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No single optimal strategy in poker

  1. Jun 17, 2008 #1
    This is probably a stupid question, but I couldn't find it through searching on google. I understand that there is no single optimal strategy in poker, and the only way to maximize your expected value is to try to exploit patterns in your opponents play, but why are pot-odds not enough to be "invincible" at poker? That is, if you make every decision based on pot-odds I do not see how you could ever get exploited, since pot odds by definition means that you only play when you have a mathematical advantage.
     
    Last edited: Jun 17, 2008
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  3. Jun 17, 2008 #2

    matt grime

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    That assumes that opponents play with the same strategy (and then your net win will be zero anyway since you're all playing properly).
     
  4. Jun 17, 2008 #3

    nicksauce

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    "Pot odds" also assumes you have perfect information, which you never do. Example: I think I am getting good odds on my flush draw, so I stay in the pot, but in reality one of my opponent's has a better flush draw.
     
  5. Jun 17, 2008 #4
    So what would be an example of a strategy that I could use to exploit someone who was playing according to pot odds?

    Wouldn't that already be calculated in, for example you know that every so often your opponent will have a better flush draw and therefore your EV is slightly less.
     
  6. Jun 17, 2008 #5

    matt grime

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    What kind of game are you playing? One with 'perfect' information, such as Texas Hold 'Em? I.e. one where you know the best possible winning hand? Or ordinary draw poker? It's a very difficult game to model: after all if in Texas you know you have the best possible hand, then you're stupid to bet outrageously since no one will meet you, but the 'pot odds' state that you can bet an arbitrary number of dollars, since you'll win an arbitrary amount in return, but in doing so, no one else will meet your raise.

    Actually, in Hold 'Em you're probably best off playing the odds, since you know them, but if everyone played 'em you'd go home all square in the long run.
     
    Last edited: Jun 17, 2008
  7. Jun 17, 2008 #6
    The problem with 'pot-odds' is someone could just 'bluff' you every time. If you probabilistically set sum static threshold on what you'll bet for a given hand (based on it's value) then the other guy will just bet over that every time and force you to fold regardless of what he has because he knows that's how you'll bet. And similarily if he knows that you're assuming the he's playing that way he'll just represent the hand he wants you to think he has and you'll act accordingly every time.
     
  8. Jun 17, 2008 #7

    That sounds reasonable, but what about the 1 in 221 times when you get AA, and you know you have the best hand, could it be possible that they would give all of their winnings back? It seems kind of similar to if you bet $9,998 that an event with a 1/9999 probability of happening would not happen to win $1. 9,998 times you win $1, and 1 time you lose $9,998 and end up even.
     
  9. Jun 17, 2008 #8
    No because THEY'RE not doing pot odds. In fact it makes it even less likely because if they see you continuing to bet really high they're just going to go 'oh crap I guess he really must have something good' and fold. Which means you could ultimately bluff someone by making them THINK you're playing only pot odds but that'd be a deception which is really just saying the same thing (you need a dynamic intelligence to win at poker)
     
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