Number of players dealt in affecting flush draws?

[Psy_Mike]

I posted a similiar question in mid stakes, but it died out when I asked this follow up question. So I'll give it a shot here.

When you have a flush draw you usually count it as you have 9 outs out of 47 unseen cards, on the flop.

What if we would do it the other way around and discount cards. So in a 6 handed game, you are dealt 2 hearts. On average 3 hearts are dealt pre-flop in a 6 handed game. When the flop comes with 2 hearts we should therefor on average have 8 outs out of 37 cards that are left in the deck yes?

First question is: 8/37 != 9/47
Where is the error? Should they not work out to be the same? Which is correct?

I thought it would work out the same, but according to the 8/37 method the likelihood of you hitting your flush is dependent on the amount of players dealt in.

Which method is more correct, and why is one not correct?

[qpw]

Although it's a little hard to grasp at first, the cards that have been dealt (provided they remain unknown) are still, for the purposes of our calculation part of the deck!

Consider it this way:

Suppose you take a complete pack of cards, cut it in two and select a card from one half.

Has the probability of it being, say, a club, changed because you spilt the deck in two before selection?

Obviously not.

So if you consider the part from which you didn't make your selection to be those cards that were dealt to other players, it is clear that the number of players dealt in does not affect your chances of making a flush (or, indeed, getting any other card).

[Psy_Mike]

Ya I actually do grasp the concept of known and unknown cards and understand why it works that way.

But it should be possible to discount cards as well shouldn't it, and getting the same results? Or why doesn't it work to discount dealt cards? Why can you not assume that an average of 3 suits each have been dealt pre, leading to the 8/37 equation?

[qpw]

[ QUOTE ]
Ya I actually do grasp the concept of known and unknown cards and understand why it works that way.

But it should be possible to discount cards as well shouldn't it, and getting the same results? Or why doesn't it work to discount dealt cards? Why can you not assume that an average of 3 suits each have been dealt pre, leading to the 8/37 equation?

[/ QUOTE ]Sorry, I didn't address that part of your question because I wan't sure I'd properly understood it. (I'm still not 100% sure.)

It seems that you are trying to say that you know you have two hearts but are still expecting the number of hearts dealt pre flop to be 3.

You can't do that!

If you take your 2 hearts out of the pack then the remaining five players have 11 hearts available out of 50 cards, so you would actually expect, on average to have around 4 hearts dealt pre flop. Thus you are nearer to 7/37 which is pretty close to 9/47 (and if you can be bothered to do the maths accurately rather than use round figures will come out the same).

[Psy_Mike]

Ah that's true!

But how is the math done exactly?
If we are dealt 2 hearts there are 11 more out of 50 cards. So 22% of the remaining cards are hearts. When dealt out to the remaining 5 players, on average 2.2 hearts will be dealt, right?

So pre-flop you will on average have 4.2 hearts. This leaves us with 6.8 outs out of 37 cards left on the flop flush draw. But to be equivalent of 9/47 you need to have closer to 7.1 outs.

I'm assuming this is not enough though. Do I have to count on how the odds change as a heart gets dealt to another player? Or is it ok to count them it as 22% of 10 cards?

Sorry if this is all really basic, just trying to figure it out hehe.

[qpw]

Going back to the beginning, I see that I've used some of your figures where I shouldn't have done.

Let's simplify it by considering the odds of getting one heart pre-flop.

Method 1 - There are 11 hearts in 50 cards yielding p = .22
Method 2 - There are 8.8 hearts in 40 cards yielding p = .22

You can follow it on from there but you can see that if you added 5 more players you would have 6.6 hearts in 30 cards which is also p = .22

When the first card of the flop is dealt:

Method 1, there are 10.78 (11 - 0.22)/49 = .22
Method 2, there are 8.58 (8.8 - 0.22)/39 = .22

And so on.

[Mr Rat]

[ QUOTE ]
But it should be possible to discount cards as well shouldn't it, and getting the same results? Or why doesn't it work to discount dealt cards? Why can you not assume that an average of 3 suits each have been dealt pre, leading to the 8/37 equation?

[/ QUOTE ]

I thought that an unknown will count fully as an unknown (since, unless you are card tracking, there is no way to really know where any of your outs are at). I understand you are averaging how many of a suit in theory get dealt out - but since all unflipped are unknown can you use this method of discounting? You can assume by averaging that some of your outs are gone, but you truly do not know they are gone.

Has a simulation been done to compare flush draws hitting heads-up vs short-handed vs full table? I would be interested in seeing the restults.

Note I am a newbie still learning so I am just going with what I know so far...if discounting cards per number of people at a table is valid, I would like to know.

[qpw]

[ QUOTE ]
Note I am a newbie still learning so I am just going with what I know so far...if discounting cards per number of people at a table is valid, I would like to know.

[/ QUOTE ]
If you look at the post directly above yours you will see that if you use the correct numbers, you get exactly the same result with discounting - indeed, you get (by inference) the same numbers no matter how many people are at the table

Given that is the case it seems that it's a valid way to do the calculation.

[WhiteWolf]

2 hearts are in your hand. 2 hearts are on the board (this is what I think you are still missing). There are 9 left in the deck of 47, so the concentration of hearts is only 19%. On average, 5 opponents will hold 1.9 hearts, and 9-1.9 gives you the 7.1 you were looking for.

It's true that, if you asked before the flop, you would say that 5 opponents would have 2.2 hearts on average given that you have 2 of them. However, once the flop comes up with 2 hearts, that presents you with some evidence that hearts in your opponent's hands are actually rarer then your original calculations (look up Bayesian probability for the reasons). So you will have to recalibrate your assumptions, and you come up with the 1.9 number. The math to prove all this is quite lengthy, but will result in the same number as you get when you treat all unknown cards equally (which is a much easier calculation and the one used in all the odds charts).

[Mr Rat]

Thanks qpw & WhiteWolf. I will stick with the standard method but it is nice to know there are two ways to get to this and that I do not have to focus on discounting at the table.

lol on the rat highlights.