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Server Time: 11/20/2009 4:49:04 PM PACIFIC |
 Calculating Odds, Joel, 10. Dec 2002 17:17 | ||
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| I was trying to calculate the odds on flop when you have a backdoor straight and a backdoor flush draw. I got roughly 10/1 against. Does this sound right? | ||
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| Re: Calculating Odds, 2jelsky, 11. Dec 2002 07:48 | ||
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| it's best to ignore backdoor situations note that I did not say "draw" | ||
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| Re: Calculating Odds, Mark, 11. Dec 2002 08:04 | ||
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| These are rough and fast calculations but backdoor straight 50:1 Backdoor flush 16:1- 20:1 For the flush you are roughly 4:1 to get your flush card on the turn or river. So for two 4:1 chances that must happen sucessively, just multiply and you get 16:1. However your odds are not exactly 4:1, they are a little worse 4.?, so in reality you get a worse than 16:1, more like 18:1. for the straght, you have to hit one of your 8 cards on the turn, which is roughly a 5:1 shot. If you actually hit it, you only have 4 cards on the river which will help you, which is roughly a 10:1 shot. So multiply and you get 50:1. Note. this is not the "proper" way to calculate odds but it gets you VERY close to the actual answer. Mark | ||
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| Re: Calculating Odds, Joel, 11. Dec 2002 08:09 | ||
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| Thanks for replying. Maybe i wasn't that clear with my original question. I understand that backdoor draws are longshots and i figured about the same odds as you did. My question is when you are faced with both a backdoor flush AND backdoor straight draw on the flop, what are the odds of hitting at least one of them by the river. ie: Jd 10d, flop comes 9d 4c 2h. | ||
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| Re: Calculating Odds, Mark, 11. Dec 2002 09:04 | ||
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| Oh, i get 12.6:1 to make either I convert the individual odds to percentages, add the 2 percentages of making the hands together, and calculate the odds for the combined percentateges. backdoor straight (1.9%) +backdoor flush (5.4%) = 7.3% = 12.6:1 odds mark Mark > Thanks for replying. Maybe i wasn't that clear with my original question. I understand > that backdoor draws are longshots and i figured about the same odds as you did. My > question is when you are faced with both a backdoor flush AND backdoor straight draw on > the flop, what are the odds of hitting at least one of them by the river. ie: Jd 10d, > flop comes 9d 4c 2h. | ||
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| Re: Calculating Odds, Joel, 12. Dec 2002 09:50 | ||
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| Makes sense...I think I made a mistake to get close to 10/1. Possibly because I counted outs for either and then assuming you hit one card on the turn, recounted for just the 4 flush or straight on the river, maybe my odds came out better because some of the outs overlap for a flush or straight on the turn. Wouldn't having outs that go either way effect your turn % of hitting? | ||
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| Re: Calculating Odds, Mark, 12. Dec 2002 13:25 | ||
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| >Wouldn't having outs that go either way effect your turn % of hitting? yeah, you have 16 outs on the turn, which is almost a 2:1 dog. But you still have to hit on the river to make a hand. (4.5:1 for flush and 11:1 for straght on the river) Mark | ||
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| Re: Calculating Odds, NiceFella, 17. Dec 2002 16:21 | ||
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| I've been curious about this as well, so I took the time to do the calculation excatly for your specific case of holding Jd Td when the flop comes 9d 4c 2h. My apologies if I've made any gross errors here. I welcome corrections! The odds of catching any flush (including a straight flush) is straightforward to caluclate. The are ten diamonds left, so the odds are: 10/47 * 9/46 = 90/2162, or 4.2%. You're roughly a 23:1 dog to make this. Making a straight requires that you catch one of the following series of turn and river cards: K then Q (4*4 = 16 ways out of 2162 for this to happen) Q then K or 8 (4*(4+4) = 32 ways) 8 then Q or 7 (32 ways) 7 then 8 (16 ways) The straight can thus occur 96 ways out of 2162. So you have a 4.4% chance of catching a straight, or about a 22:1 dog. Now for the chance of catching either a straight or a flush. We could add the ways of making a flush (90 ways) to the ways of making a straight (96 ways), but this would count the straight flush twice. How many ways are there to make a straight flush? Kd then Qd Qd then Kd Qd then 8d 8d then Qd 8d then 7d 7d then 8d -> 6 ways to make straight flush. So we can make a straight, flush or straight flush (90 + 96 - 6) ways. Our overall probability is thus 180/2162, or 8.33%. That's almost exactly 11:1 against. Happy chasing! NiceFella | ||
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| Re: Calculating Odds, Joel, 18. Dec 2002 10:10 | ||
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| So about the same as chasing a set to the river if you don't flop it. In a loose game I guess the pot odds are there. Thanks | ||
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| Re: Calculating Odds, Andrew Wells, 18. Dec 2002 12:39 | ||
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| Not quite, because if you miss a double backdoor draw on the turn you won't be staying for the river. The pocket pair is looking to hit EITHER on the turn or river. | ||
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| Re: Calculating Odds, Mark, 19. Dec 2002 07:20 | ||
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| You made a small mistake here. It doesn't have a huge effect on the math but since I'm picky... the following calculation is wrong. It should be 10/37 * 9/36. > My apologies if I've made any gross errors here. I welcome corrections! > The odds of catching any flush (including a straight flush) is straightforward to > caluclate. The are ten diamonds left, so the odds are: > 10/47 * 9/46 = 90/2162, or 4.2%. You're roughly a 23:1 dog to make this. Happy chasing! > NiceFella | ||
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| Re: Calculating Odds, NiceFella, 19. Dec 2002 22:03 | ||
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| Can you explain why it's 10/37 * 9/36? After the flop you've seen 5 of 52 cards, leaving 47 unseen. Why 37? | ||
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| Re: Calculating Odds, Mark, 27. Dec 2002 10:00 | ||
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| the equation is the # of cards to help over the # of cards which won't help make the hand. sorry i took so long to get back to you, i've gone east for the holidays and am moving west tomorrow. mark > Can you explain why it's 10/37 * 9/36? > > After the flop you've seen 5 of 52 cards, leaving 47 unseen. Why 37? > | ||
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| Re: Calculating Odds, Joel, 30. Dec 2002 07:48 | ||
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| Yeah, That's how I understand it too. | ||
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