"Inside  every small problem is a large problem struggling to get out."

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Respond to the question: Martin Gardner Game?

 10/21/2000 06:15 PM by Yue Zhang;
You can find the proof at " A Psychological Game" by N.S.Mendelsohn (American Mathematical Monthly 53, February 1946, pp.86-88). you can get it from Jstor [Manage messages]

 10/20/2000 04:14 PM by Walter;
I dont'have the book to take a look at the proof. If you could tell more details (how many actions the players choose, what is r,i, etc.)I could check. [View full text and thread]

 10/18/2000 07:21 PM by name withheld;
Thanks a lot for providing the proof. I still have some point not very clear in the proof. Why should p(i,r) be equal to p(2r+2-i,r)? "By symmetry"? [View full text and thread]

 10/17/2000 10:57 AM by name withheld; The Solution!
Spoiler Alert! Spoiler Alert! I was looking at my library last night and pulled out 'Time Travel and other mathematical bewilderments' by M.G. In a chapter called 'the rubber rope and other problems' he describes the game. He also [View full text and thread]

 10/17/2000 09:35 AM by Yue Zhang;
Yes, your answer in the first post is an explanation of why that's a NE. But that's not what he wants I think. I don't think we're far from what he wanted. For me, the most intuitive way to prove the result he wanted is to find a easy [View full text and thread]

 10/17/2000 09:15 AM by Walter;
I think we are taking this conversation farther from the main point, which is to find an intuitive explanation to the result. This is what I tried to do in the first answer. (I didn't know how far I needed to go since I don't know your [View full text and thread]

 10/16/2000 07:53 PM by Yue Zhang;
First, the player doesn't have to be indifferent among actions 1-6. He ONLY has to be indifferent among picking the numbers which is assigned positive probability in the optimal strategy. One mixed strategy doesn't necessarily include [View full text and thread]

 10/16/2000 07:35 PM by Yue Zhang;
First, the player doesn't have to be indifferent among actions 1-6. He ONLY has to be indifferent among picking the numbers which is assigned positive probability in the optimal strategy. One mixed strategy doesn't necessarily include [View full text and thread]

 10/16/2000 04:20 PM by Walter;
The previous answer was an intutitive way to show that any player wouldn't deviate from choosing the equilibrium action when the other didn't deviate. Here you have the steps you must follow to proof the Gardner statement for a [View full text and thread]

 10/16/2000 01:39 PM by Yue Zhang;
I think you only explained why choosing numbers greater than 5 is not in the optimal stategy when the opponent ONLY randomize among 1-5. That's not enough for his question. He's asking why any strategy including picking numbers greater [View full text and thread]

 10/15/2000 07:58 PM by Walter;
After working on the exercise I reached the following conclusion. I'm using a recursive argument. The result holds for a game with finite number of actions N. Start with a game where players choose 1,2,3. Conclude that the only [View full text and thread]

 10/13/2000 05:42 PM by doug bennett;
Your question takes me aback. Is there a different theory for what a 'solution' is when a game is played once rather than a finite number of times? And what does it mean to play a game an actual (countable? aleph-1?) infinite number of [View full text and thread]

 10/13/2000 03:53 PM by Walter;
I need some extra information. Is this game a single infinite game, or a sequence of finite games? [View full text and thread]

 10/12/2000 02:42 PM by Doug Bennet; Martin Gardner Game
I believe I remember this game from one of Martin Gardner's books of problems from Sci. Amer. Two players pick positive numbers and the higher pays the lower \$1, unless the higher is exactly one more, in which case the lower pays [View full text and thread]