Outs and odds I'm a bit confused
I keep seeing it listed as such: On the turn with a straight and flush draw you have 17 outs, 9 to make the flush plus 4 at each end of an open ended straight. this doesn't seem right to me since on a full table of 10 there are 18 other cards you aren't even taking into account any of which could be a card you are counting towards your outs.
Shouldn't the number of outs be 0 to 17? That is at any time all the way from all of the cards needed to 0 of them could be in opponents hands.
Secondly wouldn't it be more accurate to figure an average number of possible outs rather than the maximum. That would give you a better picture of what percentage of the time you'll hit without actually using a calculator to figure the average percentage.
If my math is faulty please let me know. My strong suit is analysis not math.
Shouldn't the number of outs be 0 to 17? That is at any time all the way from all of the cards needed to 0 of them could be in opponents hands.
Secondly wouldn't it be more accurate to figure an average number of possible outs rather than the maximum. That would give you a better picture of what percentage of the time you'll hit without actually using a calculator to figure the average percentage.
If my math is faulty please let me know. My strong suit is analysis not math.
Comments
In you're example of having 17 outs, if you've seen the flop at this point then u know 5 cards, and there are 47 left in the deck, so the chance of hitting an out is 17/47. thinking that 17 outs means that your odds are 17/33 (since 16 cards are dealt, 8 hands, and 3 are on the flop) is wrong and is not the intended way to interpret "outs"... does that help?
hopefully you see why this question is irrelevant now...
To better answer your question, shouldn't the other X number of hole cards be taken into account, you need to remember that those hole cards are just as random as the cards left in the deck. It's really hard not to say duh
Also, a flush and an open ended straight draw is only 15 outs, not 17. 9 for the flush, plus 4 x 2 for the straight, however that double counts the cards that make both your flush and your straight, so subtract 2.
/g2
1 Except in the case where you are chasing the nut flush and you have reason to believe another player 2 in the hand 3 is chasing the same flush.
2 or multiple players
3 or formerly in the hand
so youre saying that if theres 2 other people in the hand with you then you should assume you only have 8 outs to draw to a flush?
If i'm analysing your post right, then your logic is as poor as the OP's.
I think that-that is definitely faulty because I can't see a correct and balanced way in which you would eliminate potential outs that would effect the outcome.
(overlooking human errors like an opponent you believe saying "I'd like to draw to my flush,but I fold" when you are yourself also drawing to the flush.)
For example, if I am understanding correctly and using a flush draw as an example- you take a hand like ah-kh to a flop of 9h-jh-2c against two opponents at a ten handed table.
YOU are suggesting that you consider SOME of the hearts to be folded by the 7 opponents who didn't make it to the flop?(ignore the overs for a sec..this is hard enough for me..I suck at math)
and how would you determine the average (x)? I guess by figuring out- number of hearts remaining:the complete unseen deck=X:the deck not dealt to players
or
9:47=X:29...x~5.55..which is just the written form to show that determining 'average' of folded hearts demonstrates that any mathematical calculation would have to, in equal balance take into account the PROPORTIONATE amount of folded non-hearts and this calculation, therefore; makes no difference at all
would it?
I think to check and see we would compare [(9/47)+(9/46)]X100 to [(5.55/29)+(5.55/28)]X100?
(rounded to the one hundredth) [(0.19)+ (0.20)]X100 to [(0.19) +(0.20)]X100
[0.39]X100 to [0.39]X100
39% to 39%
As for calculating the range of potentially folded outs (you used 0-17, mine has 0-9 if we are considering ONLY the flush) wouldn't that negate the idea of counting outs at all?
This one gets even farther over my head because I'm not certain if the likelyhood of specific numbers of hearts being folded would actually carry weight here when averaged?
What I mean is out of the potential ways that the seven folded hands can be dealt and folded- do the large number of instances where smaller numbers of hearts are dealt and folded outweigh the small number of times that-as an extreme- all or most of the hearts are folded, in an average?
I'm interested to see one of the more competent math minds take this one on..though my gut says that if we haven't already heard about this concept the answer is probably that they somehow balance or the return on doing such a calculation is insignificant.
let me take one more stab at it too in case some are still not getting it
you can't reliably tell what cards your opponents are holding. the probability that one of your opponents cards is an out for you is the same as the probability that a card in the deck is an out for you.
even if you're opponents have folded a few of your outs, they've also folded a few of your non-outs, and the expected ratio of outs:non-outs already folded is the same as the outs:non-outs remaining in the deck, thus it does not affect the likelihood of you sucking out.
I've seen the topic discussed before. Generally, you consider all your outs 'live' and all unknown cards. just like g2 said (he is smart)
Otherwise, you'd have to look at all the cards dealt, apply a factor for number of your outs (i.e. 18 dealt, maybe 5 hearts, you have 2 so there are 6 left in the deck of 31. Â >> 6/31 is awefully close to 9/47 don't you think?) I would follow the simpler method as you can easily use this for shorthanded or full table without changing your calcs.
As an aside, I'd recommend Theory of Poker for those who aren't sure about all this stuff. As I recall, Sklansky's Hold'Em for Advanced Players was pretty good with this too, although it discusses limit (ugh!) - the math remains the same. There are other books out there on odds in poker (something by Yao?). I haven't looked at them but I'm sure others have. Please post if you have found them helpful.
You mean you shouldn't?
Plus it was also at 2 am and I had just finished drinking my FAVOURITE Cpt Morgans spiced rum.
K so lemme think about this again. You're holding 109h, board is QJ6 with 2hearts. Two other players. You got 9 hearts and 6 cards for a straight. 15 outs. There are 5 cards out that you don't know (2 players and a burn) and 5 that you do know. So 42 cards left. You then say I have 15/42 cards to win. But isn't that saying then that there are no hearts of cards for your straight in the 5 cards out you don't know? and aren't the odds that there is at least 1 of your outs in those 5 cards so it'd be say 14/42 cards to win? (Just got up still groggy so sorry if i'm trying to argue a point that really makes no sense.)
No..
He's saying you have 15 outs of the 47 cards you're unaware of (you don't discount the burn / other hands). So, 15/47 = 32% chance of hitting.
The best explanation for your argument is "If it's likely they tossed one of your outs, they likely tossed non-out cards" stated above, so it's not worthwhile to try and figure out "estimates"
Mark
But if you looked at the 25 that you're tossing before you toss them, and confirmed that there were no aces in that half of the deck, then tossed them...NOW your odds are 2/25 because those cards have lost their "unseen" status, and can now be factored in.
That was one of the easier ways I've seen it explained.
I haven't really heard the idea that "if your opponents folded some of your outs, they also folded some of your non outs, so it evens out in the wash" idea before, but it makes sense. Glad I read this thread.
sstar
/g2
Thank you. Guess I sometimes go through some not too bright moments. Simply put I had been looking at it the wrong way and trying to figure the odds on the remainder of the deck not the full deck minus known cards. There are still a LOT of books I need to read. ;-)
I may die before I know even a small portion of what's to learn about this game. I've played some chess and it seems to be simpler in that you know or at least can know the whole situation at all times.