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 Jun 8 comment Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? But not winning more than my share does not mean I have negative EV, does it? Jun 8 revised Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? added 81 characters in body Jun 8 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 :) Jun 8 revised What is the use of the Dot Product of two vectors? added 13 characters in body; deleted 1 characters in body Jun 8 awarded Teacher Jun 8 answered What is the use of the Dot Product of two vectors? Jun 8 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. Jun 8 asked Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? Mar 29 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. Mar 29 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. Mar 29 comment Cocktail bar problem very nice approach, thanks! Mar 29 accepted Cocktail bar problem Mar 29 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? Mar 29 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. Mar 29 revised Cocktail bar problem made question formulation more precise Mar 28 awarded Yearling Mar 28 asked Cocktail bar problem Mar 13 revised What is the order of $2$ in $(\mathbb{Z}/n\mathbb{Z})^\times$? edited title Mar 13 awarded Scholar Mar 13 accepted What is the order of $2$ in $(\mathbb{Z}/n\mathbb{Z})^\times$?