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Server Time: 11/20/2008 9:26:36 AM PACIFIC |
 Implied odds question, mrbippy, 7. Sep 2003 17:49 | ||
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| This is pretty dumb, I should know this but apparently I don't. I apologize in advance. I'm reading the chapter on implied odds in Sklanskys Theory and there's something I don't understand. I understand implied odds just fine, what I don't get is how he calculates the odds of a card hitting the turn. The 1980 wsop example he gives has Ungar needing a 3 to make his straight to beat Brunsons two pair. OK no big deal 47/4 to make it is reduced to 11.75:1 but he has it as 10.75:1. Now I know I'm wrong not David so where's my mistake? Other than skipping math classes way too often. Also the implied odds he gives. Brunson has $232500 in chips with $30000 in the pot. Brunson bets $17000 to Ungar. Stu figures if his 3 hits he could bust Brunson and win the match. So the pots $47000 plus Brunsons $232500 = $279500. Ungar calls $17000. 279500/17000 = 16.44:1. Sklansky gives the implied odds as 14.5:1. Is my mistake adding Stus half of the pot? ($15000) When I figure the odds without Stus $15000 the odds come to 15.5:1. Now I notice a parallel here. With both of these calculations (the latter in respect to the implied odds) I seem to be off by 1. eg 11.75:1 as opposed to 10.75:1 and 15.5:1 and 14.5:1. Is this where I'm goofing? Any comments appreciated MrB | ||
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| Re: Implied odds question, Cockpit, 7. Sep 2003 18:35 | ||
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| on 7. Sep 2003 17:49 mrbippy wrote:. > > The 1980 wsop example he gives has Ungar needing a 3 to make his straight to > beat Brunsons two pair. OK no big deal 47/4 to make it is reduced to 11.75:1 but > he has it as 10.75:1. > > Now I know I'm wrong not David so where's my mistake? Perhaps a simpler example will show you the error in your method. If you flip a coin there are two outcomes, heads or tails. But the odds are not 2:1 of a tail coming up. It is 1:1. You have to compare the number of losing possibilities to winning ones. In the example above that would be 43:4 or 10.75:1 Hope that helps. | ||
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| Re: Implied odds question, Easy E, 7. Sep 2003 18:51 | ||
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| Basically, it's the fractional chance of hitting, subtract one from the denominator and make it a odds formula. so, 1/2 heads = (2-1):1 1/4 chance = 3:1 => 3 chances to lose : 1 chance to win | ||
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| Re: Implied odds question, mrbippy, 7. Sep 2003 18:55 | ||
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| *Dunce cap on* Thanks! Where am I going wrong on my Implied odds calculations? Anyone? Anyone? I played poker with tarot cards the other night. I got a full house and four people died.-Steven Wright | ||
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| Re: Implied odds question, Easy E, 8. Sep 2003 19:56 | ||
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| Brunson has $232500 in chips with $30000 in the pot. Brunson bets $17000 to Ungar. Stu figures if his 3 hits he could bust Brunson and win the match. So the pots $47000 plus Brunsons $232500 = $279500. *** DING DING DING! You double-added the 17K from Doyle. After the bet, pot has $47K, Doyle has $215,500 = 262,500 = 15.44/1 or 14.4:1 (rounding somewhere, perhaps?) I was first- what do I win? | ||
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