9 outs for a flush, 8 for a straight.


At a 9 person table, 18 hole cards, 35% of the deck. On average, ~4.5 cards of each suit in the hole cards, 8.5 in the rest of the deck.


You get 2 suited of the 4.5 available.


Flop comes with 2 flush cards of the 8.5 left.


There are now 6.5 outs to make the flush.


Same procedure with straights gives 5.3.


Can somebody point out the flaw in my flawless reasoning?

am I drunk?I feel like I get hit in the head and I know ALOT about poker.I dont understand the question sry could you rephrase it?

Better chance of getting a flush. So, who won the pot?

Originally Posted by arq

amI drunk?I feel like I get hit in the head and I know ALOT about poker.I dont understand the question sry could you rephrase it?

At a nine person table, 2 hole cards per person is 18 hole cards. On average, those 18 hole cards will contain 4.5 cards of each suit, with the other 8.5 cards in the deck.


Say somebody gets 2 spades in the hole. The flop comes with 2 more spades. We know that on average, there were 8.5 spades in the deck after the hole cards were dealt, and two of them came on the flop.


That leaves 6.5 outs to make the flush. On average. Not 9.


Open-ended straight gives 5.3 outs instead of 8.


What am I doing wrong?

9 outs. You count them all unless you know some of your outs are gone. 2 in your hand, 2 on the board.

You are mixing averages and knowns.

You should think of it what you know and what you don't know. You don't know what the other players are holding. So whether you are playing against 9 people, or headsup...the outs stay the same.

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.

Your mistake is to asume 4.5 cards are no longer in the deck.
On average 4.5 cards are no longer in the deck, maybe they are available for you.
Check the chance calculation and you'll see they use 9 outs and do not remove the 16 cards you don't have.
This way should give the same result as yours, but is more simple.

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. 13-4=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= (A-1) x A : 2

So in relation to this topic we calculate the odds of someone else being on a flush draw:

C = (9-1) x 9 : 2 = 36 possible flush draw combinations
C = (47-1) x 47 : 2 = 1081 total possible combinations (by the way there are (52-1) 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.

Originally Posted by bbennie1

Okso you want to know the odds someone else is on the same flush draw.

No. Say our hole cards are spades, nine-handed. I want to know why we assume that when we get spades, that nobody else got any spades, when that is not the most likely scenario.


The average hand has 4.5 spades as hole cards (18 hole cards total / 4 suits) and 8.5 in the rest of the deck. (Right?) Then the mostly likely scenario is that either 2 or 3 spades are in someone else's hand, not available to us. That leaves 8 or 9 spades.


2 more spades on the flop for the flush draw leaves 6 or 7 spades/outs.


What is wrong with that reasoning?

Originally Posted by 1Player

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No. Say our hole cards are spades, nine-handed. I want to know why we assume that when we get spades, that nobody else got any spades, when that is not the most likely scenario.


The average hand has 4.5 spades as hole cards (18 hole cards total / 4 suits) and 8.5 in the rest of the deck. (Right?) Then the mostly likely scenario is that either 2 or 3 spades are in someone else's hand, not available to us. That leaves 8 or 9 spades.


2 more spades on the flop for the flush draw leaves 6 or 7 spades/outs.


What is wrong with that reasoning?

It's true that it is likely that other(s) have gotten a spade, but as it has been said before, you calculate what you can see and what could be left in the remaining deck. If you wish to come up with a different formula and tell the pokerworld how it should be done I wish you all the best

Originally Posted by bbennie1

It'strue that it is likely that other(s) have gotten a spade, but as it has been said before, you calculate what you can see and what could be left in the remaining deck. If you wish to come up with a different formula and tell the pokerworld how it should be done I wish you all the best

I know the rule. I've seen it and heard it 1,657,932 times. 33 now.


What I'd like to know is why. Why is that the rule? Why, in a game about playing the odds, do you assume a less likely scenario instead of the two most common ones?


There's my different formula: On a flush draw, you're more likely to have either 6-7 outs than 9.


Someone please tell me why that's not the rule.

I figured it out.


I'm dumb.

Originally Posted by 1Player

I figured it out.


I'm dumb.

You're not dumb.
There are different assumptions and follow up calculations.
Most importent, now you know more than before, maybe the road was not the prettiest, but it's the result that counts.

You get it? Even if my opponentis also on a flush draw, there are 9 outs but he's got 2 of them.

you figured it our? I think you're a bit off on your calculations, you may have 9 outs for a flush, but at most, you will only have 6 for a straight while having a flush draw. Think about it, 2 of those 8 you count = Flush cards. now if your cards are on the top end of the draw, you may have an additional 4 outs for over pairs, but hitting one of those would open up a possibility of villain getting their possible straight.