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Game theory: value of a game

  1. Feb 26, 2006 #1
    The problem:
    "Player I can choose l or r at the first move in a game G. If he chooses l, a chance move selects L with probability p, or R with probability 1-p. If L is chosen, the game ends with a loss. If R is chosen, a subgame identical in structure to G is played. If player I chooses r, then a chance move selects L with probability q or R with probability 1-q. If L is chosen, the game ends in a win. If R is chosen, a subgame is played that is identical to G except that the outcomes win and loss are interchanged together with the roles of players I and II"

    *whew*

    Now the question is... if the value of the game is v, show that v=q+(1-q)(1-v)

    Now the game tree is so complicated... I really have no idea how to get the value of the game. Is there any easy way to do this that i'm missing?
     
  2. jcsd
  3. Feb 26, 2006 #2

    Hurkyl

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    I don't understand the statement of the game.

    What happens when player I picks `l', and `R' gets chosen? Is it now player II's turn? Does "win" always mean a win for player I? et cetera.

    If I sat down and tried to teach this game to someone else so we could play, I'd have no idea what the rules are. :frown:


    Anyways, the analysis should be straightforward.

    What is the expected value of the game if player I picks `l'?
    What is the expected value of the game if player I picks `r'?
    What is the expected value of the game if player I picks optimally?
     
    Last edited: Feb 26, 2006
  4. Feb 27, 2006 #3
    are L,R the nodes and l,r are the branches??
     
  5. Oct 11, 2009 #4
    The Possible answer is v=(1-1)(-1=1)
     
  6. Oct 11, 2009 #5
    The posible answer is v=(1-1)(-1+1)
     
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