Question about flush draw outs

This is probably a dumb question, but if you're on a flush draw, is counting 9 outs for that draw something that really makes sense? Because that really assumes that no one else in the game was dealt a card of that suit. Wouldn't it make more sense to find the EV of the amount of, say, clubs left in the deck, and use that?

Say there are 6 people. I have two clubs and two are on board. Should I take outs to be 9 - (Expected # of clubs in 10 cards dealt)?

no, because statistically there is an equal chance of all suits being out of the deck, so the calculations work out the same when you use a full deck (i.e. 9/46) as when you assume hands are out (i.e. 6:38).

yeah but we would expect at least one of the suit in question to be dealt out, wouldn't we? can't we calculate the probability of that?

The mathematics of probability is closer to Rocket Science than one might suspect. In fact, you can find variance in the statistics cited by many poker books. Some round the figures to "easy to remember" lists. Some provide raw data and formulas that resemble calculus text books.

However, the one constant that most poker statisticians seem to agree is that all questions of probability are based purely on the known (visible) information at the time each calculation is made.

[Xhad]

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yeah but we would expect at least one of the suit in question to be dealt out, wouldn't we? can't we calculate the probability of that?

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The pot odds calculations you see thrown around include your opponent's cards in the "unknown" portion of outs/unknown. Meaning it already takes this into account. The only reason you would need to adjust is if you have evidence that a club in someone's hand is more or less likely than usual.

but counting 9 outs we are already making an assumption -- that there are no clubs in anyone else's hand. how is this any better? yes we don't know what they have, but that's just what EV is for. what our best guess is. why shouldn't we go with that, rather than just proclaiming zero?

[jackblack73]

You're not making an assumption that there are no clubs in anyone elses hand. You're assuming other's have clubs in the same ratio as would be otherwise distributed in the deck.

Let's say there are 8 other players, so that's 16 cards out. It's reasonable to assume that there are 3 clubs among those 16 cards. So whether you do 9/47 or 6/31, you get essentially the same probability.

If we follow your logic, we shouldn't count any card that has no chance of being dealt to us.

You're thinking that it's impossible to dealt a card that one of your opponent has. That's true, but it's just as impossible to dealt a card that's below the river. (cheating not withstanding)

i already said i got it, why are you posting this?