Problem Detail: I’m doing a self-study of Game Theory Evolving by Gintis, and am stuck on problem 4.16 “Poker with Bluffing”. The first question asks “Show that Ollie has 64 pure strategies and Stan has 8 pure strategies.“. But no matter how I try to approach this, I can’t get more then 4 strategies for Stan! Here is the game tree for the game in question: 
The question marks (?) in the figure mean that the payoff depends on who has the higher card. Here is the game description:
The question marks (?) in the figure mean that the payoff depends on who has the higher card. Here is the game description:
- Two players, each with a deck of three cards: H (high), M(medium) or L (low).
- Each puts $1 in the pot, chooses random card
- Ollie (P1) either stays or raises
- Stan (P2) simultaneously also stays or raises
- If both raise/stay – highest card wins the pot (tie – they take their money back)
- If Ollie raises, Stan stays Ollie gets the $3 pot.
- If Stan raises and Ollie stays – Ollie gets another chance:
-> Drop – Stand wins the $3 pot
-> Call – add $1 to the pot.
Why does Stand have 8 pure strategies?
Asked By : drozzy
Answered By : Yuval Filmus
Stan has two pure strategies for each type of card. Since there are three types of cards (H, M, L), in total he has $2times 2times 2 = 8$ strategies. Here is a list of them:
- Always stay.
- Always raise.
- Stay only if H.
- Stay only if M.
- Stay only if L.
- Raise only if H.
- Raise only if M.
- Raise only if L.
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