# Game Theory - Bayes Rule, Sequential Game

I am trying to solve the following model, but I get a few weird results. Sorry if it is too long...

Nature moves first and with probability $p$ assigns player's 1 type to be High ($1-p$ for Low) [edit: Player 1 knows his own type]; Player 2 does not observe Player's 1 type but knows p.If player 1 is High type he may either Quit (at a cost $C$ which is paid to Player 2) or Play. If he is Low type, then he necessarily plays.

Assuming Player 1 had not quit, Player 2 moves. He may choose to either play Passive or Aggressive. If player 2 plays Passive, both he and Player 1 get $0$, regardless of player 1's type. If player 2 plays Aggressive, he pays a price $x$, but can win a prize $D,(D>C)$ with probability $q$, to be paid from Player's 1 pocket, but only if player 1 is High type; if Player 1 is Low, Player 2 pays the price $x$ but gets $0$ with certainty. What Player 2 wins, Player 1 loses.

I am looking for a solution with mixed strategies. Let $\delta$ be the probability that Player 1 chooses Play if he is High type and let $\lambda$ be the probability that Player 2 chooses to play Aggressive, conditional on Player 1 playing.

Player 1's payoff if he is Low: 0. Player's 1 payoff if he is High and Quits: $-C$; High and Plays: $-\lambda D Q$. In equilibrium, $\lambda$ is such that $-C=-\lambda d Q = \lambda=C/DQ$

If Player 2 sees the Player 1 had not quit, he forms the (correct) belief that the probability that Player 1 is High type is: $p \delta/ (p \delta + 1- p)$. That means that if he plays, he can expect a payoff of: $-x+qD p \delta/ (p \delta + 1- p)$ and if he quits, a payoff of $(1-\delta)C$. In equilibrium, Player 1 will choose $\delta$ such that: $(1-\delta)C=-x+qD \frac{p\delta}{(p \delta + 1- p)}$. Hence, $\delta =\frac{\pm\sqrt{(Dpq-px+C-2Cp)^2-4Cp(Cp+px-x)}+2pC-C-Dpq+px}{2pC}$.

Questions:

1. Is my application of the Bayes rule above correct? What about use of mixed strategies?
2. If so, when I assign the following values to the parameters, I get $\lambda>1$, which does not make sense: $x=300, C=1000, D=1300, p=0.5, q=0.5$. Should I interpret that as indicating a pure strategy with $\lambda=1$? What about the possibility of negative values of $\lambda$?
• Can player 1 see his type? – Juanito Jul 1 '14 at 22:21
• Yes, player 1 knows his own type. – Puzzle_riddle Jul 2 '14 at 5:07