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This is wrong. If a strategy is actually GTO, then it is unexploitable. Meaning that if you’re playing GTO poker against a super loose player, you may not be ma

This is wrong. If a strategy is actually GTO, then it is unexploitable. Meaning that if you’re playing GTO poker against a super loose player, you may not be maximally exploiting them, but you should certainly not be losing money (in expectation). See Nash Equilibrium

skulk 3y ago
This is also wrong. Unexploitable strategies can still be -EV. This video does a good job of breaking it down: https://www.youtube.com/watch?v=HfuXD0YADIM https://www.youtube.com/watch?v=HfuXD0YADIM
- highfrequency 3y ago
  Video sets up a situation where GTO player is bluffcatcher vs nuts or bluff. He has negative EV with those cards in that spot, not negative EV marginally. If a GTO player has negative EV in heads up poker, he is not playing GTO.
antisthenes 3y ago
No, this isn't wrong. You don't understand multi-player poker and I see this meme repeated on many poker forums. It's possible for GTO to be worse than other strategies, based on suboptimal strategies of other players at the table.
- yunwal 3y ago
  You’re not disagreeing with the parent. They admitted GTO does not maximally exploit suboptimal strategies in their comment.
Kranar 3y ago
This is untrue. Nash equillibriums in general, and especially in poker, can be exploited in situations where one player is playing a negative EV strategy which happens at loose games. Morton's theorem can be used to derive examples where a player playing optimally under the assumption that all players are acting independently loses out to another player playing suboptimally under that same assumption due to the presense of yet a third player playing a negative EV strategy. These kinds of scenarios are known as implicit collusion, basically the two players who are implicitly colluding together have an advantage over the single player who is making optimal decisions. Ignoring the fact that there is no known GTO solution for more than 2 players, GTO only applies under the assumption that everyone is acting independently for some fairly formal and strict notion of independence. This requirement is broken in many real world games with more than 2 players potentially playing loose and wild, where they may not actually be colluding or coordinating with one another, but they behave as if they were.
- petesergeant 3y ago
  If one player can exploit the situation with a unilateral strategy, wouldn’t that mean it’s not a Nash equilibrium?
  - Kranar 3y ago
    Depends what you mean by unilateral. You can have a Nash equilibrium and also implicit collusion, for example if one player is playing GTO, that is they are playing optimally under the assumption that all other players are acting independently, and two other players are implicitly colluding with one another, you can still have a Nash equilibrium so that none of the players stand to gain anything by changing their strategy, and also the player playing GTO will have a negative expected value. Imagine a situation where the collusion was explicit, you are playing GTO against two opponents who are colluding with one another. A Nash equilibrium will still exist in that situation but your EV will be negative. Now explicit collusion is considered cheating but you can imagine a scenario where two players play in such a way that even though they are not explicitly colluding with one another, they behave the exact same way as if they were. This happens among a group of loose players where their actions as a group rewards the group as a whole against players who are acting independently. In other words, just because a strategy is optimal does not mean that it results in a positive expectation. You might be in a situation where playing perfectly just minimizes your losses, but you will still lose anyways.
- ilc 3y ago
  When I was playing I called this the "School of Fish" phenomenon. If there are weak players they can effectively end up colluding against a stronger player who is playing "More Optimally". Understanding the table, is a key skill. There's still a human element in the ring game.
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rerdavies 3y ago
It's not at all clear that there is a Nash Equilibrium in N-player poker, N > 2.

skulk 3y ago

This is also wrong. Unexploitable strategies can still be -EV. This video does a good job of breaking it down: https://www.youtube.com/watch?v=HfuXD0YADIM https://www.youtube.com/watch?v=HfuXD0YADIM

highfrequency 3y ago
Video sets up a situation where GTO player is bluffcatcher vs nuts or bluff. He has negative EV with those cards in that spot, not negative EV marginally. If a GTO player has negative EV in heads up poker, he is not playing GTO.

antisthenes 3y ago

No, this isn't wrong. You don't understand multi-player poker and I see this meme repeated on many poker forums. It's possible for GTO to be worse than other strategies, based on suboptimal strategies of other players at the table.

yunwal 3y ago
You’re not disagreeing with the parent. They admitted GTO does not maximally exploit suboptimal strategies in their comment.

Kranar 3y ago

This is untrue. Nash equillibriums in general, and especially in poker, can be exploited in situations where one player is playing a negative EV strategy which happens at loose games. Morton's theorem can be used to derive examples where a player playing optimally under the assumption that all players are acting independently loses out to another player playing suboptimally under that same assumption due to the presense of yet a third player playing a negative EV strategy. These kinds of scenarios are known as implicit collusion, basically the two players who are implicitly colluding together have an advantage over the single player who is making optimal decisions. Ignoring the fact that there is no known GTO solution for more than 2 players, GTO only applies under the assumption that everyone is acting independently for some fairly formal and strict notion of independence. This requirement is broken in many real world games with more than 2 players potentially playing loose and wild, where they may not actually be colluding or coordinating with one another, but they behave as if they were.

petesergeant 3y ago
If one player can exploit the situation with a unilateral strategy, wouldn’t that mean it’s not a Nash equilibrium?
- Kranar 3y ago
  Depends what you mean by unilateral. You can have a Nash equilibrium and also implicit collusion, for example if one player is playing GTO, that is they are playing optimally under the assumption that all other players are acting independently, and two other players are implicitly colluding with one another, you can still have a Nash equilibrium so that none of the players stand to gain anything by changing their strategy, and also the player playing GTO will have a negative expected value. Imagine a situation where the collusion was explicit, you are playing GTO against two opponents who are colluding with one another. A Nash equilibrium will still exist in that situation but your EV will be negative. Now explicit collusion is considered cheating but you can imagine a scenario where two players play in such a way that even though they are not explicitly colluding with one another, they behave the exact same way as if they were. This happens among a group of loose players where their actions as a group rewards the group as a whole against players who are acting independently. In other words, just because a strategy is optimal does not mean that it results in a positive expectation. You might be in a situation where playing perfectly just minimizes your losses, but you will still lose anyways.
ilc 3y ago
When I was playing I called this the "School of Fish" phenomenon. If there are weak players they can effectively end up colluding against a stronger player who is playing "More Optimally". Understanding the table, is a key skill. There's still a human element in the ring game.
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rerdavies 3y ago

It's not at all clear that there is a Nash Equilibrium in N-player poker, N > 2.