What is the formula for the value of a game in game theory?

mathlete
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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?
 
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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:
are L,R the nodes and l,r are the branches??
 
The Possible answer is v=(1-1)(-1=1)
mathlete said:
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?
 
The posible answer is v=(1-1)(-1+1)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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