[Solved]: Poker with Bluffing (game theory)

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:

• 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 $2\times 2\times 2 = 8$ 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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Question Source : http://cs.stackexchange.com/questions/10818

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