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Server Time: 12/2/2008 5:45:19 AM PACIFIC |
 Specific probabilities on deal, Fraser, 23. Sep 2003 18:08 | ||
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| Hey, I'm pretty new at poker and new to this site. In getting myself aquainted with odds I am trying to figure out the chances of facing various pockets before the flop from scratch. It's been slow going, as I have had to teach myself probabilities and combinations, but I have made a lot of progress. Now however, I am stuck. Specifically, I am having problems in determining the number of combinations in all the situations where you (the player) have no aces, but your opponents may or may not have aces in a three-player game. My basic formula is: [card combos in first opponent's hand x card combos in second opponent's hand] / [equivalent player combinations for the two opponents] or [C(50,2)xC(48,2)]/2! I am checking my work by adding up the number of combinations I get for each of these situations, and comparing that number to the total possible combinations: Between them, opponents have: 1) one Ace only 2) two single Aces 3) one pair of Aces 4) three Aces 5) two pair Aces 6) no Aces But to get them to add up to my number for the sum of all combinations, which is [C(50,2)xC(48,2)]/2! = 690,900 I have to double some of them, specifically situations 1), 3) and 4). Why is this? If this makes no sense at all, please let me know and I will try to clarify. Any help from probability whizzes is greatly appreciated!! | ||
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| Re: Specific probabilities on deal, GambleAB, 23. Sep 2003 18:34 | ||
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| If I understand what you are trying to figure out correctly, you have to double those because there could be more than one occurance. For example, with 'one ace', Player A could have the ace OR player B could have the ace. With 'one pair', Player A could have the pair, OR player B could have the pair...with 'three aces', Player A could have AA/Player B A/x, or Player A could have A/x, Player B having AA. All of the others can only happen once....for example with 'two single aces', Player A has A/x and Player B has A/x so every occurrence will be the same, thats the same with the other two times that you wouldn't have to double (because each occurrance is the same). | ||
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| Re: Specific probabilities on deal, Fraser, 24. Sep 2003 12:37 | ||
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| That's exactly the answer I was looking for... I knew it was something obvious like that. I just couldn't wrap my brain around it for some reason. Thanks! | ||
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| Re: Specific probabilities on deal, kennycatkiller, 23. Sep 2003 19:14 | ||
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| I am certainly no expert at statistics, although I did pull a "B" at Univ of Md some years ago. Unless you are an inverterate mathematician, I suggest you get Brunson's book, "SuperSystem" in which Mike Caro worked up all kinds of tables for the various combinations--not just for Hold-Em, but for Stud, Draw, etc. Good Luck | ||
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| Re: Specific probabilities on deal, ADAM THE EXPERT, 23. Sep 2003 20:12 | ||
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| Adam the expert, is no probability expert, he just makes a nice, six-figure income, playing on the internet. Here's my advice to your question: STOP WORRYING ABOUT THAT, STOP THINKING ABOUT THAT. YOU are dangerously close to an "egghead" mentality. I won't mention any names, but the so-called "top" experts, who can figure things out to the forth decimal point, are Missing the picture, and just want to show you how smart they are, so you will buy their books. As another poster said, Yes, after 26 years, the mike caro tables in super system, are very good, and they alone, are worth the price of the book. But, you can get into a lot of trouble, if you focus too much on getting exact odds, and not onthings you need not worry about. Especially if it detracts from the time you need to spend, studying and practicing ALL OF THE OTHER SKILLS, YOU NEED TO BE SUCCESSFUL. They are very simple formulas, for all the poker odds you'll ever need, unless you hire 100 people to play for you, THEN yo;u might need to know whether something is 2-1, or 1.97-1 ! ! ! ! Complicated math, just does not have a place, in poker. Comclicated thought process, DOES. You need to be able to get into the Sklansyish, Malmuthian concept of: What do you think, they think, you think, they think. You need to be creative. You need to know what people think, and learn to use those thoughts against them. You need to learn to be a good manager, keeping excellent records of your opponents' tendencies, the games you play (including day and time) and your results. And, once you have it all together, you need to PLAY, and Play, and play, and play, and play, and work your butt off, putting your hours in, and EARNING your pay. So, for now, here is a very simple formula, that may not be 100% accurate, buy at least 95% accurate: If you are playing a hold em game, and It's passed around to you and you only have to worry about the blinds, a Q 8 is the cuttoff hand. That is the hand, that will beat both of the remaining players, just slightly over 50% of the time. To determine if someone has a bigger pocket pair, than you, another simple formula, 95% accurate 220 divided by remaining players, divided by number of higher pairs possible. So, if you have JJ under the gun, in a ten handed game, it is about 8 to 1, against ANY of the players having a larger pair. BUT, when someone makes a big raise, you cannot just fall back on your math, and think "well, It's eight to one" This is why the "math" guys, CANNOT make it in poker. The are overfocused, on a small element of the game, and NOT on the more importent things. Do you think I need to know, if something is 23 to one, or 22.678 to one??? As long as you know simple, basic odds, you will be ok, IN THIS AREA. If you think that becomming a math expert, will automatically make you a winner, you will be VERY dissapointed | ||
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| Re: Specific probabilities on deal, Schuster, 23. Sep 2003 21:00 | ||
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| So Mr. Expert, would you say that it is fair statement that of the the players who are otherwise identical but one can figure out the math to 2 decimal places, the mathematically inclined one will do better in the long run? I can't imagine you saying disagreeing with this statement. Lee | ||
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| Re: Specific probabilities on deal, Brian462, 23. Sep 2003 23:22 | ||
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| I think he's saying it's worth more to focus on general poker skills instead of spending time figuring odds at each stage of the game. If instead of trying to figure a number to 2 decimal places you figure it to a whole number and then spend the remaining time trying to figure out what might fold your opponent or what might get you that free card or extra bet you might want you will be way better off. For most of us, there is only so much we can think at one time and figuring a complicated equation is just bad business. | ||
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| Re: Specific probabilities on deal, MozMan, 24. Sep 2003 17:55 | ||
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| hmm... well, as this particular forum IS about odds, I think that the question is appropriate. And as UPF as a whole is about being helpful, I suspect that Adam's response may not be appropriate. Maybe it's not inappropriate, but it certainly wasn't very helpful. -Moz "You can see your reflection in the luminescent dash." | ||
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| Re: Specific probabilities on deal, Fraser, 25. Sep 2003 10:29 | ||
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| I think it was good advice. But what do i know, I just got here. | ||
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| Re: Specific probabilities on deal, Antonio K, 1. Oct 2003 13:19 | ||
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| Adam, Do you have a rule like that, getting aproximate odds of making a hand with more than one card to come ? for example 8 outs with 2 cards to come or 2 outs with 3 cards to come ? Thanks a lot | ||
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| Re: Specific probabilities on deal, TrippH, 24. Sep 2003 06:57 | ||
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| on 23. Sep 2003 18:08 Fraser wrote: > Hey, I'm pretty new at poker and new to this site. > > In getting myself aquainted with odds I am trying to figure out the chances of > facing various pockets before the flop from scratch. It's been slow going, as I > have had to teach myself probabilities and combinations, but I have made a lot > of progress. Now however, I am stuck. > > Specifically, I am having problems in determining the number of combinations in > all the situations where you (the player) have no aces, but your opponents may > or may not have aces in a three-player game. > > My basic formula is: > > [card combos in first opponent's hand x card combos in second opponent's hand] > / [equivalent player combinations for the two opponents] > > or > > [C(50,2)xC(48,2)]/2! > > I am checking my work by adding up the number of combinations I get for each of > these situations, and comparing that number to the total possible > combinations: > > Between them, opponents have: > 1) one Ace only > 2) two single Aces > 3) one pair of Aces > 4) three Aces > 5) two pair Aces > 6) no Aces > > But to get them to add up to my number for the sum of all combinations, which > is > > [C(50,2)xC(48,2)]/2! = 690,900 > > I have to double some of them, specifically situations 1), 3) and 4). Why is > this? > > If this makes no sense at all, please let me know and I will try to clarify. > Any help from probability whizzes is greatly appreciated!! > > Since no one else was terribly helpful, some quick probability magic gives: 1) one A only: 26.37% or 1 in 3.793 2) two single A: 1.798% or 1 in 55.63 3) one pair A: 0.899% or 1 in 111.3 4) three A: 0.0799% or 1 in 1252 5) two pair A: 0.000434% or 1 in 230298 6) no A: 70.86% or 1 in 1.411 | ||
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| Re: Specific probabilities on deal, Fraser, 24. Sep 2003 12:36 | ||
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| How did you get those numbers? Thanks | ||
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| Re: Specific probabilities on deal, TrippH, 25. Sep 2003 18:15 | ||
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| 1) one single ace: (4/50)(46/49)(45/48)(44/47)(4) 2) two single aces: (4/50)(46/49)(2)(3/48)(45/47)(2) 3) one pair of aces: (4/50)(3/49)(46/48)(45/47)(2) 4) three aces: (4/50)(3/49)(2/48)(46/47)(4) 5) four aces: (4/50)(3/49)(2/48)(1/47) 6) no aces: (46/50)(45/49)(44/48)(43/47) | ||
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| Re: Specific probabilities on deal, Fraser, 24. Sep 2003 09:49 | ||
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| Wow, thanks for all the responses so fast! I am going to print them out and read them. Thanks for the advice Adam, but believe me I am not trying to memorize all the probabilities and base my game on that. This is just a warm-up for my brain so that I can learn better concentration, and an exercise to get myself really familiar with the basic ballpark probabilities. I am in no way going to rely on this information. I am going to figure it out, then put it on the back burner and focus on strategy and psychology. And playing experience! I'll read the math info and get back if it doesn't answer my question fully. Thanks again! | ||
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