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Server Time: 12/1/2008 11:31:14 PM PACIFIC |
 odds theory, nickdel, 21. Oct 2003 13:34 | ||
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| Would anyone be able to explain odds theory, or perhaps direct me somewhere online where I might find it illustrated? Situations come up in my playing that I often want to calculate percentages/odds either at that moment or just after, but I am unsure how to do it. I'd really appreciate any help offered, thanks! Nick | ||
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| Re: odds theory, modestmice, 21. Oct 2003 13:42 | ||
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| on 21. Oct 2003 13:34 nickdel wrote: > Would anyone be able to explain odds theory, or perhaps direct me somewhere > online where I might find it illustrated? Situations come up in my playing that > I often want to calculate percentages/odds either at that moment or just after, > but I am unsure how to do it. I'd really appreciate any help offered, thanks! > > Nick there is an odds table at the bottom of the browser window under "poker info, odds and dictionary" | ||
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| Re: odds theory, pt_Gatsby, 21. Oct 2003 15:31 | ||
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| I'll give you the easiest way to calculate probabilities - but this is a mathematical model, its not meant to be used at the table (I'll give you a little of that after). Basic probability goes: ---1--- To figure out a single chance for something to occur, take the number of desired outcomes out of all available outcomes. For example: You are dealt AA. What are your chances that the flop card is going to be an A? There are 2 desirable outcomes and 50 possible outcomes - this gives you 2/50 chances, or a 4% chance. Simple enough. ---2--- To figure out multiple chances for something to occur, add all of the single chances together. For example: You are dealt AA. What are your chances that the flop (3 cards) is going to be an A? The chances are: 2/50, 2/49 and 2/48 - or 12.25%, give or take a bit. ---3--- To figure out dependent chances for something to occur, multiple the two chances together, assuming the first one was succesful. For example: You are dealt AA. What are your chances that the turn and the river will be A's? The chances are 2/47 and 1/46 - or 0.1% (roughly). ---4--- So, to figure out multiple conditions, for example, an open straight with a flush, you would go something like this: 4 cards open flush and straight draw after the flop, giving you: 9 flush cards and 8 draw cards, with 47 cards available. You have: 9/47 + 9/46 chances to draw your flush 8/47 and 8/46 chances to draw your straight Together, you probably have less chances (some cards will be double counted ie: the 5 straight may also be a flush, so...), but this will give you roughly the number of 'outs' you have. ---*--- To calculate odds on the fly, you should know the following bits of information: * How good your odds are for 2 outs, 4 outs, 8 outs and 9 outs with 1 and 2 cards to come... and 5 cards for the 2 outs. * How good your odds are for a pair hitting the board with no pair showing (with 5, 2 and 1 card to come) With those combinations, you can calculate your odds of making the majority (90%+) of the hands that will pop up. Have a pair? Think you need a set? 2 outs. Have a flush draw? 9 outs. Have fliped a set of three, but think you may need the full house? Need a pair to come on the river and flop. Open straight? 8 outs. Inside straight, 4 outs. I hope that helps... the odds are at the bottom, but it's actually better if you calculate them with pen and paper (and calcualtor). It seems to make more of an impact. One thing that helped me was 'rounding' inclusive events (need two cards) - basicly you roughly estimate the 1/x chance (ie: 10/47 becomes 1/5), then squaring it (1/25)... that gives you the chance of a runner flush, for instance. Just remember, odds are one thing, the right play is different. There is a large difference between the three 'runner flushes' - 3 suited on the flop is not the same as 2 suited in the hand with one matching. Odds go both ways! Ask if you have any questions... and I hope it was all right. Did it from memory and through my nice advil haze... | ||
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| Re: odds theory, Marshall Farrier, 24. Oct 2003 00:34 | ||
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| Excellent description on the odds, but a further question: From a purely mathematical standpoint, when is it "correct" to call with a (presumably winning) flush draw on the flop? For simplicity's sake, let's say it's down to two players (or that the flush draw is last to bet) so as not to have to guess what the other players in will do, and just leave the odds at 1/5 for each draw. If the pot is, say $60 and I need the draw to win, then the break-even is a $15 bet for each card (I lose my $15 four times and win $60 once). On the other hand, I have about a 40% chance of making my flush by the time the river comes--i.e., on two cards. So, if Joe bets $20 to weed out my flush, what happens if I call? 80% of the time I don't get my flush on the turn. Now the pot is $80; Joe bets $25; and I call again. Playing this through 25 times, these are the results: I made $60 five times (on the turn). I made $105 four times (on the river). So my total winnings were $300 + $420 = $720. The other 16 times I lost all the way and put in $45 each time. So my losses were exactly $45 x 16 = $720. Hence, I broke even over all 25 hands. My conclusion: All else being equal, you should really fold a flush draw (on the flop) only if it costs more than 1/3 of the pot and should call it if it costs less than 1/3 of the pot. | ||
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| Re: odds theory, pt_Gatsby, 24. Oct 2003 13:29 | ||
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| Excellent description on the odds, but a further question: From a purely mathematical standpoint, when is it "correct" to call with a (presumably winning) flush draw on the flop? I think this depends a lot on the cards. For example, having 23s is not the same as AKs, for example. Also, if a pair hit the flop, your odds are greatly reduced (ok... somewhat reduced) since it reduces your 'outs' by 2 (ie: they overcome your flush with a FH), not to mention, if they have trips, is reduced by a great deal more (often 7-9 outs). My conclusion: All else being equal, you should really fold a flush draw (on the flop) only if it costs more than 1/3 of the pot and should call it if it costs less than 1/3 of the pot. I'm not very good at pot odds, so take this with a grain of salt. I use a different sort of measurement stick... I look at it from the point of view of tournament odds, if such a thing exists. I also look at new money coming in: and on a flush draw, you want to have at least 2 other people in the pot calling any draw. That way for every bet you put in, you'll win their 2 40% of the time (ie: -2 *0.6 + +4 * 0.4 = +0.40). When you add the pot value, its almost always worth calling. The pot would have to be really really small and their bet really large compared to it... and even then, if there are more than just you and him, its probably worth calling. Do you pull out on the river? That's a tough one. You should always use your most current odds, and 9/46 (say, 20%) isn't great. That's when pot odds matters most - if fixed, you should be pretty close to getting pot odds (1:4), though he could easily drive you out in NL. Overall though, I don't think it needs to be calculated that percisely. Games are too variable to lock in such a strategy - your overcards, level of flush, their potential hands, their style, the game type, your position, your chip count... they all play a huge role. I think that understanding that you have a 40% chance, roughly, to win with that flush (A high) is more important. Then you can look at the table, evaluate his strength and your position and make the best bet available to you. Normally this will involve putting as much money with as few players as possible into the pot, but not always. It is positive EV, but you are still the underdog. And in a tournaments (especially NL), where I tend to reside, underdog is more important than pot odds. You can't afford to go all in on a draw, most of the time... you want to be the 60%, not the 40%. Unless, of course, you have the chip lead... And if he didn't actually have a pair, and you have overcards... then vs random cards, you actually have a 70%+ chance of winning... and if he did have the pair, you have about a 43% chance of winning with an overcard (pair higher than yours)... and if he has a higher heart... or a pair in hand... or if the board pairs against a random hand... *erf* My head is starting to hurt again. | ||
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| Re: odds theory, Aisthesis, 24. Oct 2003 15:14 | ||
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| Agreed, the situation always has so many factors that there's never a "hard and fast" rule. Clearly, if you have at least 3 callers in front of you, you've always got a good bet. Also if your made flush is weak, a good reason to fold. An anecdote from a 9 player sitngo I played yesterday: With three players left I was in the pleasant position of having around 10k to the other two players with less than 2k each. But both players were not going down easy. One was playing very cautiously, I guess shooting for second, the other was making some clever all-ins in the hope of getting back in the running for first. With Ad-6d in my hand and two diamonds on the board, I decided it was time to call one of the all-ins and try to put him out of his misery. Sure enough he had a good pair and beat me that hand since I made neither the flush nor the ace. But I still had enough going to end it with him a few hands later. (if the ratio hadn't been so enormous I really would have hesitated to call the all-in and lose my big advantage--say, if I had had 6k to his 3k, he would suddenly be in the driver's seat). Then it was really rough taking out the tight player, who did manage at one point to pull himself up to almost 4k at one point. But the amusing thing was that in the final hand, he went all-in on a good flush draw and missed it against my good pair. :) | ||
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| Re: odds theory, Candide, 24. Oct 2003 09:54 | ||
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| Good post, but one error. In calculation 4 I think it was, you have the straight flush draw odds (which is 15, 9 from the flush, plus 6 for the straight, 2 cards are counted in the flush for the straight). The way you say do it, is to say 15/47+ 15/46, which is 65%. This is wrong... How you calculate is to reverse the process. See, what you really want figure out is the % of NOT hitting a card on either one. So you want calculate this (47-15)/47 * (46-15)/46. this is 45.9%. Thus the odds of you hitting the card on one of those two cards yet to come, is 100-45.9, or 54.1%. Hard to grasp, but think about it and you will realize it is correct (you can also compare it to Sklansky and Malmuth's book on holdem to see the correct odds for that). A general rule of thumb someone once posted here, that works up to 13 or so outs, is to multiply your % of outs by 4 if you have 2 cards to go, and 2 if you have 1 card to go. So if you had a open ended straigh, which is 8 outs, it would be 8*4 for 32% (real probability is 31.5) or 8*2 for 16% (really 17.4). But it gives you a quick estimate. BTW, that figures the probability, or percent likely, of something occuring. What you often want to do is then figure the ODDS of that happening. This is important to factor against pot odds. To take a probability into an odd, you simply do (100-%)/%...so back to the straight draw above. Using the quick math method, we figure we have a 32% chance. The odds for this would be 100-32=68. 68/32= 2.1. So it is 2.1 to 1. To call a bet on the flop, you would need pot odds of greater than 2 to 1. This is why it is almost always correct to call with open ended straight draws, 4 to a flush draws, and higher (you can sometimes count hitting your over card as well if you are sure your opponent has a smaller pair). BTW, pot odds are simply how much you have to bet, vs how much you can gain. strict odds would be, say you have to call a 10$ bet to win 100, well that is 10 to 1. (100/10). But in reality you need to consider future bets as well you may have to call. This is why it is often better to figure your odds on hitting it on the turn, as it is a cheaper bet on the flop, where you may have to call a 20$ bet on the turn if you miss, to draw to the river. Hope that makes sense. | ||
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| Re: odds theory, Aisthesis, 24. Oct 2003 11:15 | ||
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| Nice post! If you count your outs and then take a few shortcuts on you probabilities, it's usually pretty easy to get pretty easy to figure out in your head the odds of your getting the draw. Important however, is to realize that if you have, say, a 30% chance of getting a winning draw, the break-even point is not 30% of CURRENT pot, but of pot PLUS your bet: For example, with a $100 pot and a $30 call bet, you should definitely call if you have a 30% chance of a winning hand. The break-even is just under $50. Your second point, on additional winnings is also extremely important. Just guessing from practical play, I'm thinking 50% of pot is just enough to make it unprofitable to go for a winning flush or straight draw. Any less than that one should probably call, and 50% or more of pot, you should probably fold. By the same token, 50% of pot should be just enough to keep rational players from attempting the flush or straight against your good pair--unless there are multiple players likely to call, in which case it might be a good idea to bet the full pot on your good pair. | ||
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| Re: odds theory, pt_Gatsby, 24. Oct 2003 12:37 | ||
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| I didn't actually do the calculations for #4... but you are correct. (which is 15, 9 from the flush, plus 6 for the straight, 2 cards are counted in the flush for the straight). Quite true - which is why I listed them seperately. To combine them would cause you to have 15 outs, not 17... It was only my intent to How you calculate is to reverse the process. I always tend to think of it as a binomical tree rather than estimates... and then its just the other side either way (ie: I know that for 9 possible outcomes, 3 are positive on the first iteration (1/3) and then 1/3 of the two combinations (2/9) are positive... so you end up with 5/9 positives, roughly). Funny that I can do that at the table... but not pot odds! A general rule of thumb someone once posted here, that works up to 13 or so outs, is to multiply your % of outs by 4 if you have 2 cards to go, and 2 if you have 1 card to go. So if you had a open ended straigh, which is 8 outs, it would be 8*4 for 32% (real probability is 31.5) or 8*2 for 16% (really 17.4). But it gives you a quick estimate. While I agree that this works fairly well, I think people still should understand the theory. Sometimes you need to know how valuable top pair with a runner flush is, or something similar. To do really well - winning by the numbers, anyway - you need a more complete understanding of the theory... IMO, anyway! You can do quite well without it... I think its just different personalities, generally. This is important to factor against pot odds. Now that's good advice! I am awful at pot odds. Just can't seem to get them to work for me... Probably why I do so well at tournaments and so bad at the tables, actually. | ||
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| Re: odds theory, Blue Sky, 22. Oct 2003 12:17 | ||
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| nickdel, I am going to cut and paste some odds info that I had posted on another poker web site, so if there are names or references that don't make sense it is because I just didn't want to re-type everything here. If you have any questions after you read this reply back and I will see if I can help out. I'll try to get you started in the right direction the best I can and Angel and a few others here can hopefully help out if I miss any major points. Odds 101: To calculate odds you subtract the known cards from the total number of cards in the deck to equal the total cards left to come. You then subtract the cards that help you(your outs) from the total cards left to come to equal the number of cards that don't help you. You then divide the cards that don't help you by the cards that help you to equal your odds of hitting your hand. IE - Flush draw (after the flop with 2 to your suit on the flop and 2 of the suit in your hand) 52 total cards - 5 know cards = 47 unknown cards. 47 unknown cards - 9 remaining of your suit that give you the flush = 38 cards that don't help you. 38:9 (38/9) = 4.22:1 Odds of hitting your flush on the turn card Further for the same hand - you hold KQ of hearts and the flop comes Ah 8d 3h. You now know 5 cards Kh Qh Ah 8d 3h. There are 47 remaining unknown cards. There are 9 hearts that help you still available to come on the turn or river. You subtract the 9 hearts from the 47 remaining cards and get 38 cards that don't help you and 9 that do giving you 38:9 or 4.22:1 odds to make your hand. The pot must be giving you atleast 4.22:1 or better odds to call to make your flush. To calculate pot odds you divide the total amount of the pot by the amount of money you must call. IE. There is $32 in the pot and you have to call $6 = pot odds of 32:6 or 5.33:1 Any money already in the pot when it is your turn to act is considered the pot's money, forget about how much you may have already put in b/c its the pot's money now. So to answer your question no you do not count the cards that are dealt to other players or the burn cards when calculating odds because they are unknown cards. If you think you understood this take a shot at figuring out the odds for an open ended straight. You hold 89o in the blind and everyone calls the blind and you check. The flop comes 7 6 2 rainbow. What are your odds of hitting the straight on the turn and river? I have learned a ton on these boards and hope you do also, keep asking questions | ||
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| Re: odds theory, nickdel, 23. Oct 2003 14:15 | ||
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| Thanks! I really appreciate the help. Let me copy this stuff down and learn it and if I have any more questions, I'll post them. Thanks again!! Nick | ||
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