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Skoochy Skoochy
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6 years ago
Which is NOT an axiom put forth by Lloyd Shapely?
  a.If x is the Shapely value vector for game v and y is the Shapely value vector for game , then game (v + ) has a Shapely value vector of x-y.
  b.If v(S-i) = v(S), then for all S xi=0.
  c..
  d.Relabeling of players interchanges the player's rewards.
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MacdonaldcMacdonaldc
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6 years ago
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Skoochy Author
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6 years ago
Smart ... Thanks!
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Yesterday
Just got PERFECT on my quiz
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2 hours ago
Brilliant
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