A question of outs

IrishDave

IrishDave

A Member
You're on the button at a full 10 seat table with Ah-Kh. When you're up, you post the appropriate raise and all but one person folds. The flop is 8h-6h-3c. You need the flush to assure (reasonably) victory so the question is - how many outs do you have.

Here's my answer - I have no idea. 23 cards were dealt to the table to the flop, you've seen 5 (your 2 and the flop). Of the 18 cards dealt that you haven't seen some or all of them may be hearts. The number that I hear on TV in this situation is 9 which is the maximum, assuming no hearts were dealt to the other players.

I grew up playing poker long before tables, charts, and software to track the hands so my answer and assumption is based on very simple reasoning. Be glad to hear what other folks have to say...
 
soadwes

soadwes

Guest
9 outs is the correct number... i think roughly 20% chance of hitting the flush?
 
robwhufc

robwhufc

Cardschat Elite
Yeah, 9. You know what 5 of the cards are (the 2 you have and the 3 showing). You know there are 13 hearts and you can see 4 of them which leaves 9 remaining. Of course those 9 could have already been dealt out to opponents so in reality you may not have any outs, but you cant know that.
 
IrishDave

IrishDave

A Member
So when we're talking about outs, the number quoted is the maximum? Not as scientific as some folks tend to make it is it? This was just something I was thinking about as while playing with some "scientists" yesterday they kept referring to how many outs they had...
 
C

chicubs1616

Guest
You have 9 outs for this hand, the fact that other players may hold some of your outs is irrelevant.

When computing pot odds you do not factor in your opponents cards, you only factor in those that are visible.

So after a flop with two diamonds, with you holding 2 in your hand, there ar 9 other diamonds in the deck (unseen cards).

Your chance of hitting the flush is # of outs / # of unseen cards or in this case:

9 / (52 - 2 in your hand - 3 on board = 47)

So you have 19.15% chance of hitting your flush, or just under 5:1 odds.

Do not worry about some of your outs being in other peoples hands.

As to your example Dave, I would argue that an A or K would also win you the hand, also AK might be ahead with a flop like that (you are only fearing a pocket pair here).

I would add an additonal 6 outs here for you to win, and maybe even think my ace high is best here.

So in actuality if I was in this hand I would give myself 15 outs so I would have a 15/47 chance of hitting with the possibility of having the best hand at the moment (If your opponent has AQ, KQ, AJ, etc.).

So instead of a 19% chance of hitting, you would probably have around a 32% chance of hitting.
 
IrishDave

IrishDave

A Member
I made this an artificial hand as I know that the AK itself is capable of winning the pot. It interests me that you say the number of outs already dealt is irrelevant, true to a point as you don't know/can't control. Just thought this would make a good topic and get some interesting discussion going - and it has...
 
S

StackThemUp

Guest
chicubs1616 said:
You have 9 outs for this hand, the fact that other players may hold some of your outs is irrelevant.

When computing pot odds you do not factor in your opponents cards, you only factor in those that are visible.

So after a flop with two diamonds, with you holding 2 in your hand, there ar 9 other diamonds in the deck (unseen cards).

Your chance of hitting the flush is # of outs / # of unseen cards or in this case:

9 / (52 - 2 in your hand - 3 on board = 47)

So you have 19.15% chance of hitting your flush, or just under 5:1 odds.

Do not worry about some of your outs being in other peoples hands.

As to your example Dave, I would argue that an A or K would also win you the hand, also AK might be ahead with a flop like that (you are only fearing a pocket pair here).

I would add an additonal 6 outs here for you to win, and maybe even think my ace high is best here.

So in actuality if I was in this hand I would give myself 15 outs so I would have a 15/47 chance of hitting with the possibility of having the best hand at the moment (If your opponent has AQ, KQ, AJ, etc.).

So instead of a 19% chance of hitting, you would probably have around a 32% chance of hitting.
You've missed calculated the % of hitting a flush. If you have 9 outs remaining and still the turn and river to come you are 35% to hit and roughly 1.9/1 to hit.

This is worked out as following: (looks complicated but isn't)
(47-number of outs) /47 x (46-outs) / 46

So in this case you do: 38/47 x 37/46
Then minus the total from 1 to give your % which is near enough 35%

You worked out the % and odds for a single turn of a card not both.

Hope this helps
John
 
IrishDave

IrishDave

A Member
Hopefully our newbies will drop in to read threads like this one as there's some good info here. Since the word free isn't in the title I don't really think so but I'm a pessimist anyway...
 
Crippler450

Crippler450

Guest
IrishDave said:
So when we're talking about outs, the number quoted is the maximum? Not as scientific as some folks tend to make it is it? This was just something I was thinking about as while playing with some "scientists" yesterday they kept referring to how many outs they had...
There is no way you can even try to guess how many cards dealt have been hearts. It would be stupid to try and figure out how many hearts there are left in the deck, so you need to calculate odds based only on cards you have seen. If there are 35 cards left in the deck, you dont base your % on the 35, you use the 47 that you have not seen. If you wonder why you use this 'maximum' number, its because there is no way to see if the other players have been dealt 8 hearts or no hearts. Unless you are going to make everyone show their hand, there is no way to know. You calculate the % based on how many hearts there would be left on average, not how many are ACTUALLY left in the deck. If you have an inside straight draw but all of the 9s have been dealt to other players, you can only assume that those 9's are still possible to show up, unless you know otherwise somehow.
 
soadwes

soadwes

Guest
I'm still new to the math that goes in calculating these percentages... Can someone check my math on this? If you get 10sJs suited and the flop shows 8s 9s 3c, what is your chance of hitting a flush or a straight? (I didn't calc. chance of hitting straight flush...)

(47-number of outs) /47 x (46-outs) / 46
(47-17) / 47 x (46-17) / 46 = 59.7% of hitting straight or flush?

Please correct me if i'm wrong! I got this hand the other day and went all in on the flop... Didn't know the percentage then, but seemed like i had a bajillian outs. I hit a spade on the river winning the hand...
 
twizzybop

twizzybop

Legend
The 4 and 2 rule is even simplier.. after the flop calculate your # outs X 4 and after the turn calculate your # outs x 2.

In your scenario you have the open ended straight flush draw, giving you 15 outs.
15 X 4 is roughly 60% on the turn card while 30% on the river.

Now the straight flush is allready accounted for but it does make for the absolute NUTS if you hit it. So 2 cards to come for that would be 2 X 4 is 8% and 2 X 2 for 4%.
 
Dennis C

Dennis C

Guest
When your talking abuot outs I think were all screwed. Fold that's the best pot odds because we live to see another hand. Now, We all know this is a crock of sheet so to think about what they say is th max # of pos. and what we all think with 20some cards out. I would give it a 10% shot if the bet was high and 25% if the bet is low or checked. And that's my two cents. LOL:icon_joke
 
~~Shelynn~~

~~Shelynn~~

Legend
OK IrishDave help me out on this,how would you bet it? Then what if the other player goes all in,what do you do,it feels good to you but you've got this? in your mind. this has happened to me several times.and mainly get burnt. Any advice from you poker players. Thanks
 
bigjace

bigjace

Guest
IrishDave said:
Hopefully our newbies will drop in to read threads like this one as there's some good info here. Since the word free isn't in the title I don't really think so but I'm a pessimist anyway...
Most of our newbies can't count past 15 Dave!!The maths side of the game is very handy to know but sometimes gut instinct is the greatest gift we have.:icon_thum
 
HoldemChamp

HoldemChamp

Rock Star
Cubs,

You forgot the possibilty of the other guy having KK as well.

Ok. Let's break this down for the turn alone to make it easy and worry about only the odds of you getting the flush.

9 (cards to make your flush) DIVIDE that by 47 (the number of unknown cards in the deck EQUALS 19.148%

Now take 47 and DIVIDE that by 9 to come up with the odds. 5.222 to 1.

OK. Let's take it to the last and most crucial step.

You figured the odds now what?

Call this a heads up match to make it easy. Make the pot 1000 chips. Also to make things easy.

OK. So here is the question. How much of a bet can you call and still have proper odds to play your flush draw?

I will only give you a hint. If the guy makes a pot sized bet you fold.

Ok. Get those pencils ready. What is the answer. Remember I am not going to tell you. You have to come up with it on your own.
 
V

venturer49

Rising Star
Don't know how accurate you want the answer, HoldemChamp, but I'd say I needed 6:1 at least to call the bet, which (I think I've got this right) means calling a bet of not more than 200. I'm pretty new to this, but am I close?
 
HoldemChamp

HoldemChamp

Rock Star
Wow,

Only one answer so far. Come on all you Sklanksy followers. Answer this question. It's an easy question. Help out the newbies with the odds calculations. Remember I am not going to answer this question. So, it is up to the odds pros to answer this question and help out those new to calculating odds start to improve their game.
 
S

StackThemUp

Guest
Hi,

Holdem you've worked out the odds for hitting the flush on the turn of one card not 2, the odds of hitting your flush by the river with 2 cards to come is roughly 1.95/1 so you need 2/1 or higher to make it worth you while to call. However implied odds changes things a lot as rarely you'll be able to see the river card for free,but implied odds are for another time. I explained in my other post how to work out the odds for 2 cards to come,take a look at it..



John
 
V

venturer49

Rising Star
StackThemUp said:
Hi,

Holdem you've worked out the odds for hitting the flush on the turn of one card not 2, the odds of hitting your flush by the river with 2 cards to come is roughly 1.95/1 so you need 2/1 or higher to make it worth you while to call. However implied odds changes things a lot as rarely you'll be able to see the river card for free,but implied odds are for another time. I explained in my other post how to work out the odds for 2 cards to come,take a look at it..



John
My calculation was based strictly on Holdem's stipulation that we were ignoring the river odds, and for the same reason I left out any implied odds, since they belong to a calculation where there is a future betting round, which Holdem also excluded for simplicity's sake. Basically, the way I see it, if the card odds are slightly worse than 5:1, the pot odds need to be 6:1 or better to make the bet worthwhile, though I don't know how much better Sklansky et. al. recommend. If the pot is 1000 and your oppo bets 200, that makes it 6:1 for you to call, hence my 200 estimate. I don't know if I've missed something - I've just assumed all the complications, like implied odds, counterfeit outs, etc. are out of the picture in this case. Looking forward to finding out the right answer...:)
 
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