Possible Strategies in Extensive Form - Game Theory

Question

I have just started a course and had a question that I may be thinking too much of. Basically, I need to find all the strategies Person 3 has.

My reasoning was that person 1 has two choices and then person 2 has five choices based on what person 1 has made. Person 3 has 11 choices. So I thought, in order to find all possible strategies of person 3 I simply need to multiply the options which would be 2X5X11 = 110 possible strategies.

Picture of the extensive form

Is this correct or is there something I am overlooking?

Answer 1

From the figure, I would say that person 1 has 2 possible choices of actions $A$ and $B$; person 2 has 3 actions $C$, $D$ and $E$; and person 3 has also 3 actions $F$, $G$ and $H$. Thus, person 1 has 2 strategies, person 2 has 3 strategies and person 3 has 3 strategies. You may want to take a look at the definition of strategy.

The number of cases is the number of endpoints of the branches in the figure you posted: 11, in this case, meaning that not all combinations are possible. Hence, to determine the best action of each player, you'll have to study these 11 cases:

• The pay-off of player 3 playing $F$ given that player 2 played $C$ and player one played $A$;
• The pay-off of player 3 playing $G$ given that player 2 played $C$ and player 1 played $A$;
• The pay-off of player 3 playing $F$ given that player 2 played $D$ and player 1 played $A$;
• ...
• The pay-off of player 3 playing $G$ given that player 2 played $E$ and player 1 played $B$.

With this information, you can observe which is the best response that player 3 can choose in each case when player 2 chooses $Y\in\{C,D,E\}$ and player 1 chooses $X\in\{A,B\}$.

Answer 2

Generally, in an extensive form game, if a player moves at $M$ information sets and each information set $m$ has $n_m$ available actions, then this player has $n_1\times n_2\times\cdots\times n_M$ number of pure strategies.

Therefore, in the game you showed,

• Player 1 has $2$ pure strategies
• Player 2 has $2\times3=6$ pure strategies: $\{CC,CD,CE,DC,DD,DE\}$, where the notation $CD$ for example denotes the strategy "play $C$ if player 1 chooses $A$ and play $D$ if player 1 chooses $B$.
• Player 3 has $2\times3\times2\times2\times2=48$ pure strategies. One of these strategies could be $FFGGG$, which means "play $F$ regardless of player 2's choice as long as player 1 chooses $A$, and play $G$ regardless of player 2's choice as long as player 1 chooses $B$".