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Server Time: 8/30/2008 12:06:11 AM PACIFIC |
 Calculating Odds, David Erickson, 21. Aug 2003 13:17 | ||
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| I'm a bit unsure how Sklansky comes up with his percentages when calculating odds after the flop. One source I've read states 2 outs = 4.3% (2/47), whereas Sklansky (in "Holdem Poker for Advanced Players") states that it's 8.4%. Does anyone know his formula for coming up with that and it's logic? | ||
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| Re: Calculating Odds, Jav, 21. Aug 2003 13:31 | ||
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| The first stat (2/47) = 4.26% is the percentage that you will hit one of your two outs on the turn. If you want to figure out your chance of hitting one of your two outs on the turn or the river (before the turn has happened), then it is (2/47 + 2/46) = 8.60%. That assumes that you didn't hit one of your outs on the turn... | ||
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| Re: Calculating Odds, David Erickson, 21. Aug 2003 13:33 | ||
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| I figured that, except he's got 8.4% in the book. Hence my confusion. | ||
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| Re: Calculating Odds, David Erickson, 21. Aug 2003 13:37 | ||
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| Or to give a better example: Your formula for 8 outs = 34.5% Sklansky for 8 outs = 31.5% on 21. Aug 2003 13:33 David Erickson wrote: > I figured that, except he's got 8.4% in the book. Hence my confusion. | ||
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| Re: Calculating Odds, Jav, 21. Aug 2003 13:48 | ||
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| His percentages might be based on simulations that actually play out every single hand. In that case the chance of hitting an out on the turn and on the river will skew the percentages a little bit. | ||
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| Re: Calculating Odds, David Erickson, 21. Aug 2003 14:03 | ||
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| hmmm, yeah I guess so. He referenced a technique for doing this. Guess I was hoping someone knew what it might be. on 21. Aug 2003 13:48 Jav wrote: > His percentages might be based on simulations that actually play out every single hand. In that > case the chance of hitting an out on the turn and on the river will skew the percentages a little > bit. > | ||
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| Re: Calculating Odds, tpir90036, 21. Aug 2003 14:02 | ||
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| actually, that is because the actual math has to include the possibility of hitting the out on the turn *and* the river. here is an example using 2 outs: in english: hit on turn OR hit on river OR (hit on turn AND hit on river) in math: (2/47) + (2/46) + (2/47 * 1/46) in situations like this it is actually easier to determine the odds of *not* hitting an out on *either* card and taking the inverse of it: english: 100% - (not hitting on turn AND not hitting on river) = odds of hitting math: 1 - ( (45/47) * (44/46) ) = .0842 = 8.42% hope that helps!! | ||
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| Re: Calculating Odds, David Erickson, 21. Aug 2003 14:18 | ||
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| ok, now we're getting somewhere. this makes sense. if you could please, for my own edification, clarify where you get the 45/47 and 44/46 figures. I presume we're referencing outs of course, but relative to what? on 21. Aug 2003 14:02 tpir90036 wrote: > actually, that is because the actual math has to include the possibility of hitting the out on the > turn *and* the river. here is an example using 2 outs: > > in english: > hit on turn OR hit on river OR (hit on turn AND hit on river) > in math: > (2/47) + (2/46) + (2/47 * 1/46) > > in situations like this it is actually easier to determine the odds of *not* hitting an out on > *either* card and taking the inverse of it: > > english: > 100% - (not hitting on turn AND not hitting on river) = odds of hitting > > math: > 1 - ( (45/47) * (44/46) ) = .0842 = 8.42% > > hope that helps!! | ||
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| Re: Calculating Odds, tpir90036, 21. Aug 2003 14:33 | ||
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| no problem! the 45/47 and 44/46 are for a two out example. i was just showing that most of the time it is quicker and clearer to calculate *not* hitting at all and take the opposite of that. so if we have 2 outs there are 45 cards that *don't* help on the turn and 44 cards that *don't* help on the river. so we want to know what the chance of *not* hitting on the turn AND *not* hitting on the river is, and then take the other side of that: english: 100% - percentage of not hitting at all = percentage of hitting *somewhere* (we don't care where exactly) again: 100% - (MISS turn AND MISS river) = hit somewhere in math (repeat from previous post): 1 - ( (45/47) * (44/46) ) = .0842 = 8.42% you can do it the longer way too and the answer is still the same. i was just trying to show a different way of solving it that is both easier and hopefully reinforces why there was a small discrepancy in the way you originally did it. good luck! | ||
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| Re: Calculating Odds, David Erickson, 21. Aug 2003 14:41 | ||
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| lol, ok....duh. it's all clear to me now. For some reason I couldn't see what the 47/46 figures were :) Thanks! on 21. Aug 2003 14:33 tpir90036 wrote: > no problem! > > the 45/47 and 44/46 are for a two out example. i was just showing that most of the time it is quicker and > clearer to calculate *not* hitting at all and take the opposite of that. > > so if we have 2 outs there are 45 cards that *don't* help on the turn and 44 cards that *don't* help on the > river. so we want to know what the chance of *not* hitting on the turn AND *not* hitting on the river is, and > then take the other side of that: > > english: > 100% - percentage of not hitting at all = percentage of hitting *somewhere* (we don't care where exactly) > > again: > 100% - (MISS turn AND MISS river) = hit somewhere > > in math (repeat from previous post): > 1 - ( (45/47) * (44/46) ) = .0842 = 8.42% > > you can do it the longer way too and the answer is still the same. i was just trying to show a different way > of solving it that is both easier and hopefully reinforces why there was a small discrepancy in the way you > originally did it. > > good luck! | ||
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