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Server Time: 11/22/2008 12:33:12 AM PACIFIC |
 again with odds.., cold_cash, 18. Jul 2003 12:21 | ||
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| Hey all. I think I've finally got a handle on a lot of this stuff, but there's one thing I'm still not sure about. Let's say for example I'm holding a drawing hand in which I'm a 6:1 underdog to make. I know that for me to correctly call a bet at this point, the pot should be laying me at least 6:1. So far so good. My new problem, however, is this: what if there is a bet and a raise in front of me? Now it will cost me 2 bets (at least) to keep going. Let's assume for now that there will not be another raise behind me. How do I know figure the right play according to the pot odds? (Calling one bet here would be okay, but how do I re-calculate to figure if calling two bets would be okay?) Thanks again. (I'll get this stuff eventually, I swear.) | ||
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| Re: again with odds.., ezcheese, 18. Jul 2003 12:58 | ||
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| I'm by no means an expert on odds... but this is what I know... What you are talking about is called "implied odds"... you have to take into account if you are going to be raised behind you as well as if you think that raise will be re-raised in front of you.... if you are getting 10:1 from a pot ($100 pot $10 bet) you can call just about anything... if you get raised in front of you you are gonna be getting 5:1 ($100 pot $20 bet)... it doesn't matter whether or not you had good odds to begin with... the one thing I hate about playing odds (which I try to do) is that you may make the $10 bet based on odds but then get raised and re-raised and now you don't have the odds so you just wasted the bet. | ||
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| Re: again with odds.., Roy Cooke, 18. Jul 2003 13:50 | ||
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| You ask a very good question.....There is no exact method, as you can never know for certain what your opponents are going to do... ....That said, you just need to be conceptually aware that it MAY get raised behind you (and rerasied ect) and you need to adjust your odds based on the propensity of that event to happen......Your best answer will be a logical educated guess based on past experience knowledge of your opponents and flop texture.......The greater the chance of it being raised behind you the greater odds the pot needs to be laying you. Roy Cooke on 18. Jul 2003 12:21 cold_cash wrote: > Hey all. I think I've finally got a handle on a lot of this stuff, but there's > one thing I'm still not sure about. > > Let's say for example I'm holding a drawing hand in which I'm a 6:1 underdog to > make. I know that for me to correctly call a bet at this point, the pot should > be laying me at least 6:1. So far so good. My new problem, however, is this: > what if there is a bet and a raise in front of me? Now it will cost me 2 bets > (at least) to keep going. Let's assume for now that there will not be another > raise behind me. How do I know figure the right play according to the pot odds? > (Calling one bet here would be okay, but how do I re-calculate to figure if > calling two bets would be okay?) > > Thanks again. (I'll get this stuff eventually, I swear.) | ||
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| Re: again with odds.., cold_cash, 18. Jul 2003 15:21 | ||
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| Thanks for all the replies to my question -- I think I might be sneaking up on this problem, slowly but surely. One more thing I thought about, but don't know if it's right or not -- let's say I'm drawing to a hand that I'm a 5:1 underdog to make, and there are 5 bets in the pot (for simplicity's sake).. The first player to act bets, and the next player raises. Now there are 8 bets in the pot, but it will cost me 2 bets (at least) to keep playing. In my mind, this means the pot as of right now (excluding any re-raises behind me) is laying me 8:2 (or 4:1), so my pot odds have dropped from 5:1 to 4:1 because of the raise. Is this the right way to be thinking about this, or is there a serious flaw in my reasoning? If I could just get a "You're on the right track here..." or a "You're completely wrong about this..", I would be very appreciative. Again, thanks a ton. | ||
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| Re: again with odds.., ezcheese, 18. Jul 2003 19:19 | ||
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| that sounds right to me. | ||
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| Re: again with odds.., Schuster, 18. Jul 2003 22:50 | ||
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| > One more thing I thought about, but don't know if it's right or not -- let's say I'm > drawing to a hand that I'm a 5:1 underdog to make, and there are 5 bets in the pot (for > simplicity's sake).. The first player to act bets, and the next player raises. Now there > are 8 bets in the pot, but it will cost me 2 bets (at least) to keep playing. In my mind, > this means the pot as of right now (excluding any re-raises behind me) is laying me 8:2 > (or 4:1), so my pot odds have dropped from 5:1 to 4:1 because of the raise. You nailed it right on the head. The other players raise has caused you improper odds and you can no longer call the bet. However, if you anticipate the original bettor calling the raise AND at least one other player, you will have the correct odds to draw. Be careful in these situations though, as it might get reraised, and then you have to keep calling as you'll have good odds the next time it comes back to you but you've had bad odds as applied to this betting round as a whole. Generally you should fold unless you have a really good read. Lee | ||
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| Re: again with odds.., cold_cash, 18. Jul 2003 23:33 | ||
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| Cool. Thanks a lot. I'm reading a few books right now and I was puzzled when the authors would say "by making such and such a bet, you will now make it improper for your opponents to call..." I kept thinking to myself, "If I'm putting MORE money in the pot by raising, I'm giving my opponents BETTER pot odds to chase a hand...", but I knew that just couldn't be right, (after all, this guy wrote a book so he must know at least a little about what he's talking about, right?). t I was just having trouble wrapping my mind around the concept. I think I"m getting it now though, thanks to you guys. Thanks again. | ||
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