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Server Time: 12/1/2008 5:04:42 PM PACIFIC |
 Another math question, Robert M, 5. Dec 2003 19:34 | ||
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| Remember I am a little inept in math, and this question has very little practical application because it refers to a non-playable hand, however it is a good excersise in figuring odds. I was trying to calculate by myself the odds of making a straight when you only have 3 consecutive intergers after the flop. I thought that the chances of improving on 4th st would be 16/47 and on 5th st would be either 8/46 (if left outside) or 4/46 (if inside). So Do I average out the last ratio? In other words (16/47)*(6/46) = 96/2162 or 22.5 to one? That doesn't sound right it should be much more of a longshot than that. Oh, I would like to learn the math by the way, I can find this on the odds charts I'm sure. TY in advance-----------------------R | ||
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| Re: Another math question, PairTheBoard, 5. Dec 2003 23:47 | ||
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| It sounds about right to me. I would guess it's similar to the chance of making a backdoor flush which is pretty close to what you're getting. | ||
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| Re: Another math question, ekim2000, 6. Dec 2003 00:15 | ||
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| open (.17) * .17 = .0289 gut (.17)* .086 = .0146 -------- .04 Looks right to me. And whadda you mean it's not playable! ;-) | ||
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| Re: Another math question, Aisthesis, 8. Dec 2003 08:52 | ||
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| The averaging out won't as a rule give you a correct answer (it might in some cases--not sure about this one--actually turn out to be right, but it's not the way to figure the right answer). Basically, you would have to distinguish the various cases and add the probabilities of each. I am also assuming there is no way at all for the other two cards to give you any help. For example, you have 87 in your hand. Flop shows K92. With two cards left, you can now make 3 straights: 5-9, 6-T, 7-J. Or, as you mentioned, the 6 or the T on the turn will give you open draws for the river. So that scenario gives you chances of (8/47)*(8/46) chances. The 5 or the J on the turn each give you 4 outs for the gutshot. So that case has a probability of (8/47)*(4/46). Since the two cases are mutually exclusive, you can just add the two probabilities to get the correct answer: (8/47)*(12/46) I think that actually is the same answer you got by the averaging out method, but I'd really suggest not using that as I'm pretty sure it will get you in trouble in some calculations. | ||
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| Re: Another math question, Robert M, 8. Dec 2003 13:56 | ||
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| Okay, Thanks alot. | ||
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| Re: Another math question, jajo, 9. Dec 2003 16:11 | ||
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| The correct way to answer this question is by using combinations. If you hold 78 and the flop is K92 then you will make a straight if the last two cards are one of the following: JT,T6,56. Each of these two-card combinations can be selected in 4x4=16 ways. The number of ways of catching one of these combinations is 16+16+16 = 48. The total possible combinations for the last two cards is 47 C 2 = 1,081. Therefore 48/1081=.0444. Chances of making a straight is therefore 4.44 percent | ||
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| Re: Another math question, Aisthesis, 9. Dec 2003 19:21 | ||
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| yep, also correct. | ||
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| Backdoor straight, Banning, 10. Dec 2003 00:27 | ||
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| So, the possibility of making a backdoor straight are about 1/21. That could actually be something that might tip the scale. Imagine holding a single overcard plus a backdoor straight. Or two overcards plus backdoor straight possibility. I recall reading that a backdoor flush is worth adding a single out to your outs. I would imagine doing the same with the straight asuming that there is no obvious flush draw would be good math. If there is a flush draw alot of the cards that give you improvement also give opponent a flush so your outs fall down the tube. It is risky playing both of ones overcards as quality draws because if the overcard that is part of the 3 straight gets drawn it can give someone else a straight. Although tricky it is more math that a good player should be aware of. | ||
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