#11
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Re: Odds for Hitting Outs on Turn AND River
[ QUOTE ]
The doubling method works like this; take your outs (4 in the case of a gutshot) and double it for your chance to make it on the next card (the turn - if your on the flop OR the river if your on the turn) which would give us an 8% chance to make our straight on the next card. If you are paying to see two cards then double it again 4 doubles to 8 and that doubles to a 16% chance. This method is a quick and dirty method of calculating your pot odds at the table and it is resonably reliable within a certain range i.e. from about 4 to about 14 outside of this range it begins to lose accuracy. [/ QUOTE ] Not really doubling. Just multiplying your outs by 2% for 1 card to come, and 4% for 2 cards to come. The rule of 4 (for 2 cards to come) actually works pretty well up to 11 outs. After 11 outs, each successive out is worth 3% (until you get way up there in outs). 13 outs twice would be 11*4 + 2*3 = 50% |
#12
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Re: Odds for Hitting Outs on Turn AND River
The 2 basic probability formulas for independent events are as follows:
(1) P(A and B) = P(A)*P(B) (2) P(A or B) = P(A) + P(B) - P(A and B) It is the latter that you need to use. You can't just add them together, because the events A and B intersect each other. Example: If P(A) = 10%, P(B) = 20% are viewed as "lengths," they cannot simply be added together, because the "lengths overlap each other. If the events are independent, then the amount of overlap is P(A)*P(B), which must be subtracted from the sum. Therefore P(A or B) = P(A) + P(B) - P(A)*P(B), which combines the 2 formulas (1) and (2) given above. As it turns out, if the values for P(A) and P(B) are small, then P(A or B) = P(A) + P(B) can be a very decent approximation. So for the example given above: P(A) + P(B) = P(A) + P(B) - P(A)*P(B) [approximately] 10% + 20% = 10% + 20% - 2% [approximately] 30% = 28% [approximately] which may be close enough for practical purposes. In poker, with 2 cards to come, as an approximation, use: 4*outs, for 8 or fewer outs 3*outs + 8, for 9 or more outs to compute your chances of hitting your with 2 cards to come. If you want to be accurate to within 1%, for 11 through 16 outs, use: 3*outs + 9 instead. These approximations are more than enough for practical purposes and are what I use at the table. |
#13
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Re: Odds for Hitting Outs on Turn AND River
I guess I'll have to spend some time rereading this thread. My rough method has always been something like this:
Say I have a flush draw on the flop: 47/9. I treat it as 9 outs twice, so it's 47/18, so roughly 2.6:1 that it comes by the river. If I don't hit on turn, it's 46/9 or 5:1 that it hits on the river. I've never given it much though past this method. I guess I set aside a few hours tonight to decipher all the gibberish I couldn't internalize in my quick read through this thread. |
#14
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Re: Odds for Hitting Outs on Turn AND River
If X=outs on flop/turn and Y=outs on the turn/river and Z=probability of making your hand, then
Z = X/47 + Y/46 - (X/47 * Y-1/46) |
#15
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Re: Odds for Hitting Outs on Turn AND River
[ QUOTE ]
i might be wrong, since i'm new to poker as well, but i believe that on the turn,it's the number of outs X 4 = %odds and on the river it's the #outs X 2... therefore 10 outs on the turn = 40% and on the river 10 outs = 20% i guess that means you're 40% and 20% to win the hand... i'm sure i'll be corrected if i'm wrong... [/ QUOTE ] this is deffinitely correct to some extent. this method is used by many poker players as the best ESTIMATION of your odds to win the hand. however i have found that multiplying by 2.5 is a much more accurate estimation with just the river card to come as opposed to multiplying by 2 keep in mind that these are merely estimations and will be off from the true odds by as much as 6 or 7% |
#16
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Re: Odds for Hitting Outs on Turn AND River
w0rd
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#17
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Re: Odds for Hitting Outs on Turn AND River
2.5 is--less--acurate that using 2 i.e. 100/46 = 2.1739130435 or about 2.2
I use 2.2 at the table. Example: 21 outs * 2.2 = 46.2 The true value is 45.652173913. A difference of < 1% The mechanics of multiplying by 2.2 is no more difficult than multiplying by 2.5 if not even easier e.g. I multiply the number of outs by 2, then add 1 tenth of the result. |
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