Why do the guru's talk out of both sides of their mouth?

I don't want to belabor the fact, but I am going back to my "outs" and "pot odds" theory.

The guru's say to only use the known when figuring outs....but when it comes to pot odds, they say to only use the unknown!

My point is this, the guru's say...in a flush draw, where you hold suited pockets and two of the same suited are on the flop...then figure only 4 suited when doing your calculations, because they are the only "for sure" cards known. However I say, with 13 suited in a deck and 21 cards dealt the probabilities are that 5.25 of that suit have been dealt.

Now when they say to figure pot odds at the flop, they say to use the unknown....use 2 cards to be dealt whenever you figure flop pot odds...You can't be sure that you are going to see two cards, so why would you use two cards....the pot odds might be correct for the flop call...but they may be off for the turn call....so you would only be seeing one card not two and figuring for a two card draw when you only have a one card draw.

They need to make up their minds and be consistant!

Now I know that in my future play, I am going to use the 5.25 card probability thereby leaving me 7.75 outs not 9 and I am going to use 1 card draws on both the flop and the turn individually....thereby giving me 3/1 pot odds at the flop and 2.88/1 pot odds at the turn instead of their figures of 1.9/1 at the flop and 4.1/1 at the turn.

I might be in less flop pot then they will be in...but I will be a lot more flop/turn pots then they will be in.

what's important when calculating outs is to count everyone's hole cards as part of the deck. Rather than count the number of (let's say hearts) likely to be left in the deck because some hole cards may be hearts, simply pretend their cards are on the bottom of the deck. if 4 hearts are out in 5 cards (hole and flop), there will always be a 9/47 chance of the next one being a heart, regardless of how many players may be playing. To make a simple example, if i had a deck of cards and said guess where the A of spades is, your chances would be 1 in 52. Alternatively, If I dealt out half the deck face down on a table in lots of two, and then got you to pick from the remaining deck, no adjustment in calculation is required, you still have a 1 in 52 chance.
Your point about calculating two cards worth of outs into one call on the flop is a valid one, and not nearly enough players factor in the likelihood of the need for another call to see the river into their initial call.
Sorry, if I have misunderstood the point you were making about the outs calculations.

Well wobrien, let me answer in this way...if all the suited cards were dealt out and the remaining deck had 0 cards of that suit...you would have a 0 chance of then dealing another card of that suit, since there are none left in the deck.

Since I am talking about probable averages one out of every 13 cards is of a paticular suit...so on probable average, 5.25 were dealt in the 21 cards and since the probable average of the 21 cards is 5.25 suited then 15.75 are not of that suit, therefore, when we figure the cards remaining in the undealt portion of the deck, we would not count the 5.25 suited or the 15.75 unsuited cards....in figuring probable averages, we dont want to count the cards remaining in the deck that we say are no longer there.

do the maths, im telling you with 2 hearts in your hand and two on the flop, the chance of the next one being a heart is 9/47. Ok, (just to make this simpler) lets say you deal the flop and your hole cards. Every time you deal another players hand, yes you are probably dealing out some hearts, reducing your outs - BUT YOU ARE ALSO REDUCING THE DECK SIZE. Your chance of hitting a heart is not just about how many hearts are left - its about how many hearts are left as a proportion of the deck.

If you still disagree with me - then think about this. A bag with rocks in. 4 white, 4 black. Lets say have one try to pick out a rock and you want it to be white. Lets call getting a white rock an out. You have 4 outs of a possible 8 rocks which equals a 50% chance of hitting one of your outs. If you throw out four random rocks (like dealing out hands) on average you have thrown out 2 whites. Yes, you now have less outs on average - only 2 now. Have your odds become worse? No, because the outs as a proportion of the rocks in the bag is now different. You have 2 from 4 instead of 4 from 8. Always 50%.

Your starting to get through to me wobrien...if we have 7.75 outs and 31 cards left in the deck, we still have a 1 in 4 shot of hitting one...let me think about it a little more.

What threw me off was the standard formula 1-[(47-9)/ * (46-9)/46], 1-[(38/47) * (37/46)], 1-{0.81 * 0.80], 1-0.648 = 35% where I substituted 31 and 30 for the 47 and 46 and 7.75 for the 9, which gave me a higher percentage.