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Festus22
06-30-2003, 12:12 PM
What are the probabilities of actually making a straight, flush or straight flush with all 5 board cards revealed with starting hands:

Suited connector (8-7s)
Suited one gapper (8-6s)
Suited two gapper (9-6s)
Suited 3 gapper (10-6s)

I know these are can be played profitably in loose, passive games but how many callers should there be to justify playing solely from a pot odds perspective.

Copernicus
06-30-2003, 02:18 PM
The probabilities of straight or better (I assume you wont object to quads or a full house) are 17.66/16.20/15.95/13.94 for the hands you list.

The number of callers to make it profitable varies based not only on loose vs passive, but blind structure, NL vs Limit etc.

The flush equity in having them sooted is less than most think, as well. the 13.94-17.66 range goes down to 10.45-14.40. I prefer them unsuited because its much harder to get away from the flush than the straight, but Ive seen those low flushes cracked by Ax and Kx limpers too often. Those hands are dead when its the straight, not the flop that comes.

Slowplay
07-06-2003, 04:06 PM
Here's a curve ball question for the math guys (seeing as I have been crucified by a higher flush at least a dozen times in the past month):

What are the odds that another player has a suited pocket the same as yours (e.g. you hold 67d, they hold J9d)AND two hit on the flop?

Also, can anyone calculate a staggered set of odds that address a higher flush draw than 67s? (e.g. A2, T4, KJ, etc.)

Thanks if you can help at all.

SP

asdf1234
07-07-2003, 12:50 AM
These are only approximations using random hands, which is not necessarily the case in real life.

If you have something like 7 /forums/images/icons/spade.gif 6 /forums/images/icons/spade.gif and the board is
K /forums/images/icons/spade.gif J /forums/images/icons/spade.gif 4 /forums/images/icons/diamond.gif
then the chances of a single opponent holding a flush draw as well is 3.33%. If you extend this to 9 opponents, then it increases to 26.27%. It decreases 2-3% for each opponent you lose. Of course, the likelihood is probably less, because the king and jack are both out there, and people are less likely to play trashy suited cards, but we're not taking that into account anyway.

As far as the number of flush draws that will beat you, I'm sure you know, that if there is one out there, it's probably beating you. There are 36 other combinations of hands that can be drawing for a flush and 30 of them beat you. The chance that there is another flush draw out there and that it beats you is 2.77% for a single opponent up to about 23.4% for nine. Again, this decreases 2-3% for each opponent you lose.

These calculations are not exact because they don't take into account situations where multiple opponents hold flush draws, some players have one of the suit, but this is relatively close.

Slowplay
07-07-2003, 12:24 PM
Hey ASD--Thanks a million for replying! I have been attempting to figure this one out for a while now, and it seems that most probability charts rarely list practical odds of this nature (I mean, sh**t, we're faced with this situation a helluva lot more often than flopping quads!).

So, basically, when your dealt suit-cons in a regular ring game and you flop a four-flush there's roughly a 30% chance that someone else has the same draw (assuming everyone calls). So it's 3-to-1 that you're the only one on the draw and these odds increase in your favor with less callers?

Great to know. Can I call upon your fountain of mathematics in the future? :O)

SP

Festus22
07-07-2003, 03:44 PM
Just another angle on this and these are approximations:

3-Flush Board
Kx beat by Ax 6.2% of the time
Qx beat by Ax or Kx 11.5% of the time
Jx beat by Ax, Kx or Qx 15.9% of the time

Basically, the formula amounts to subtracting 0.9% from each previous result and adding that number to the total. So for 8x to lose = 6.2 + 5.3 + 4.4 + 3.5 + 2.6 = 22%.

Since there's already 5 of the suit out (your 2 + the board's 3), there's only 2 more possibilities to go meaning that 3-2s wins 74.5% of the time on a 3-flush board.

DONKEYJAWS
07-07-2003, 06:36 PM
How does one calculate the odds of hitting a hand with 5 cards to be drawn? Do you add the total outs over the total blanks? I have absolutley no idea.

asdf1234
07-08-2003, 02:11 AM
It's somewhat brutal, but you essentially have to calculate the number of possible boards that give you the hand out of the total number of possible boards.

For example, if you hold K /forums/images/icons/spade.gif K /forums/images/icons/diamond.gif and want to calculate the probability of hitting quads by the river, you need to find out exactly how many boards give you quads.

Obviously, to have quads, the other two kings out need to show up, so these are "locked in", and there are 48 remaining cards to show up on the board. Now just figure out how many ways out of 48 3 cards can show up C(48, 3) = 17296.

So there are 17296 boards that give you quads. Now just calculate the total possible boards C(50,5) = 2118760.

Dividing 17296/2118760 = .00816
So you get quads about .8% of the time with a pocket pair. (This is 121.5:1 against when you hold a pocket pair).

You can calculate C(n, r) with the formula

C(n,r) = (n!)/(r! * (n-r)!)
Most calculators that have factorial will probably have the combination function as well (usually called nCr).

Going back to the flush stuff, the numbers that you should apply to an actual game situation are probably a few points less than what I calculated. These are theoretical and assume that nobody ever folds (well, this isn't a far cry if you play on party). Most reasonable players will fold 82s from early position while you might be able to limp in from late position with 76s. It depends on the number of opponents you have and their caliber.

DONKEYJAWS
07-08-2003, 03:40 PM