AceHighSteve

Hi I'm a tournament player from a community near Seattle Wa. I have a question, can anyone think of more "outs" you can have than: "TWENTY FIVE", and I can think of 36 ways, out of the roughly 2.8million hands you can get in Texas holdem to end up with 25 outs.

Can anyone think of anymore?

Here are the 36 ways of ending up with 25 outs on the river:

You Hold: Ks,Kd
challenger holding 8a9a(a=any suit w/no spade)
Flop: 10s,Js,Qs

Turn Card: Kc

Outs on the river, you have
1 - K = Quads
3 - Q = FH
3 - J = FH
3 - 10 = FH
9 - spades = Flush
3 - A = Str8 (or str8 flush)
3 - 9 = Str8 (or str8 flush)

25 - Total Outs
(These outs of course could be repeated with any pairs, K’s,Q’s,J’s,10’s,9’s,8’s,7’s,6’s and 5’s, with any suit, for a total of 36 hands.)

If anyone can think of a hand with more……well, I can’t!
Oh, and the odds you will hit the river….
The Straight is a slight favor...45.45% to 43.18%,
But depending on the size of the pot, this hand would be almost impossible to fold with the implied odds… (if you knew the other player had the low end straight!), the problem is, after the Turn card you HAVE to figure that the challenger could have the Ace of spades (or Any other Ace) which changes your percentage of winning the hand to just 25%, and the number of “outs” to 10, (if he holds the As), so calling a large bet or raise in that position would be VERY difficult!

Although, if the challenger does not have the Ace of spades, and has the low end straight, your card will come 69% of the time in a ten handed game to win the hand. I hope you got all that…Good Luck!

I have KsQs on a JsTsTh flop, and I'm facing 6h6d. Turn is the 8c.

I now have 4 aces, 4 nines, 7 remaining spades, 3 kings, 3 queens, 3 jacks, and 2 eights (the 8 of spades is already accounted for).

That gives me 26 outs.

You're welcome, if this helps you win a prop bet or some sort of trivia contest. :-D

Sorry, but if you had KQ you were NEVER BEHIND, against 66!! So you didn't NEED ANY outs! So you have NO outs if you're ahead!

THE 6h6d, has "outs" but only if the board pairs or he gets a 6 on the turn, and even then not many!

I still can't come up with MORE than 25 outs, can ANYONE....remember inorder to have "outs" you HAVE to be BEHIND in the hand with still a chance to win it!

Uhhh, pretty sure JMcCabe is right, 66 is a pair, while KQ is just high card... Sure, on that flop KQ has more equity, but if the hand were to conclude and go to showdown on the flop (just pretend those are the rules) then 66 would win, not KQ.

On the turn we gain 2 more outs, but I think at that point our equity is actually less than 50%, so we will be losing the hand the majority of the time, assuming we always go to showdown.

So the KQ has to hit one of its "outs" in order to win the hand at showdown. This will happen more than 50% of the time, and i would rather have KQ in that spot, but as is, 66 is the made hand.

YOU Have ZERO outs when your ahead!! LOL the 66 has the "outs"! But NOT MANY! If he gets a 6 on the turn that would give him TEN! If the Board pairs it gives him 2, the 6's!

TRY AGAIN! LOL, If your going to play professionally, you have to know what "OUTS" are, and you can ONLY have "outs" if your BEHIND in the hand! LOL

LOL, KQ FLOPPED the STR*!!! So he is AHEAD already!!! the 66 has the "outs"!! LOL You have to be BEHIND to have "OUTS!"

Read more closely... the flop is JsTsTc, so KsQs has four cards to the straight, KQJT. He needs an ace or a 9 to complete his straight.

correct me if I'm wrong, but the 3 A's and 3 nines are really only 2 each because the spades are already counted in the 9 spades. Only gives 24 outs and not 26.

I'll keep my eyes peeled to see if I'm right or not

Thanks and gl

It seems that JMcCabe already accounted for that, there would be 9 total spades, which includes the ace and 9 of spades. Then there are 3 more aces, 3 more spades, 3 more kings, 3 more queens, which brings us to 21. We can also pair the jack on the board (another 3), then after the turn we can pair the eight (another two, since we already counted the eight of spades). That makes 26.

Ahh got ya. That's why I asked. LOL Thanks

OOPS, just noticed to remember the other player has one of the 9's, do we count that one?

Well villain in the example has pocket 6's, but if he did have 9's then no, we wouldn't count it. Same thing if he held a spade. Good question.

I don't get it, I thought the challenger had 89, no spade, with flopped st to put us behind with KK.

Sorry to harp, just trying to understand the scenerio.

Thanks again, Yota

oh, sorry. I was looking at JMcCabe's scenario. Didn't mean to confuse you.

Wow, JMcCabe is right, cjatud2012 is right for backing him, and reinforcing the point.

I apologized to McCabe, he was spot on correct, I read it WRONG!
MY BAD!
I must be getting OLD! Can't see anymore!
Sorry guys, but I screwed the pooch!

McCabes scenerio is correct, also there are 168 ways of making 26 outs!...(I think)... you can do it of course with any suit, and you can switch the turn pairs, which also doubles the number. Then you can also do it with Q-J,J-10,10-9,9-8,and 8-7, for a total of 168 ways of getting 26 outs. Actually more than the 36 ways of getting 25 outs. I can't think of a way to get 27...If you can, let me know....once again I'm sorry to McCabe, I should have read it more thoroughly...Kind of like the legislators that voted on the 2300 page Health Care Bill without reading it, but that's another subject! LOL

No worries.

This is a pretty common question that I thought about quite a bit when I started playing Hold'em (a long time ago).

As far as I'm aware, there is no way to get 27 outs. The scenario I outlined is the most you can have (24 on the flop, 26 on the turn).

This can occur with a number of suited connectors or gappers when you flop on open-ended straight flush draw on a paired board vs. an underpair to the board and your cards.

For example, if you had 8sJs on a flop of 9sTsTd against 2h2d, and the turn was the 3c, you would again have the 8 straight outs, 7 remaining flush outs, 3 jacks, 3 nines, 3 eights, or 2 treys.

Glad to help.

BTW, thanks for backing me up Cjatud2012.

Really enjoyed railing you in that Rush tourney.

waoh never thought of poker that way. thanks for the info though

WoW, I've been working on this almost all day now, just because it got me going and I wanted to know how many different ways that you could end up with 26 outs, and I believe, I was right, it is 168 different ways of ending up with 26 outs with suited connectors...such as (K,Q)(Q,J)(J,10)(10,9)(9,8)or(8,7)(below or above these connectors wont work, such as (A,K), or (5,6)). BUT..., I didn't take into account "gappers, as you discussed, and WOW, is all I can say. That changes the total A LOT!

And the example you give using Js,8s, also is a good example, however, there can be a total of 80 different ways of getting 26 outs with Js8s!!

I have figured out that a suited (5,8), will give you 8 different ways of getting 26 outs, and (6,9), will give you 24. With (7,10), you can get 26 out's 48 different ways, and (9,Q), equals 120, "26 outers", and you end up with the most ways of getting 26 outs with a suited (K,10)...168 ways! If you add all those up, with suited gappers you could expect to have 26 outs, 448 different ways.

If you add that to the 168 ways of getting 26 outs with suited connectors, you end up with 616 different ways of ending up with "26 outs on the river"!

(I think that would be a good title to my book....

"26 outs"

(If I can't get rich playing, I'll have to write a BOOK!) LOL
TY JMcCabe, (I'll put you in it for some credit...LOL)
Steve

hahah, no problem, and thanks for the rail! It felt really good to hit for that much, I think I'm gonna play it again tonight. No moving up in stakes for me

No worries Steve.

You've still forgotten other hands though, such as 1 gappers (Js9s on a Ts8s8d flop versus 3d3c for example), or other gappers that can flop double belly busters (Jh5h on a 7h8d8c9h board versus 4d4s).

While the possibilities aren't exactly endless, there are far more than it's worth thinking about. :-D

Holy Crap, Screwing around with cards and numbers, I just came up with a way to have "27 OUTS" on the RIVER !!!! Here it is:

You Have: (9s,3s)

Challenger has: (2h,2d)


FLOP: 5s,6d,6h


Turn: 7s

At this point for the river you have 27 OUTS!

4-4's
4-8's
7-spades
3-3's
3-5's
3-7's
3-9's

= 27 outs

Nope, you don't win with the 6 of spades, because it gives your opponent a full house, so there are only 6 spade outs (A, K, Q, J, T, [8], [4], 2). [ ] counted in your straight outs.

Actually, you can't win with the 2s either, because it gives your opponent a full house, so you're back down to 25 outs.

Uh-oh.....does that screw up all the other ones as well?

I think it does....sigh.

Dang you're right, gotta go back to the drawing board! We may be back down to 25 outs as the highest # of outs that you can get.

But there is a heck of a LOT more ways of getting 25 outs than the 36 I had already stated!

This all started because I was curious as to what was the "most" amount of outs on the river one could have and still be in the hand and what were the odds that card would come on the river.

Being that there are roughly 2.8 MIllion hands possible in Texas Holdem, I thought that information may be useful, but now it has become my OBSESSION! LOL