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.

but if i expect 3 clubs, then i only have 6 outs, not 9.. what am i missing here?

[jackblack73]

Using my example, that's 6 outs out of 31, not 47. You have to count the cards that aren't clubs also. 6 out of 31 is the same probability as 9 out of 47.

yeah i get that part. but when we count outs we don't factor in the amount of total cards left in the deck. that's why i'm saying it should be 6 outs, giving us 25% equity, not 9 outs giving us 35% equity.

[jackblack73]

You can't count what you don't know, so that's why you use 9/47. Obviously, if you know a club is burned you can adjust your probability to 8/46.

[Xhad]

Here's a simpler example. I randomly give you one card out of a full deck. The probability that that card is the A /images/graemlins/spade.gif is 1/52 right?

Now what if I deal myself five cards, then deal you a card. The probability that the card I give you will be the A /images/graemlins/spade.gif is still 1/52 because you don't know what my cards are, so you're still randomly getting one card out of 52. Get it?

[WhiteWolf]

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yeah i get that part. but when we count outs we don't factor in the amount of total cards left in the deck. that's why i'm saying it should be 6 outs, giving us 25% equity, not 9 outs giving us 35% equity.

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The odds -> equity tables assume you are using the standard way of counting outs, ie treating all other player's cards as "unknowns".

well, i guess so.. still seems strange to me though.

yeah but you still have to assume there are no clubs it seems.. i mean i get what the other poster said but it still seems odd.

[Xhad]

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yeah but you still have to assume there are no clubs it seems.. i mean i get what the other poster said but it still seems odd.

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No you don't. The part you're not getting is that when you say "I have 9 outs on the river out of 46 cards," that is counting the cards in the other players' hands. If you were assuming no one else was dealt a club, or that none of the burn cards were clubs either, then you would have 9 outs out of 25 cards (assuming a 10-handed game).

Going back to my single-deck game example, if I deal the cards to myself face-down, then your probability of drawing the A /images/graemlins/spade.gif is still 1/52 from your point of view. If I show you the five cards and none is the A /images/graemlins/spade.gif, your odds have improved to 1/47. If I show you the A /images/graemlins/spade.gif from my pile, then obviously your odds of drawing it have decreased to zero.

ok, i think i get it now. just seemed counterintuitive initially. thanks.

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?