Suited Domination

[hurlyburly]

Can anyone post the math used to determine the average likelihood that a non-nut flush is in danger of being dominated?

EX: 10-person table. I'm BB with Q5 Spades, 3 people see the flop, which comes Ac Xs Xs. I get 3.11 odds to see the turn after checking and Xs comes. No Ace or King of spades on board. What are my odds of being dominated in this situation?

[hurlyburly]

Ok, I think I figured it out:

10 ppl = 20 cards; 20/52= ~38% of the deck
5/20 = 25%; # cards for each suit for a 25% chance of catching any particular suit
6.25% = # of times you get 2 cards of the same suit.

This is where I get lost. Now there are 3/18 cards left in the suit you have, giving a second person a 2.77% chance of having a hand of the same suit. Now since we don't take 2 off the top, how do the numbers work together to determine the chance of both people getting suited hands of the same suit? Averaging it? Average is 4.51% so I guess it's safe to assume that ~1/20 times I get a suited hands, someone else has a same suited hand? Then since I'm queen high, I fear 2 over cards, so is it (2/13)*(1/20) for a .7% chance that I'm dominated?

That seems both overly simple and wrong, so I'd sure like some help here. If that's the case, then i'm not playing my weak flushes strong enough.

[binions]

I have always heard that for every 2nd nut flush, there is a 6% chance the nut flush is out. For the 3rd nut flush, there is a 12% chance the nut or 2nd nut is out. And so on, at 6% intervals, down the line.

This is of course using 2 cards in your hand and 3 on the board. When a miracle flush is out, all bets are off.

Of course, I have never bothered to see if this is accurate, and frankly wouldn't know where to begin.

[Siegmund]

17 of the 1081 possible two-card hands out there are Axs or Kxs in spades.
The chance of one opponent having your flush draw dominated is,then, 17/1081; the chance of one of nine opponents having been dealt this hand is roughly 1-(1064/1081)^9 = 13.3%.

It's a safe bet that the suited ace would surely have seen the flop. The suited king might not have -- Kxs might play from the button but anything worse than KTs is folding in early seat, unless you have loose opponents. Against sound opponents your chance your flush is dominated is probably in the 10-11% range.

[binions]

[ QUOTE ]
17 of the 1081 possible two-card hands out there are Axs or Kxs in spades.
The chance of one opponent having your flush draw dominated is,then, 17/1081; the chance of one of nine opponents having been dealt this hand is roughly 1-(1064/1081)^9 = 13.3%.

It's a safe bet that the suited ace would surely have seen the flop. The suited king might not have -- Kxs might play from the button but anything worse than KTs is folding in early seat, unless you have loose opponents. Against sound opponents your chance your flush is dominated is probably in the 10-11% range.

[/ QUOTE ]

Run that by me again?

There are 1326 possible hands. Assume you have KQs spades. Now there are 1225 possible hands - 10 of which can be Axs in spades. How did u get 1081??

Assuming you have KQs, what are the chances that one of the 9 other players will have Axs in your suit? And what if one of them has a spade, or even two lower spades? Seems like you will have to make a lot of assumptions along the way.

And if he has Axs, what are the chances of you both making a spade flush with 4 spades taken out of the deck?

If we assume everyone has folded, and you have KQ spades in the SB, then the BB has 10/1225 chance of having Axs. If he does, then there is 48 cards in the deck, and 9 spades left.

What are the chances of making a flush by the river in this case?

[elitegimp]

[ QUOTE ]
[ QUOTE ]
17 of the 1081 possible two-card hands out there are Axs or Kxs in spades.
The chance of one opponent having your flush draw dominated is,then, 17/1081; the chance of one of nine opponents having been dealt this hand is roughly 1-(1064/1081)^9 = 13.3%.

It's a safe bet that the suited ace would surely have seen the flop. The suited king might not have -- Kxs might play from the button but anything worse than KTs is folding in early seat, unless you have loose opponents. Against sound opponents your chance your flush is dominated is probably in the 10-11% range.

[/ QUOTE ]

Run that by me again?

There are 1326 possible hands. Assume you have KQs spades. Now there are 1225 possible hands - 10 of which can be Axs in spades. How did u get 1081??

[/ QUOTE ]

After the flop, you have seen 5 cards (the 2 in your hand and the 3 on the board). So there are 47 cards left to deal to your opponent, or C(47,2) = 47*46/2=1081 possible hands for him to have. Personally, I would count after hitting the flush on the turn, so C(46,2) = 46*45/2 = 1035 possible hands. Since you can account for 5 of the spades, there are 7 Axs hands the villian can have.

[Siegmund]

I thought the original poster was asking about whether he was drawing dead on the flop -- rereading the post, yes, he was asking about his chances after the turn.

You can redo the previous calculations with 8 spades and 38 nonspades / 15 of 1035 hands can beat you, and come up with 12.4% instead of 13.3%, maybe 9-10 instead of 10-11 after making adjustments for hands that would fold.

But the third spade on the turn has hurt you, in that an unsuited ace or king now has a chance of outdrawing you. Hard to quantify this one: as a shot in the dark I'd say your chance of winning is reduced by another 3% or so - figure between 10 and 20% that either the SA or SK was dealt to a playable unsuited hand, and between 15 and 20% that a fourth spade will hit on the river.

[Cobra]

I created a excel spread sheet that calculates the probability of loosing a flush. It assumes that only three of your suit come on the board. The information was taken from an article written by Brian Alspach called "loosing flushes". He has a lot of good probability articles search for Alspach + poker and you will get his web site.

The Probability that anyone at your tabel was dealt the A or K in a ten person game is 11.6%. The mathematics used for this include determining how many of the suit may be in each persons hands and then partitioning them into hands.

Cobra

[binions]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
17 of the 1081 possible two-card hands out there are Axs or Kxs in spades.
The chance of one opponent having your flush draw dominated is,then, 17/1081; the chance of one of nine opponents having been dealt this hand is roughly 1-(1064/1081)^9 = 13.3%.

It's a safe bet that the suited ace would surely have seen the flop. The suited king might not have -- Kxs might play from the button but anything worse than KTs is folding in early seat, unless you have loose opponents. Against sound opponents your chance your flush is dominated is probably in the 10-11% range.

[/ QUOTE ]

Run that by me again?

There are 1326 possible hands. Assume you have KQs spades. Now there are 1225 possible hands - 10 of which can be Axs in spades. How did u get 1081??

[/ QUOTE ]

After the flop, you have seen 5 cards (the 2 in your hand and the 3 on the board). So there are 47 cards left to deal to your opponent, or C(47,2) = 47*46/2=1081 possible hands for him to have. Personally, I would count after hitting the flush on the turn, so C(46,2) = 46*45/2 = 1035 possible hands. Since you can account for 5 of the spades, there are 7 Axs hands the villian can have.

[/ QUOTE ]

OK. Assume you have KQ spades, and you flop a 2nd nut 4 flush.

1081 starting hands left, 8 of which can be Ax spades. Assume 9 other players at the table

1 - (1073/1081)^9 = 6.47% chance the Ax spades is out.

So, if you make your flush, there is a 6.47% chance you lose to a higher flush.

This corresponds to the 6% rule that I had always heard.

Now, assume you have QJ spades, and flop a 3rd nut 4 flush with 9 foes. There are 1081 possible hands, and 15 of them can be Kx or Ax spades (we don't count AKs twice).

1 - (1066/1081)^9 = 11.82% chance that a higher flush draw is out, and if you make a flush, you will lose 11.82% chance.

Again, this confirms the 6% rule.

Note, if you assume only 7 Axs hands in the KQs example because if a flush comes, that's one less possible Axs hand, then the chance of drawing dead is actually less than 6.47%.

It's also less than 6.47% if there are less than 9 foes.

[Siegmund]

Now that we've settled down on roughly how often a non-nut flush is going to beat it may be worth taking a minute to think about how that effects your play of it.

A nut flush draw wins a large pot when it makes and (usually) folds when it doesn't. A second-nut flush is winning 94% and third-nut maybe 90% (some of the Kxs will fold), but when it loses it is probably losing anything from two to four big bets on the late streets.

It would be a little harsh to treat these as being only 88 and 80% as good as the nut draw (.94-.06 and .90-.10), but they are certainly a lot less than 94 and 90% as good.

Looks like a good rule of thumb is to treat a second nut flush draw as 8 outs and a third nut flush draw as 7 outs. Smaller draws are probably still worth around 7 outs unless you have a game where everyone plays all sorts of suited hands, in which case a a baby flush draw might only count as 6.

Anyone have better suggestions for how to treat this in the betting?

[binions]

Big difference between limit and pot/no limit. At limit, when Kxs makes a flush using both its cards, it will win whole pots 94% of the time, and will lose a couple of extra bets 6% of the time. Hardly enough to discount the outs.

But pot/no limit, one of the ways you double up is flush over flush. I am very reluctant to play any suited card that's not also connected less than Kxs. The big problem with suited connectors, however, is making an underflush. So you have to be extremely careful.