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Problem_Set_1 - That is there is some play of the game for...

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Kaushik Basu Spring 2008 Econ 367. Game-Theoretic Methods Problem Set 1 1. In the game of Hex we proved that Black does not have a winning strategy. The proof took the form of showing that if Black has a winning strategy then White can always win (which is of course a contradiction). Why can we not construct a similar proof to show that White does not have a winning strategy? Being able to answer this question to your satisfaction will help you understand properly the original proof (that is, of why Black does not have a winning strategy). 2. Try to write down in words why in chess you cannot use the kind of argument used in Hex to show that Black does not have a winning strategy. 3. Consider an n × n Hex board, where n is an odd number ( ). 3 From this board block off the hexagon in the middle, so that no one can place counters on it. Now suppose the game is played in the usual way – White moves first and tries to link the top edge to the bottom and Black ….. (a) Show that this game can end in a draw.
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Unformatted text preview: That is, there is some play of the game, for which the outcome is a draw. (b) Show that in this game either White has a winning strategy or both players have a strategy to ensure a draw (or, said differently, if both players were perfect, then the game will either always end in White winning or always in a draw). 4. Consider a two-player game in which there is a table with a round top and a diagonal line in red drawn on it. Two players play, in turn, by placing a coin on the table so that the coin does not touch the red line and does not touch another coin. The last player who is able to place a coin in this manner wins. Assume that the table is big enough so that (*) a coin can be fit into one half of the surface without touching the red line. (a) Would the first mover or the second mover win this game and what is his/her winning strategy? (b) Does your answer change if condition (*) is not satisfied....
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