n)$ is equal. You only need to check if $(s,(y,1-y),n)$ is an equilibrium for $0<y<1$. Given that $3$ plays $n$, $s$ is a best response for player $1$ for any $y$ of player $2$ such that $$(-1)y+(1)(1-y)\ge (1)y+0(1-y)\implies y\le \frac13$$ Now, given that $1$ plays $s$, $n$ is a best response for player $3$ for any $y$ of player $2$ such that $$(-2)y+0(1-y)\ge(1)y+(-1)(1-y)\implies y\le \frac14$$ Hence, any strategy profile of the form $$(s,(y,1-y),n)\quad \text{with } 0<y\le \frac14$$ is a Nash equilibrium where only player 2 uses a mixed strategy.
To see that there are no other such equilibria:
Assume that player $3$ plays the mixed strategy $(z,1-z)$ where $0<z<1$ is the probability of playing $s$. If players $1$ and $2$ play the pure strategy profile $(s,s)$ then player $3$ has an incentive to choose $z=1$, hence this is not an equilibrium. If players $1$ and $2$ play any other pure strategy profile, then player $3$ has an incentive to choose $z=0$, hence there is no equilibrium where $3$ uses a mixed strategy and $1$ and $2$ use pure strategies.
Repeat for player 1: Assume that player $1$ plays the mixed strategy $(x,1-x)$ where $0<x<1$ is the probability of playing $s$. If players $2$ and $3$ play the pure strategy profile $(s,s)$ then player $1$ has an incentive to choose $x=1$, hence this is not an equilibrium. Similarly, if players $2$ and $3$ play any other pure strategy profile, then player $1$ has an incentive to choose either $z=1$ or $x=0$, hence there is no equilibrium where $1$ uses a mixed strategy and $2$ and $3$ use pure strategies.
Best Answer
Your reasoning is correct, the player can deviate to any strategy. Thus, your suggested strategies and randomization device do not constitute a correlated equilibrium.
The way you try to construct it, the correlated equilibrium cannot get more than Nash equilibrium. The main point is that you observe your recommendation but not the recommendation to the other player. However, you construct it in such a way that the players should play either $(T,L)$ or $(B,R)$. Thus, if the a player is assigned a strategy, she knows the assigned strategy of the other player. Let me illustrate this point by constructing a correlated equilibrium in your example.
Consider the following recommendations. With probability $x_1=\frac{1}{4}$ the players are assigned the strategies $(T,L)$, with probability $x_2=\frac{3}{8}$ the strategies $(M,L)$, and with probability $x_3=\frac{3}{8}$ the strategies $(M,R)$.
Now, if Player 1 observes the recommendation $T$, she is sure that player 2 was recommended $L$ and $T$ is a best reply. If player one observes the recommendation $M$, she is not sure whether player 2 is supposed to play $L$ or $R$. However, she knows that both strategies are equally likely given her own recommendation. Thus, she is indifferent between playing either of her strategies. In particular, $M$ is a best reply.
If Player 2 observes the recommendation $L$, she knows that Player 1 is recommended $T$ with probability $\frac{2}{5}$ and $M$ with probability $\frac{3}{5}$ (Bayesian updating). In this case, she is indifferent between playing $L$ or $C$, so $L$ is a best reply. If player 2 observes the recommendation $R$, she knows that player 1 is recommended $M$ and $R$ is a best reply. So overall you have a correlated equilibrium.