Can Nash Equilibrium strategy be exploited, if another player deviates?

Question

Consider a 3 player game (e.g. poker for 3 people).

Nash equilibrium is a set of strategies $s_1, s_2, s_3$, such that the expected profit value for each player $s_i$

$E_i(s_1, ..., s_i', ..., s_N)$

is maximized when $s_i'= s_i$

If I understand correctly, this applies when all other players are playing according to the equilibrium strategy.

However, suppose that the 3-rd player deviates from the equilibrium strategy $s_3' \neq s_3$. Could the first player choose a strategy $s_1'$ to exploit both the 3-th and the 2nd player, such that

$E_2(s_1', s_2, s_3')$ < $E_2(s_1, s_2, s_3)$ for the second player?

Answer 1

Here's an example from the book Naive Decision Making by Tom Körner. We have a 3 player game in which each player has two strategies A and B. The payoffs are that if everyone plays the same strategy they each get 2, otherwise the majority players get 3 and the minority 0. "All choose A" is a Nash equilibrium, since any individual player loses out by switching to B (they would get 0 instead of 2). On the other hand if player three deviates then player 1 does best by also deviating, assuming player 2 stays with A.