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Server Time: 11/20/2009 6:41:03 PM PACIFIC |
 Statistical poker question, Banning, 8. May 2003 17:05 | ||
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| There was alot of brain teasers flying around and thinking about them made me think about poker and odds. As such I have come up with a minor flaw in the current statistics that alot of people use for drawing to a 4 flush. There is a decent chance that my math may be flawed and I am open for criticism. Here goes. In the Monty Hall dilemma the algorithm to maximum sucess is to first pick a door then when a door is exposed as goated you then switch...this gives you a 2/3 chance of sucess. Now Let's apply this to poker. On the flop with a 4 flush. Under the assumption that a flush will bring a win. The standard doctrine is to say that now I have a 1.86 to 1 chance of sucess asuming that I will stay until the river. If a turn comes a blank now you have a 4.11 to 1 chance of sucess. I agree with the 4.11 part. I am going to contest the 1.86 to 1 part. I think that it is flawed because it is not fair to always assume you will stay all the way to the river. The pot odds may force you to fold on the turn against a raise or reraise etc. Thus the odds of 1.86 to 1 are flawed because it uses the assumption that we will stay all the way to the river. Thus against a very very agressive field on the turn your implied odds of making your flush on the flop should not be 1.86 to 1 because you know that more frequently will there be a more dramatic increase on the turn which may justify a fold. In the donkey door problem you can create an algorithm based on foreseable events that always come to pass. In the poker situation one cannot make the same algorithm. In math this is referred to a "adaptive" and a "non-adaptive" algorithm solution. As such i hope this brings some discussion. | ||
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| Re: Statistical poker question, Schuster, 8. May 2003 17:41 | ||
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| > I am going to contest the 1.86 > to 1 part. I think that it is flawed because it is not fair to always assume > you will stay all the way to the river. The pot odds may force you to fold on > the turn against a raise or reraise etc. Thus the odds of 1.86 to 1 are flawed > because it uses the assumption that we will stay all the way to the river. The odds of your flopped 4 flush completing, assuming you stay to the river are always going to be 1.86 to 1. You are working in the assumption that you are not going to always stay for the river. If that is the case, you would have to adjust your odds. Say, for example, you were going to see the river 80% of the time. 1.86 to 1 is 35%. So the odds of you both staying to the river and making your flush are 80% * 35% = 28%, or 2.57 to 1. It is good to identify situations where you probably will, might not, or definately will not have odds to call on the turn and adjust your pot odds accordingly. For instance, you might be getting 11 to 1 on the flop to chase a gutshot against 1 other player. However on the turn, your pot odds will be down to 7 to 1, and you clearly can't pay to see the river and if your gutshot hits. Sklansky talks about this in a lot more detail in TTOP. It's definately a good read if you haven't checked it out yet. | ||
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| Re: Statistical poker question, Schuster, 8. May 2003 17:47 | ||
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| Hmm I think I goofed on that calculation, since if you make it on the turn, you are clearly going to stay to the river. I finished my last final today and my brain is fried, I think! Anyway... You get the idea that you should adjust your odds if you think the turn betting could be high. I'll think about it more and figure out the adjusted odds for my example. Lee | ||
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| Re: Statistical poker question, shorn, 9. May 2003 04:58 | ||
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| I tend to agree with Banning here that using 1.86 to 1 with your 4 flush on the flop as the odds you need to stay in the hand is not correct. I NEVER assume that if I don't hit on the turn I will stay to the river. Since the turn and river are independent trials, there is no way that I will call a bet on the flop with less than 3 to 1 (I give myself 1 bet for implied odds thinking that i will win at least that if i do hit). If I miss on the turn, then I need to have 4 to 1 again to call on the river. Also, it depends on the type of game you are in. I am more apt to call with slightly less than correct odds if the game is passive because it is likely that I won't have to pay another bet to see the turn. Additionally, If I have backdoor outs, like I hold AKs so i have (potentially) overcard outs as well, then i might call. In my experience, using the 1.86 to 1 to "talk yourself into" a loose call is a leaky play with -EV long run. | ||
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| Re: Statistical poker question, Snorbolus, 9. May 2003 05:14 | ||
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| I think that the problem here lies not so much in accepting the 1:1.86 odds of making your flush by the river. It is in forgetting to imply the cost of getting there. You will almost always have to pay at least one more big bet on the turn - regardless of wether you hit or miss. If you are planning to stay to the river every time, then you should consider how much it will cost to get past the turn. Rounding to whole numbers, out of each 3 tirals the expectation is that 2 times you will loose 3 small bets each (assuming you don't call on the river when you miss), and 1 time you will win the pot plus whatever extra bets you can confidently imply. If you think that your opponents are likely to raise the turn then you need to consider this too. Snorbolus on 9. May 2003 04:58 shorn wrote: > I tend to agree with Banning here that using 1.86 to 1 with your 4 flush on the flop as > the odds you need to stay in the hand is not correct............. > there is no way that I will call a bet on the flop with less than 3 to 1...... | ||
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| Re: Statistical poker question, Schuster, 9. May 2003 11:21 | ||
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| > Since the turn and river are independent trials, > there is no way that I will call a bet on the flop with less than 3 to 1 (I give myself 1 > bet for implied odds thinking that i will win at least that if i do hit). If I miss on > the turn, then I need to have 4 to 1 again to call on the river. I usually do this as well, but I use the 1.86 to 1 figure to pump up the pot when I do have a draw and there's a lot of people around. If I'm drawing to the nut flush, the board isn't scary, and an early position bettor gets 3 callers, I'm surely going to raise. They will likely all call, and I'm getting 1.86 to 1 on my 4 to 1 pot investment. I think that's the general use of the 1.86 to 1 figure. Lee | ||
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| Re: Statistical poker question, shorn, 9. May 2003 11:26 | ||
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| on 9. May 2003 11:21 Schuster wrote: > I usually do this as well, but I use the 1.86 to 1 figure to pump up the pot when I do have a > draw and there's a lot of people around. If I'm drawing to the nut flush, the board isn't > scary, and an early position bettor gets 3 callers, I'm surely going to raise. They will > likely all call, and I'm getting 1.86 to 1 on my 4 to 1 pot investment. I think that's the > general use of the 1.86 to 1 figure. > > Lee Lee- This is a good point. I agree to use this play when in late position in a passive game with a not scary board. However, I would view it more here as a value play and not a straight decision to call or not. If you have a bet and two callers than you are right to raise unless you fear a re-raise somewhere. This also has the benefit of potentially buying you a free card if you need it on the turn. Steve | ||
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| Re: Statistical poker question, Banning, 9. May 2003 16:18 | ||
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| I somewhat disagree with you calling it as a "value play" I've read every single one of Mike Caro's online articles twice and if my memory serves me correctly he says that in late position with a drawing hand it is "sometimes" worth it to raise for deception or in the hopes of stealing the pot right there. Thus; if I understand everything correctly that means that your raise from late position in an unraised pot with a draw to the flush is a semi-bluff in an attempt to win the pot right there. Mike Caro also says that generally one should NOT raise from this position unless trying to set up a free card. The smartest play is one in which you invest zero dollars to get the winning hand. He then did a bunch of math which i kinda understood to prove this. Of course now that I evoked the Mike Caro reference I hope i'm not invoking his power to come down and destroy the likes of me. All that I'm trying to say is that it isn't really a value bet but moreso a semi- bluff. on 9. May 2003 11:26 shorn wrote: > on 9. May 2003 11:21 Schuster wrote: > > > I usually do this as well, but I use the 1.86 to 1 figure to pump up the pot when I do have a > > draw and there's a lot of people around. If I'm drawing to the nut flush, the board isn't > > scary, and an early position bettor gets 3 callers, I'm surely going to raise. They will > > likely all call, and I'm getting 1.86 to 1 on my 4 to 1 pot investment. I think that's the > > general use of the 1.86 to 1 figure. > > > > Lee > > Lee- > > This is a good point. I agree to use this play when in late position in a passive game with a not > scary board. However, I would view it more here as a value play and not a straight decision to call > or not. If you have a bet and two callers than you are right to raise unless you fear a re-raise > somewhere. This also has the benefit of potentially buying you a free card if you need it on the > turn. > > Steve | ||
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| Re: Statistical poker question, Ron, 8. May 2003 19:43 | ||
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| the odds have nothing to do w/ how one plays a hand. You might fold b/c of the raising on the turn, but the guy sitting next to you might stay. With each turn of a card (or flop) the odds change b/c less cards (some you know & some you don't). When you read stat page of anything they are for the long hall. Example: baseball. Ted williams hit .409 but he didn't get 4 hits every 41 (approx) AB. There were times he got 7 straight hits & times he went 0-5. The average (or in poker the odds) is on the long hall, which is the entire game. | ||
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| Re: Statistical poker question, Ron, 8. May 2003 19:46 | ||
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| Sorry 2 hits out of every 5 (approx) at bats. I was typing something else. Sorry for the typo. | ||
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| Re: Statistical poker question, stdioh, 9. May 2003 08:54 | ||
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| One more piece of information is missing in this calculation and that is implied odds. If you're playing against opponents who will pay you off if you make your hand on the turn, then that counteracts a lof of the cases where you will be forced to fold the turn if you do not hit. | ||
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| Re: Statistical poker question, shorn, 9. May 2003 09:06 | ||
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| Hence the need for a more passive game. If you are in an agressive game, you won't be able to draw again on the turn without putting more bets in, so your implied odds are worse than what the pot is laying you. | ||
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