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Prove that if a player in an extensive-form game has only one information set, then his set of mixed strategies equals his set of behavior strategies.

This is the exercise $6.4$ on page $246$ in Game Theory by Michael Maschler, Eilon Solan and Shmuel Zamir.

But I think the example game $A$(absent-minded driver game) in $6.3$ invalidate this claim, since Player I can't reach the payoff $10$ by a mixed strategy but it's not the case for a behavior stategy. What's wrong?

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I think your counterexample is spot on. You might want to contact Eilon Solan, who manages a list of typos. –  Michael Greinecker Nov 4 '13 at 11:28
    
@MichaelGreinecker Not so fast. :) It is conceivable that "extensive-form game" has actually been defined (in the book) as "finite extensive-form game with perfect recall". –  Glen The Udderboat Nov 4 '13 at 12:03
    
@aufkag No. Exercise 6.3, the one that comes before, deals explicitely with cases of imperfect recall. Also, the chapter is Behavior strategies and Kuhn's Theorem. –  Michael Greinecker Nov 4 '13 at 12:07
    
@MichaelGreinecker OK. But (for my information): What then is their formal definition of a behaviour strategy? Wikipedia speaks of assignment at nodes, whereas you seem to think of assignment at information sets. (Which does make more sense.) –  Glen The Udderboat Nov 4 '13 at 12:16
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@Transmissionfrom, Hi, maybe you have noticed that Solan gave his correction on our website. This exercise is in fact a step in a theorem, but I haven't thought of it throughly. I think I will have time tomorrow to add some relevant information to make this question more "complete". –  Metta World Peace Nov 11 '13 at 10:59

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@EilonSolan Many thanks for drawing our attention to the way this exercise was phrased. We should have been more careful. Here is the new version of the exercise. I hope now it is clear. The new Exercise 6.4

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