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Do odds change when short handed as opposed to when you start the action 10 handed?

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  • Do odds change when short handed as opposed to when you start the action 10 handed?

    Someone asked this:

    “If the action folds to me on the button in a 10 handed game, is there any reason to think that the SB and BB might have a better hand than if I am first to act, on the button, in a three-handed game? "

    I had to work this type of thing out in my head when I played bridge.

    Assume you deal the SB and BB first, and haven’t dealt any other hands yet.
    (The order that the cards are dealt doesn’t change the odds of what they get dealt)

    The odds that the SB has JJ+, AJs+, KQs, AQo+ is 4.83 % and the odds that the BB has JJ+, AJs+, KQs, AQo+ is also 4.83 %

    Now deal out 8 more hands, but don’t look at them yet. Have the odds changed? (No...)

    Now look at all 8 hands (but not the SB or BB). Now the odd do change – for instance if all 4 aces are visible then the odds for the SB or BB having this range drop to 1.95%
    (Since there are no A's left)

    So if 7 of the people fold to you, it does change the odds slightly (If we assume that none of the folders had JJ+, AJs+, KQs, AQo+, then more of these cards may already be in the SB or BB) But we would have to run a lot of scenarios to determine how much the folded hands increased the odds of the SB or BB having a premium hand.

    However, if we were just 3 handed, and no one had folded, then the odds are still 4.83 % until we look at our hand.
    (When we look at our hand, the odds change a bit depending on what we have. If we have AA, then it is a lot less like that one of the blinds has an A)

    But the odds of running into a premium hand to your left, are the same whether 10 handed with 7 folds, or 3 handed and no one has acted yet.

    Basically, calculate the odds of running into a premium hand on your left as 5% times the number of players left whether the other players have folded or are sitting out.

  • #2
    I'm sure i'll butcher this but here goes nothing.

    Your odds do not change. If you have the nut flush draw it's still 25% to make thee flush.

    The odds though that your AX and draw is the best hand does improve with two players compared to 8 because your odds are competing against the odds of the other players exact holdings which increase (or in your case decrease) for every player in the hand.

    The reason for this is that with every live hand the number of cards you have to beat (put another way, the number of cards that hurt you) increases.

    This is variance. Let's says you are in a game where no one folds. The absolute odds of your holdings in a single hand are are really meaningless because in only one hand we are gambling. You have to play 200k hands to overcome variance.

    Let's say you change tables where all hands isolate and each hand is played out heads up. One hand is still meaningless in the greater scheme of things but it may only take you 20k hands to overcome variance.

    Is that clear?


    • #3

      Variance is an explanation for why things do not always work the way probabilities would predict. But still if you make the right play based on probability, you will profit in the long run. And it is usually the best way to go.

      But the specific question here was,.for example, if I push with J9 suited, and will only get called in the SB or BB have a Premium hand, JJ+, AJs+, KQs, AQo+, does the fact that it is 3 handed, rather than 10 handed change my chances of winning or my expected return?

      Here is some math:

      If we assume that the SB or BB will only call 5% of the time, then 10% of the time, one will call and I will win the antes 90% of the time when no one calls, and double up 31% of the time when they do call, and lose 59% of the time.

      So my expected return for shoving with 10 BBs is:

      10 Handed - 25/100/200 blinds:

      90% of the time, win the antes and blinds 550 chips, 10% of the time get called and win 31% of the time - 2000 chips + 450 chips for antes and the other blind =
      550 * 0.10 + 2450 * 0.10 * 0.31 = 130 chips.

      3 handed - 25/100/200 blinds:

      90% of the time, win the antes and blinds 375 chips, 10% of the time get called and win 31% of the time - 2000 chips + 275 chips for antes and the other blind =
      375 * 0.10 + 2275 * 0.10 * 0.31 = 108 chips.

      So you should push with a tighter range 3-handed, rather than 10 handed - but not because there is a higher probability of running into a premium hand.

      Float the Turn App

      By the way J9 suited is a solid push - 3 handed or 10 handed - based on the Float the Turn Push/Fold App:
      Click image for larger version

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      • #4
        The first paragraph is incorrect. When we get aces cracked, it's not variance, that is the normal reality of the math. Aces have to lose against specific hands X number of times on order for chance to be a variable in the game.

        I understand the question in paragraph 2 and your conflating odds with, let's say something else for clarity. The concepts you demonstrate in the subsequent paragraphs are EV which connect the odds of a series of specific hands and stacks/bet sizes.

        Try this kind of like your example
        P1 has XX and 70% card odds to hold
        P2 has YY and 60% card odds to hold

        How can this be, that's 140%? It's because we're not comparing the holdings to each other. Player 1 v Player 2 are now (say for example) 80/20 due to holdings on a single runout.

        So next hand the same preflop hands are out there but this time..
        P3 has ZZ and also has 60% card odds to hold.

        So the money goes in and based on the exact holdings it's now 60/15 15 with some assumptions. Because now the top hand instead of having 2 outs to fade, now has 4 outs to fade and due to the new player the best hands odds diminish. This is the source if variance.


        • #5
          Mike Caro says yes, due to card removal. I haven't done the math to figure this one out as I think the difference would be marginal at best. His theory was that since the first 7 players folded, there was a low % of the combinations A, K, and Q, among the first 7 players. Leaving a slightly higher frequency of high cards and less low cards than would be if the hand was dealt 3 handed.


          • jpgiro
            jpgiro commented
            Editing a comment
            I tend to concur that any effect would be marginal at best. While there are plenty of Ax/Kx/Qx combinations that can (and should) be folded in earlier positions, I cannot imagine that those card removal effects should play a significant role in our decision making.

        • #6
          “If the action folds to me on the button in a 10 handed game, is there any reason to think that the SB and BB might have a better hand than if I am first to act, on the button, in a three-handed game? "

          If one wants to answer the question asked then the answer is YES due to card removal. By how much, I haven't a clue. Seven payers folded because they did not have premium hands so there is more of a chance that the players left to act have. The difference might be marginal but the answer is yes.
          If the game was a turbo and the average stack was shallow , players would be shoving with any ace .


          • #7
            Here is a variation on the question. Suppose that you are 9-handed and everyone has over 200 BB.

            The odds that any particular player gets dealt a pocket pair are about 1/17. You are on the button. The odds that either of the blinds is dealt a pocket pair are about 2/17. The odds that one of the 9 players at the table gets dealt a pocket pair are about 9/17.

            Action folds to you on the button. Assume that this is a typical group of poker players and no one would open fold any pocket pair with a stack size of more than 200 BB (obviously, some guys always open fold small pairs but they are not at this table).

            You look down at your hand on the button, and you don't have a pocket pair. So the odds, before the deal, of one of the other players at the table having a pocket pair were 8/17. Now that you know that the first six players to act did not have a pocket pair, are the odds any higher that one of the blinds does have a pocket pair, or are they still about 2/17?


            • #8
              Fun logic question: here's my shot at it. In general, this is a question about variance. The specific answer to your question is that each player behind has a 1/17 chance of a pair. ( the 8/17 is incorrect it's 1 in 17, 8 discrete times.) My answer consequently is the same as Al's conceptually. It may be more probable the short run that the 1/17 could hit on this round but in reality the odds in the short run are 1/17 twice.

              Let's reverse the thought discussion. Let's say we see a hand where 3 players had pairs and they got drawn out on by AX flush. Should we no longer be concerned that in the next orbit or 6 that we won't see a pair? No, the odds are still 1/17 that each player may be dealt a pair on the next hand.

              We know that cards havve a funny way of running. One day we could see trips and sets taking down pots other days one pair is good enough. In order to make any assessment of increasing or decreasing probability you'd have to play a long session at a single sitting with no deck changes. over 170 hands you could generally say 'lots of pocket pairs today' over 1700 the wheel should have come and gone and the short term variance gets balances the equation through long term optimal play.

              This is the cornerstone of playing like a gambler, if you only play short sessions with the different groups of people, your play will be subject to more variance and playing optimally may or may not be the best way to play. In other words better to gamble. If you are serious about the game and play longer sessions and regularly with the same group, optimal play over time should yield better results vs gambling.
              Last edited by XBobLove; 06-14-2017, 08:03 PM.


              • #9
                In my question I made a dumb mistake in not distinguishing between probability and expected value. I recognized this when I considered that If you had, hypothetically, 17 players at the table, the expected number of pocket pairs would be about 1 (maybe it is affected slightly in some way by the removal issue) but of course the probability of someone having a pocket pair could not be 100%, since it would be entirely possible to deal 17 hands without a pocket pair. The probability that there would be at least one pocket pair would be 1 minus the probability that each player did not get a pocket pair. That equation correctly maintains the possibility that you might deal 500 hands without a pocket pair, which would be rare but not impossible. So the probability that at least one player out of 9 players gets a pocket pair is (1 minus 16/17 to the 9th), which equals about 42%, which is a lot lower than 9/17.

                I bother posting this only because I think I am not the only one who sometimes confuses EV with P. The difference can be pretty significant.