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 Jun8 revised Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? added 81 characters in body Jun8 comment Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? Ok, your right, this is trivial. Let's assume collusion is not possible as it is against the rules anyway :) Jun8 revised What is the use of the Dot Product of two vectors? added 13 characters in body; deleted 1 characters in body Jun8 awarded Teacher Jun8 answered What is the use of the Dot Product of two vectors? Jun8 comment Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? Note I said "non-negative", not "positive". So if all players apply the same strategy, the expected payoff for everyone would be zero, which is non-negative. Jun8 asked Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? Mar29 comment Cocktail bar problem @Aryabhata: Of course $\frac{n-1}{2}$ are not enough, but using the construction of the book one can obtain $\frac{n+1}{2}$ paths that satisfy our requirements. Mar29 comment Cocktail bar problem Ok the construction described in the book works as well for the case of $n$ being odd, for which it yields $\frac{n+1}{2}$ paths where the $\frac{n-1}{2}$ pairs $\{1,n\},\{2,n-1\},\dots$ are covered exactly twice, all others being covered exactly once. Mar29 comment Cocktail bar problem very nice approach, thanks! Mar29 accepted Cocktail bar problem Mar29 comment Cocktail bar problem I understand the proof for the case of $n$ being even, and it is indeed the minimum as $|A|\geq \frac{n}{2}$ which Greg Martin pointed out in his answer. But in case of $n$ being odd, we need at least $\frac{n+1}{2}$ paths, and I don't see how to construct them from the $\frac{n-1}{2}$ cycles. Am I missing something trivial? Mar29 comment Cocktail bar problem @hardmath: I'm not looking for the number of swaps that have to be made, I'm only interested in $A$ or its cardinality. I reformulated the question in order to make it more clear what I mean. Mar29 revised Cocktail bar problem made question formulation more precise Mar28 awarded Yearling Mar28 asked Cocktail bar problem Mar13 revised What is the order of $2$ in $(\mathbb{Z}/n\mathbb{Z})^\times$? edited title Mar13 awarded Scholar Mar13 accepted What is the order of $2$ in $(\mathbb{Z}/n\mathbb{Z})^\times$? Mar12 awarded Supporter