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Suppose two players A and B remain in the game of poker and they both have the same number of chips.

Suppose player A follows the Nash equilibrium strategy.

  1. Is the player's A expected value guaranteed to be greater or equal to 0?

  2. If player B does not follow the Nash equilibrium strategy, will player A have a strictly positive EV?

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  • Nash equilibrium is not necessarily a strategy in poker, but a concept. The concept is that if player A knows that player B is playing a perfect GTO strategy, player A's best possible strategy is to play perfect GTO as well because a perfect GTO strategy can not be exploited. This leaves player A and player B in a sort of equilibrium where neither player will have an advantage over the other.
    – Clarko
    May 9 at 5:52
  • @Clarko, so if player A uses GTO, in the case of two players his EV will never be negative? (I know that's not the case if there are 3 players or more) May 9 at 11:03
  • theoretically yes, averaged out over a large number of hands/decision points. This is just a theoretical yes because perfect GTO play is pretty unattainable in the real world, all human players will make some sort of mistakes at some point, small or large.
    – Clarko
    May 10 at 1:39
  • @mercury0114 Yes. The 3-player situations where a GTO strategy can become -EV cannot occur with just two players. Those are only cases of "implicit collusion".
    – David
    May 11 at 8:25
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  1. Equal, but not greater. Otherwise both players could use the same strategy and money would be created out of nowhere. The exception is of course if the game only lasts one hand, in which case the SB would be expected to win in the Nash equilibrium.

  2. Yes. Players A and B will be in a Nash equilibrium if neither player has an incentive to deviate (in other words, the player who deviates does not win). In poker, there is only one strategy per player that leads to a Nash equilibrium, so in this case, "deviating and not winning" implies "deviating and losing".

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