Game theory questions

Ex. 1

a. The extensive form and the strategic form are computed in the attachments. For the strategic form, the first letter denotes what to do if the card that player 1 gets is high, the second letter denotes what to do if the card that player 1 gests is low; for example, $PF$ means "Play if the card is high, fold if the card is low", while $PP$ means "Play in both cases". The expected payoffs are computed in the "expected payoff" file (look at the most recent version, it has slightly more details).

Since $PP$ dominates $FP$, and $PF$ dominates $FF$, we can conclude that player 1 should always play if they get a $H$, so we can simplify the table as in the picture "strategic_form_simplified".

b. If $1 < A < 2$, there is a unique Nash equilibrium in pure strategies: P1 always plays, and P2 always folds: in fact, if P1 switches to $PF$, their expected payoff goes to $0$, and if P2 switches to "Call", their payoff becomes $(A-4)/2 < -1$ for $1 < A < 2$.

If $A=2$, we have the same Nash equilibrium in pure strategies, but we also have an equilibrium in mixed strategies (see later).

If $A > 2$, there is no equilibrium in pure strategies. To find the equilibrium in mixed strategies, we proceed as follows. Suppose P1 picks strategy $PP$ with probability $p$, and $PF$ with probability $1-p$; P2 calls with probability $q$, and folds with probability $1-q$.

For P1, the expected payoff for $PP$ is \[ q \cdot \frac{4-A}{2} + (1-q) \cdot 1 \] and the expected payoff for $PF$ is \[ q \cdot \frac{3}{2} + (1-q) \cdot 0; \] $P1$ will use mixed strategies if and only if those are the same, meaning  \[ q \cdot \frac{4-A}{2} + (1-q) \cdot 1 = q \cdot \frac{3}{2} + (1-q) \cdot 0 \] which gives \[ 2q - \frac{A}{2} q + 1 - q = \frac{3}{2} q \] or $q = \frac{2}{A+1}$.

For P2, the expected payoff for "play" is \[ p \cdot \frac{A-4}{2} + (1-p) \cdot (-\frac{3}{2}) \] and the expected payoff for "fold" is \[ p \cdot (-1) + (1-p) \cdot 0, \] so again we equate them and get \[ \frac{A}{2} p - 2p - \frac{3}{2} + \frac{3}{2} p = -p \] which gives \[ \frac{A+1}{2} p = \frac{3}{2} \] or $p = \frac{3}{A+1}$.

For $A=2$, this gives the mixed strategy where P1 always plays, and P2 plays with probability $2/3$ and folds with probability $1/3$. In general, for $A>2$ this gives the mixed strategy where P1 always plays with a high card, plays with probability $\frac{3}{A+1}$ with a low card, and P2 plays with probability $\frac{2}{A+1}$.

Ex. 2

a. The proposition "Ann is in Paris" is represented by $E = \{a,c\}$. The proposition "Ann is in London" is represented by $F = \{b,d\}$.

b. If Bob knows that Ann is in Paris, then we must be in state $a$ (Bob can't distinguish $c$ and $d$), so $K_{Bob} E = \{a\}$.

c. Carla cannot distinguish between states $a$ and $b$, and between states $c$ and $d$, so she cannot determine whether Ann is in Paris or not. It follows that $K_{Carla} E = \varnothing$.

d. We have $K_{Bob} E = \{a\}$ and $K_{Bob} F = \{b\}$, so $K_{Bob} E \cup K_{Bob} F = \{a,b\}$, so \[ K_{Carla} (K_{Bob} E \cup K_{Bob} F) = \{a,b\}. \]
e. We have shown that the event "Bob knows where Ann is" is $K_{Bob} E \cup K_{Bob} F = \{a,b\}$, and we already said that the event "Ann is in Paris" is $E = \{a,c\}$, so the event "Ann is in Paris and Bob does not know that Ann is in Paris" is \[ G = E \cap \lnot (K_{Bob} E \cup K_{Bob}F) = \{a,c\} \cap \lnot \{a, b\} = \{a,c\} \cap \{c,d\} = \{c\}. \] By definition, the event "David considers $G$ possible" is \[ \lnot K_{David} \lnot G = \lnot K_{David} \{a,b,d\}, \] bud David cannot distinguish any case so $K_{David} \{a,b,d\} = \varnothing$, which in turn means $\lnot K_{David} \lnot G = \{a,b,c,d\}$, meaning that David always considers it possible that Ann is in Paris and Bob doesn't know that Ann is in Paris.