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 Jun 12 comment Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? Note that . I don't require the strategy to have a positive expectation, only a non-negative one. Jun 9 comment Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? I'm sorry but If you are sure about your statement, you should be able to at least sketch some kind formal proof. If you make statements without proving them, I cannot accept your post as an answer. This is a game theoretical question which should not be answered by referring to (unproven) "common sense". Jun 9 comment Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? I'm not sure with the second collusion example either. Could you elaborate on why exactly our EV becomes negative there? Jun 9 comment What is the use of the Dot Product of two vectors? @Muprid: Good point. I've deleted the sentence now. Jun 9 comment What is the use of the Dot Product of two vectors? @Muphrid: Of course you can define something like an "angle" in arbitrary dimensions. I've changed "geometrical" to "visual", I think that makes more sense. Jun 8 comment Game Theory: Is there an “unexploitable” strategy in No Limit Holdem? I'm not sure whether collusion matters here. In your example for instance, we're effectively playing against one other player with a very strong range, but who also pays $(n-1)$-times the blinds we do, so it's at least non-trivial that this constellation gives us a negative EV, isn't it? 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 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 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. 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 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.