Real # of outs for flush draw

[zeke1111]

I just made an interesting discovery, which must be a rite of passage of sorts for novice poker players.

You don't have 9 outs on the turn and river for a flush draw!

Or more precisely, in a 10-player game, all 9 remaining suited cards will be available for your draw only 0.7% of the time.

It's very late at night, but are my Excel formulas telling me what I think they are telling me? Is it true, given your suited hole cards and the two on the flop, that four of the remaining suited cards will have been dealt among the other players 26.7% of the time [COMBIN(9,4) * COMBIN(38,14) / COMBIN(47,18)]?

And--my hands are shaking, I feel betrayed /images/graemlins/mad.gif--do I have it right that four or more suited cards will be held by the other players 76.1% of the time???

Can someone point me to a good analysis of this problem?

Steve

[parkerren]

If were going to calculate the average number of outs that could be distributed to the other players lets also look at how the cards are distributed through out the remaining deck. Now we have 3 cards making up the flop, your two (2) hole cards, the 18 cards given to your opponents and finally the burn card (so a total of 24 cards). So there are twenty-eight (28) cards left to play. We would have the top card being a burned card, followed by the turn card, the next card would be a burned card followed by the river card. This leaves 24 cards left in the deck that will never be played and we should calculate the number of out cards that will be among the 24 cards. As a result we would have even FEWER outs. We could also calculate how many of the out cards could be among the two burned cards. This would leave us with even fewer outs for the turn and river cards. I would suspect it would calculate down to LESS THAN ONE OUT.

Its not about where the cards are distributed, because we know most of the cards we need are not going to be in position to be the turn or river card. Its about calculating the odds that the turn or river card will be one of our outs.

Understand that ONLY ONE (1) out the forty-seven (47) cards will be the turn card. Each of the 47 unknown cards has an EQUAL chance of being that turn card. If we are in need of one out, we would have a 1/47 chance of getting the card (or 46:1). If we have two outs we would have a 2/47 chance of the turn card being one of the two cards (or 45:2). Three outs would be 3/47 (or 44:3).


In looking at calculating outs the book published by two plus two “Small Stakes Hold ‘Em” has two good chapters on the subject, ‘Counting Outs’ and ‘Finding Hidden Outs’. It discusses (in ‘Finding Hidden Outs’) observing the players play and guessing what hand your opponent has. Thinking through the possible hands and then calculating the number outs you would have for each hand followed by calculating a weighted average to determine your effective outs. Its a good book to have.
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