probability of drawing

[RJT]

I assume the following is invalid logic, because I have never heard/read it discussed. What is the flaw? Or is it logical, but impractical?

I will use a flush draw for an example. I also use % for brevity.

10 player table
Player 1 pocket is suited –say hearts.
Flop is heart- heart-club

Probability to draw a heart on turn: 9/47 =19.15%
No heart on turn, then probability to draw heart on river: 9/46=19.56%

But in reality, there are 18 cards already dealt(9 other players x 2). There almost certainly is at least one heart in the batch of 18 and more probably 4.5 hearts already dealt(18 divided by 4 suits).

Why don’t we calculate chance to draw heart on turn as either:
9-4.5/47=9.57% or 9-4.5/29=15.52% (29= 47-18)

Chance to draw heart on river as either:
9-4.5/46=9.78% or 9-4.5/28=16.07%

[gaming_mouse]

Another poster asked almost the exact same question, and discussed the flaws in his logic here:

link

What it boils down to is that the (correct) method you used to calculate the odds implicitly takes into account the possibility that opponents hold cards. You are only interested in how many "unseen" cards there are. I discuss the reasons for this in the thread above, and also show how you can explicitly factor the possibility of opponents holding cards -- it just makes the computation much more difficult but arrives at the same answer.

Cheers,
gm

[memphis57]

Actually the odds are that there were 3.447 hearts in the 18, because we KNOW that 4 of the 13 hearts were set aside in special places in the deck where they would end up in player one's hand or the flop. Thus, 9/47*18=3.447. So you can take either 9/47=.1915 or 5.553/29=.1915 for your probability.

[RJT]

Got it guys, thanks.