Originally Posted by 1Player
Yeah, I know the rule.
Why is that reasoning not valid?
Or try this: What are the odds that exactly two spades get dealt in those 18 hole cards? My math (which I'm not at all confident in) says that the odds are very very small. Infinitesimally small. And yet that's what we assume happens EVERY TIME we get suited hole cards and a flush draw on the flop: oh yeah, there are 9 spades left in the deck, no one but me got any spades!
Makes no sense to me.
Please, please, make it make sense.


Ok so you want to know the odds someone else is on the same flush draw.
52 cards in the deck with 13 cards of the same suit.
2 of the same suit in your hand + 2 = 4 that we know. 134=9 unaccounted for. 9:2=4,5 means 4 other people could have the same type of flush draw.
52 minus two cards from your hand minus 3 cards on the flop = 47 unaccounted cards.
How to work out the total amount of hand combination of an unpaired hand (AK). Not important here but I mention it anyway:
C= total combinations
A¹= available cards for the first card
A²= available cards for the second card
C= A¹ x A²
How to work out the total amount of hand combinations for a paired or suited hand:
C= total combinations
A= available cards
C= (A1) x A : 2
So in relation to this topic we calculate the odds of someone else being on a flush draw:
C = (91) x 9 : 2 = 36 possible flush draw combinations
C = (471) x 47 : 2 = 1081 total possible combinations (by the way there are (521) x 52 : 2 = 1326 hand combinations in poker)
36 : 1081 x 100 = 3.333% chances of one other person being on a flush draw had you been headsup.
Against 9 others it's 9 x 3.33 =
30%
You could also use free software like
Pokerstove to figure this out, or so I've heard.