40-80 at the Commerce: J8s

I open-raise J [img]/images/graemlins/spade.gif[/img] 8 [img]/images/graemlins/spade.gif[/img] in the CO.
Button folds, SB calls, BB calls.

SB is very loose pre-flop, the loosest player at the table. He is passive post-flop, and in general very weak. However, he is capable of folding when he suspects he's beat and actually thinks logically; he's just particularly loose pre-flop.

BB is a TAG. In my last encounter with him I called his turn and river donks down and won with Ace high, and I got the feeling he wanted to get me back.

I have a very strong image at the table. I had won almost every hand I played over the previous couple of hours. No one had seen me make any moves.

POT: 6 SB

FLOP: Q [img]/images/graemlins/heart.gif[/img]9 [img]/images/graemlins/diamond.gif[/img]7 [img]/images/graemlins/spade.gif[/img]

SB checks, BB checks, I bet. SB calls, BB calls.

Pot: 9 SB

Turn: A [img]/images/graemlins/spade.gif[/img]

SB checks, BB checks, I......??

[Lmn55d]

Let's try to look at this mathematically.

Let's assume that if you bet, you will get 1 caller on average. I think this is reasonable because this board is pretty draw heavy. The TAG is also capable of folding a 7x or 9x hand I'd assume.

With 12 outs you are a 2.9:1 dog so you have about 25% equity. If you bet and are called you lose .75BB.

How often must you take the pot down to make betting better than checking?

I know you're good at math Vahe so maybe you can help me fix this if it's wrong.

Assumptions: If called, we get 1 caller on average. We do not get checkraised (this might be a big assumption).

When Betting is Neutral EV:

(x)(9 SB) = (1-x)(.25)(11 SB) + (1-x)(.75)(-2 SB)

x = percentage of the time our bet takes down the pot
1-x = percentage of time it doesn't
.25 = percentage of time we improve
.75 = percentage of time we don't improve


10 sb= pot before turn bet + 1 that our opponent calls

Solving for X:


(x)(9) = (1-x)(.25)(10) + (1-x)(.75)(-2 SB)

9x = 2.5 - 2.5X + -.1.5 + 1.5X

9X + 2.5X - 1.5 X = 2.5 -1.5

10x = 1

x = 1/10 = 10%

Ok so I'm not exactly sure if this is what we are after. THis tells us that if our opponents all fold 10% of the time, betting has a neutral EV. Now we need to compare it to checking.

[Lmn55d]

Ok so after talking with VKH I realize I need to compare the EV of betting to the EV of checking. Let's say for simplicity that my implied odds are the same if I bet or check (this may not be true, but I don't think they are drastically different).

I will set the EV of betting equal to the EV of checking and see how the result changes if I expect to be checkraised 20% of the time.

Never checkraised on turn:

EV of betting : 9x + (1-x)(.25)(11 sb) + (1-x)(.75)(-2sb)

EV of checking : (9 sb)(.25)

Setting them equal:

9x + 2.75 -2.75x -1.5 + 1.5x = 2.25

7.75x = 1
x = .129

A very interesting result! If I was making a pure bluff on the turn it would need to work 10% of the time because I'd be getting 9:1. Since I have a nonmonster draw (25%) equity, a bet needs to work more often! That seemed counterintuitive to me at first but now I understand it. I'll post a follow up in a bit that takes into account getting checkraised 20% of the time.

Hi Dave,

You made an interesting observation that given some specific assumptions, the value of betting actually decreases as you gain more outs.

A couple of things:

In the 2nd equation you should have

EV of checking : (11 sb)(.25)

since you assumed the implied odds to be equal in the cases of betting and checking.

This would lead to 7.75x = 1.5
x = .194 = 19.4%

Also for the case where you have 0% chance of winning the pot when called (i.e. a pure bluff), you need at least a 2/11 = 18.2% chance of your opponents folding (pot is laying you 9:2) to making betting better than checking (in this case greater than 0 EV).

We can generalize what you did. Instead of setting our chances of improving to .25, we can just call it y. I also think that, for the case when you bet and are called, we should increase your implied odds to 3 SB instead of 2 SB since 2 SB would imply that you are NEVER called on the river.

Then, we have:

EV of betting : 9x + (1-x)(y)(12 sb) + (1-x)(1-y)(-2 sb), 0<=x<=1, 0<=y<=1.
EV of checking : (11 sb)(y), 0<=y<=1.

Setting them equal:

9x + 12y - 12xy + (1 - x - y + xy)(-2) = 11y

or

11x + 3y - 14xy = 2

=> x = (2 - 3y) / (11 - 14y) 0 <= y <= 2/3

This gives us the required turn folding equity of your opponents such that betting is the same EV as checking, as a function of our chances of improving to the best hand.

We can differentiate this function with respect to y, and see if it has a maximum or minimum:

dx/dy = -5/[196(y-11/14)^2], 0 <= y <= 2/3.

So, it is a decreasing function on the interval [0,2/3]; it attains its maximum at the endpoint y=0 and minimum at the endpoint y=2/3.

So, the better your draw is, the less the chance you'll need of your opponents folding on the turn for betting to be correct. This is opposite the conclusion reached when you assume that the implied odds of betting and checking are exactly the same. Bottom line: math can be dangerous with simplified assumptions since small changes can drastically change results.

[Subfallen]

I would check this.

[ QUOTE ]
I would check this.

[/ QUOTE ]

[Lmn55d]

Wow that's really interesting. Thanks for doing that work.

One question:



[ QUOTE ]
In the 2nd equation you should have

EV of checking : (11 sb)(.25)

since you assumed the implied odds to be equal in the cases of betting and checking.

[/ QUOTE ]

Why does the EV of checking equation contain 11sb instead of 9sb if we assume that our implied odds on the river are 0? When I check there are 9sb in the pot, right?

You are assuming that you get 1 more bet on the river when you check the turn. Otherwise the implied odds between checking and betting would not be the same.

[sthief09]

i would bet this unless you have a trashy image. a lot of live players are weaktight and will assume you have the A. you have a lot of outs so its not a problem.