Why do we assume all of our outs are out there?

[bdk3clash]

Disclaimer: I am officially awful at probability, so forgive me if this is a really, really dumb question.

Why do we calculate our outs based on the assumption that all of the cards are still available? For example, if I flop a flush draw, I have 9 outs to complete it. But wouldn't logic dictate that some of those "outs" are already in the hands of my opponents?

[AJo Go All In]

the chance that some of your outs are in your opponents' hands is factored in when you use 47 as the total of number of cards you are drawing from, rather than say 29 in a ten-handed game. i suspect that if you assumed your opponents had random hands, and calculated the expected number of diamonds out in your opponents' hands given that you have 2 diamonds, subtracted that from your outs, and then used 29 as the number you are drawing from, the calculation of your odds would turn out the same.

[bigpooch]

Correct! Also, some of those outs may be "tainted". The
calculations are obviously also going to be approximate as
the possible hands your opponents hold will affect the
chances of completing your hand. As your opponents could
have one of many hands, to calculate more accurately what
your chances are can be much more difficult.

Yes, you're right about the logic: why are your opponents
even continuing in the hand? So in additon to the raw
mathematics, you would need to know roughly how your
opponents play and how they think to determine the play with
the highest long term positive expectation.

[spamuell]

It works the other way as well, remember that. I posted a thread (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=holdem&Number=468406&Forum =All_Forums&Words=5053&Match=Username&Searchpage=0 &Limit=500&Old=allposts&Main=468406&Search=true#Po st468406) about this in the general forum a while ago.

[BruceZ]

I think he's saying that even with no hand reading and just assuming your opponents hold random cards, that some of the outs are likely to be in your opponent's hands. That is true, and it is already taken into account when we say that we have 9 outs out of 47 unseen cards. A card in your opponent's hand has the same 9/47 chance of being an out as a card in the deck. If you thought that all your outs were still in the deck, then you would have 9 outs out of only 29 cards, equivalent to over 14 outs out of 47 cards, so your drawing chances would be much better. Since your opponents hold 18 of the 47 unseen cards, 9 outs assumes that your opponents hold 18/47 * 9 = 3.45 outs on average, and only 5.55 outs are in the deck.

If you somehow know that a particular opponent holds one of your outs, then you would now have 8 outs out of 46 cards, so your drawing chances would be reduced from 9/47 to 8/46. If you found out that 3 or 4 of your outs were in particular opponent's hands, then your drawing chances become 6/44 or 5/43 which are much worse than 9/47, even though we said that 9/47 assumes that your opponents hold between 3 and 4 outs. This is no contradiction. It is because assuming that all of your opponents hold between 3 and 4 outs in total is much different from knowing that particular opponents hold 3 or 4 outs. In the latter case, the remaining opponents will on average hold additional outs, so there will be even fewer left in the deck.

Determining that an opponent holds a non-out card also changes your drawing chances. If you know that a particular opponent holds no out, then your chances have become 9/45, since the number of unseen cards has been reduced by 2. Note that it is not necessary to actually "see" his 2 cards to reduce the number of unseen cards. It is only necessary to distinguish between outs and non-outs.

[M.B.E.]

Here (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=316008&page=&view=&sb =5&o=&vc=1) is a link to the answer.

[ctv1116]

It goes both ways, and if you average out you get the 9/46 chance to catch the flush on the river. Just like you can say that its possible that your opponents may hold a few of the 9 outs, you could also say that they DONT have the flush cards, in which your probability would be even higher than 9/46. For example, if you somehow knew that no one else (10 player table) had any of the flush cards, the probability of making the flush is now 9/28, much better than 9/46. Of course, your opponents may hold all the other 9 flush cards, in which case you're drawing dead. But on average, you have 9/46. May be better, may be worse. But if you're good enough to read suits of players who folded preflop, you won't really need odds all that much.

[RydenStoompala]

How does the location of the outs become a factor? If I'm 9 outs to the flush with a known number of unseen cards, and I can count the pot and bets, my odds (or probability if you convert it) are known and the bet is either good or bad.