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  • First subgame is a 2-person simultaneous game. The Nash Equilibria in pure strategies are: (No, L) and (No, NL).
  • Player 1 has a dominant strategy of No (so PL1 never mixes strategies in a solution).
  • For PL2, every profile (L, NL)=(q,1-q) can be part of a mixed strategy solution.
  • Second subgame is a simple 1 person decision problem with Nash equilibrium Yes. This yields a payoff of $(5,5)$

Now, I am not sure what are all pure and mixed strategies Nash Equilibrium and also all the subgame perfect equilibrium.

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  • $\begingroup$ This is not a complex game and I just want to know how to solve properly. $\endgroup$ – Melina Apr 12 '16 at 5:59
  • $\begingroup$ And to know correct results! $\endgroup$ – Melina Apr 12 '16 at 6:17
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I think the point of this problem is to illustrate the difference between a Nash equilibrium and a subgame-perfect Nash equilibrium.

For subgame-perfect equilibria, the situation is clear. Player $2$ cannot make an empty threat to play $(0,0)$ in the left-hand branch; she plays $(5,5)$, and thus Player $1$ plays $A$; there is no other equilibrium.

If the Nash equilibrium doesn't have to be subgame-perfect, the situation is quite a bit more complicated. Player $2$ can threaten to play $(0,0)$ with non-zero probability $1-p$ in the left branch, thus making this branch worth an expected payoff of $5p$ for Player $1$. In the right-hand branch, as you wrote, Player $1$ will always play "No", and Player $2$ can play L with probability $q$ and NL with probability $1-q$, giving an expected payoff of $3q+4(1-q)=4-q$ to Player $1$.

There are two types of equilibria, depending on whether Player $1$ prefers A or not.

Player $1$ prefers A: Here Player $2$ must choose $p=1$, since the left-hand branch actually gets played. $q$ can be anything, since the right-hand branch doesn't get played and any value of $q$ lets Player $1$ prefer $A$ if $p=1$.

Player $1$ doesn't prefer A: Here Player $2$ can have mixed strategies in both branches, subject to the condition $4-q\ge5p$ so that Player $1$ doesn't prefer A. In these equilibria, Player $1$ always plays B. Note that this includes (weak) equilibria with $4-q=5p$ in which Player $1$ is indifferent between A and B. Nevertheless, for an equilibrium Player $1$ must play B and cannot mix strategies, since Player $2$'s empty threat in the left branch wouldn't be in equilibrium otherwise.

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