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Server Time: 8/20/2008 3:45:38 AM PACIFIC |
 Easy Formula for Flushes?, Bart Mann, 25. Sep 2003 11:10 | ||
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| Good Afternoon, All: A few weeks ago ADAM THE EXPERT forwarded me an easy formula to use when trying to figure out the odds of someone at the table having a bigger pocket pair than you. That formula was as follows: 220 divided by the # of larger pairs possible divided by the # of players in the pot with you. So as an example if you're holding pocket 10s, there is a 1 in 27.5 chance that someone has a bigger pair in the hole (220 / 4 / 2). My question is this: does anyone have a formula like this for flushes with three on the board (not four)? For example, let's say that I am involved in a pot with 9s10s. All of the cards are out, there are three spades on the board and I've got people betting into me. What are the odds that one of them also has two spades and a bigger flush than mine? With a calculator and a pencil I can figure this one out--but does anyone have an easier formula I can use at the table when I need to make a decision in 5 seconds or less? I'm probably on a fishing expedition here, but I guess you never know. ADAM, maybe you're out there somewhere? | ||
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| Re: Easy Formula for Flushes?, stdioh, 25. Sep 2003 14:48 | ||
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| That formula is a good guideline, but it doesn't take into account conditional probabilities. For a start, since you're holding 2 of the tens it means that there are 2 less cards for the opponents to be dealt from. Not only that, but they affect eachother. If player 1 doesn't have a higher pair than you then he may have 1 or 2 or no overcards to you. If he has 1 overcard to you then that changes the odds of other players having a higher pair, etc. To be precise in the chances of somebody else having a higher pair than you would actually require quite a bit of math. But as an approximation, this formula is great for all practical purposes. | ||
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| Re: Easy Formula for Flushes?, Bart Mann, 25. Sep 2003 17:57 | ||
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| But given that you can only include "knowns" in your analysis, and the only thing you "know" for sure is the two cards you have in your hand, this formula is about as close as you're going to get--right? If you have no other information, then by definition you can't even begin to generate a "conditional probability." It's circular logic, I think. The other players could have overcards, or they could have a pair of deuces, or 2-7 of spades, or 4-9 offsuit. I don't see how you can do any better than this unless you get a look at someone else's hand. | ||
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| Re: Easy Formula for Flushes?, kimmi690, 25. Sep 2003 21:59 | ||
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| Ok, I failed math about a gizzilion times in high school and college...is there a formula I could use without having to take a calculator to the table? I am one of those with a serious mental math block, and all the talk about pot odds and formulas makes me feel like I am at a SERIOUS disadvantage. I play by my gut, I guess its why I am so affected by my emotions, but I really want to get to the heart of the game...is it really all about math? Hope I don't sound like a moron here. Thanks Kimmi :) | ||
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| Re: Easy Formula for Flushes?, Schuster, 25. Sep 2003 22:27 | ||
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| Regarding the flush question, I don't think there will be any simple formula. Just think that in general it's "unlikely". Even then, if someone 4 bets you on the river and the nuts is the ace high flush, will it matter if the odds of them having it are only 1%? Surely you will be getting at least 99 to 1 to call. As far as the math goes, there are a lot of "feel" players who have trouble calculating pot odds, and even make mistakes on them. If one of these "feel" players calls with 2 pair against a known flush getting only 9 to 1 (he is 11 to 1 to fill), this mistake only costs them a small fraction of a bet, and will only become apparent at the highest levels of play. For most all poker situations at the table, remembering basic odds of common hands (flush draw, straight draw, overcards, gutshot / two pair to fill) and the pot odds you need to call with those hands is more than fine. Also, some special situations need to be remembered, such as a straight flush draw open end draw (15 outs total - 9 flush and 6 more for the straight) is better than 50/50 to make a hand by the river and should be played as such. A flush draw with 2 live overcards (14 outs - 9 flush cards and 6 pair cards) is a favorite by the river over a made pair in most situations. Generally, 12 or more outs should be played very aggresively on the flop and get as much in the pot as possible. Just make sure all your outs are clean before going overboard. There are tons of odds tables floating around, find one you feel comfortable with and print it out if you need to. The "ODDS" link at the bottom of this forum is a good place to start. Lastly, there's no dumb questions here. If don't know something, don't be shy to ask. Lee | ||
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| Re: Easy Formula for Flushes?, Bart Mann, 26. Sep 2003 09:17 | ||
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| Just for the fun of it, let’s break this flush question down. It’s been a few years since MBA school, so please feel free to jump in if you see any mistakes. If I have two spades and there are three on the board, there are 8 spades left in the deck. Since I know the value of 7 cards in the deck (the 5 on the board plus the 2 in my hand), there are 45 cards unknown to me. So in a heads-up pot, the odds of the other guy also having two spades are as follows: (8/45) * (7/44) = 2.83%, or about 1 in 35 Now if there is more than one player in the pot with me, then I have to add those odds for each player. So if there are two other players, the odds are 5.66%, or 1 in 17.67, that one of them also has a flush. If there are three other players in the pot, the odds are 8.49%, or 1 in 11.78 that one of them also has a flush. Once I do the math on this, I see that Lee is correct. Adding the contingency of “a bigger flush than mine,” assuming I have 9s10s, would make it over twice as unlikely for someone at the table to have a bigger flush. So Lee, I think your estimate of 1 in 100 is pretty darn close. How does this look to everybody? | ||
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| Re: Easy Formula for Flushes?, kimmi690, 26. Sep 2003 10:28 | ||
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| i just got dizzy.... | ||
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| Re: Easy Formula for Flushes?, -Sammy, 26. Sep 2003 10:37 | ||
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| ....and my eyes just crossed and my brain hurts....... | ||
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| Re: Easy Formula for Flushes?, kimmi690, 26. Sep 2003 10:40 | ||
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| LOL!!!!!! | ||
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| Re: Easy Formula for Flushes?, Schuster, 26. Sep 2003 10:42 | ||
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| The point of the 1% comment was that most players won't cap the river without the nuts. Does it matter if the odds are 99 to 1 if he's telling you that he has it? In the same situation, would you expect the player to flip over the ace high flush only 1 in 100 times? Probably not. Anyway, just for fun, I'll work it out for the heads up scenario. Beyond that, it's a decent amount of math. The odds of someone holding a higher spade than yours is simple, the amount of higher spades divided by the number of unknown cards. So for example, for your 10s 9s, there are 4 higher spades, so the odds that one of them is in your opponents hand is 4/45. The odds of that spade being paired with any of the remaining spades in the deck is 7/44. (4/45)*(7/44) = 1.4%. The best odds they could have, if you are holding the 3s 2s still only gives them 8 higher spades. (8/45)*(7/44) = 2.8%. The problem with multiple opponents is that you can't simply add the probabilities (incorrect in most cases anyway) or multiply the chances of each one not holding, because they're dependent on each other. If one guy has the 4s 5s then the odds of someone having the As Xs go down because now the 4, 5, 9, and 10 are all unavailable. So, it's a bit more complicated. I remember seeing some numbers for this a while ago, and if I recall, if you have the king high flush, the odds of someone having the ace high flush at a full table are somewhere between 2% and 4%. Maybe you could try searching the archives and see what pops up. Good luck Bart! Lee | ||
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| Re: Easy Formula for Flushes?, Grateful Rooster, 26. Sep 2003 11:37 | ||
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| These questions might be fun for math geeks (and I'm one so no offense intended), but the "what are the odds of the other guy having X?" questions just aren't very useful in play, I don't think. You learn far more about the other guy's hand by studying the betting pattern in association with the board (was he betting like a draw?), past betting pattern (will he bet a draw at all?) and any other tells (detect excitement at drawing the nuts?) I would say that if a player bets you large at a flush draw, the pattern matches a draw and s/he's bet a draw that way before, you've got a 50/50 (great bluff vs. nuts or near) at best. Better if a loose player, worse if tight. If you bet those as 100 to 1 shots, you'll lose a lot of money. So kimmi, I'd say don't worry about the math on this one. GR | ||
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| Re: Easy Formula for Flushes?, Bart Mann, 26. Sep 2003 13:00 | ||
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| I don't think anyone here is endorsing a strictly mathematical approach as the only way to play poker--but I wouldn't say that a strictly intuitive approach is best either. If someone sat down and made a list, he/she would be lucky to name three professional poker players who use no form of math in their games. If you're going to risk your own money at the table, then why not sit down with as many tools as you can? It certainly can't hurt . . . can it? | ||
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| Re: Easy Formula for Flushes?, Grateful Rooster, 26. Sep 2003 14:26 | ||
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| I'm not saying that math is not important in poker -- odds calculations are critical to long term success, especially in limit games. I only meant to say that calculation of odds for figuring an opponent's hand are not very useful. Knowing that a guy has > 100 to 1 chance to hold a bigger flush than you is misleading because you are not playing against a random hand. All the non-flush, non-bluff hands have likely been folded. Put another way, do you think the same guy would stay in the pot all 99 times out of 100 when he does not hold a flush? Yes, s/he's 100 to 1 to catch a flush with you, so if you don't know anything else, I suppose you could lower the odds slightly below 50% that s/he has you beat. But all the other information (ie. card and player reads) is much more valuable. GR | ||
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| Re: Easy Formula for Flushes?, stdioh, 26. Sep 2003 15:08 | ||
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| Nope. You know the two cards you have. Now you can figure out the chances of any given opponent having a higher pair than you. But you have to multiply the chance of him not having a hand that beats you by the chance you get for a second opponent having one that does beat you and then alter the opponents chances based on the first guy having one. This is because the opponents' hands are not mutually exclusive. For instance if you're holding TT what is the chance that 5 players have a unique higher pair than you do? Obviously it is 0 since there are only 4 higher unique pairs. But if you work out the chance of each one having it and multiply that together you'll get a positive number. Yes, you have very limited information, but in any problem where you have some information it is possible to work out the precise odds *given* that information. You never need to say, "I don't have enough information so I need to round off" .. .but for all practical purposes, close enough is close enough. | ||
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| Re: Easy Formula for Flushes?, Easy E, 29. Sep 2003 13:42 | ||
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| "For instance if you're holding TT what is the chance that 5 players have a unique higher pair than you do? Obviously it is 0 since there are only 4 higher unique pairs." stdioh, "unique"- Is not AhAs and AcAd two unique hands, that are both possible, if improbable, in this situation? | ||
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| Am I missing something?, Easy E, 29. Sep 2003 13:40 | ||
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| "A few weeks ago ADAM THE EXPERT forwarded me an easy formula to use when trying to figure out the odds of someone at the table having a bigger pocket pair than you. That formula was as follows: 220 divided by the # of larger pairs possible divided by the # of players in the pot with you. So as an example if you're holding pocket 10s, there is a 1 in 27.5 chance that someone has a bigger pair in the hole (220 / 4 / 2). " Shouldn't that be at least halved? Are there 8 possible pairs at least above your Tens? And if you go possible combinations, there are 6 pair combos for each rank for one player to potentially hold (not that AcAh is really different from AcAs for this calculation)... ...? | ||
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| Re: Am I missing something?, jaustin, 29. Sep 2003 13:56 | ||
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| Yes there are 8 possible pairs, but the formula still is fine. The odds of getting dealt a specific pp (regardless of suit) is .44% or 1/225. However as you know your own two cards the odds of them holding a specific pp higher than yours is .49% or 1/204, so the 220 is a fine number to start with - not sure why its 220 vs. 200 (I assume it has to do with people's cards canceling each others odds - I didn't do a thorough evaluation). | ||
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