What is the most common expected value formula for this type of situation?

• Bet/Raise on Flop
• Bet on Flop

Example:
#1 Villain bets and Hero raises (villains option)
#2 Villain checks and Hero bets (villains option)

I'm looking for a mathematical formula

#2 is the most commonly used but both are used. You just have to use the one that fits the situation the best.

How can I calculate the EV for situation #2?

I'm not exactly sure what you mean. Are you facing a raise on the flop, or deciding if you want to raise a bet on the flop?

Generally, EV calculations are difficult when the action is still open on future rounds. Its more easily calculated with action that closes all future action, such as all-ins on any street or calling a river bet. EV calculations are the sum of all possible actions, taking into account the likelihood of each action.

For example, say you are on the SB with 10BBs and action has been folded to you. You are deciding if you want to shove all-in, so you formulate an EV equation. (We will assume blinds of 50/100, no antes, so your stack is 1000, and BB has you outstacked).

If you shove, there are three possible scenarios.

1) BB folds
2) BB calls, and wins
3) BB calls, and loses.

You must assign a likelihood to each of these scenarios (and this is where it gets subjective). In our example, assume our read is that BB will call with the top 20% of hands. Our hand is 87s, and we calculate (using pokerstove) that our hand has 37% equity (37% chance of winning) against the top 20%. So here's each piece of the equation.

1) BB folds = (.80)(150). The .80 represents the chance of occurring, and he must be folding 80% if he's calling 20%. The 150 is how many chips you'll win when he folds (the blinds).
2) BB calls, and wins = (.20)(.63)(-1000). The (.20) represents the chance he'll call, and the (.63) is the chance he'll win when he calls. Multiplying them together gives us the chance he will call AND win. The -1000 is the chip outcome for you, because you'll lose your stack.
3) BB calls, and loses = (.20)(.37)(1150). Same as above for (.20) and (.37). The 1150 is how much you will gain when he calls and you win.

So you add them all together to find the total EV:

(.80)(150) + (.20)(.63)(-1000) + (.20)(.37)(1150) =

120 - 126 + 85 = 79

So you will win 79 chips on average, each time you make this shove. If the answer is negative (-), then you will lose money on average with that play.

Thats what I'm looking for...

Total EV=
(%oppfolds)(Pot)+(%oppcalls)(%win)(Pot+Bet)-(%opp calls)(%lose)(Bet)

=FP+CW(P+B)-CLB

and then assume that
%call+%fold=1
%win+%lose=1

=(1-C)P+CW(P+B)-CB(1-W)
=P-CP+CWP+2CWB-CB

You got it.

But there are several other forms that you will see floating around the forums. Also, you need to bear in mind that this formula assumes that you will be able to realize your equity by calling when in actual play, this isn't always true. Imagine that you calculate that you have enough equity to call and then on the next street, opponent blasts very large bet 80% of the time with a balanced range that you can't call...and you are forced to fold. Just saying that be careful with this...

Aise,

Can you elaborate to how this equation might change for this scenerio?

Do you incorporate an R into the equation?

Total EV=
FP+C[W(P+B+R)-L(B+R)]

Can someone check this for me, I feel like I am doing something wrong.

We shove $50 into a pot of $100 with the nut flush draw. Assume we need to hit the flush to make a winning hand. Assume our opponent folds 50% of the time.

(.50)(100) + (.50)(.75)(-50) + (.50)(.25)(200)
50+(-18.75)+25
= 56.25

Bolded numbers have to sum up to total pot. When you win, you win 150$.

(.50)(100) + (.50)[(.75)(-50) + (.25)(150)]

Got it. Thanks.

Careful here. We must be clear what baseline we are calculating EV relative to. You must be consistent. It is possible to calculate EV relative to what you had before the hand, for example, but then you must compare that EV to the EV of other options relative to what you had before the hand.

Normally, we calculate EV for a decision relative to folding at that decision point. Thus:

Yes. The 50 chips we put in for the SB are not ours at the moment; they are part of our winnings, relative to folding, if BB folds.

No; we already have 50 chips in the pot, so we will only lose 950 relative to folding if BB calls and wins.

No; we will win the other 900 chips in BB's stack and the 150 that are already in the pot, making 1050. Double-check: if we fold, we have 950 chips; if we win, we double up and have 2000 chips.

The errors here are slight, and tend to cancel: avoiding rounding, I get a result of +78 chips EV.

What is the formula if there are n Villains?
Worse case in 6max : n=5

Start by answering the question "what are all of the possible calling scenarios?"

Looks something like this:

1. CCCCC (all call)
2. CCCCF (One player folds)
3. CCCFC
4. CCFCC
5. CFCCC
6. FCCCC
7. CCCFF (Two players fold)
8. CCFCF
9. CFCCF
10. FCCCF
11. CCFFC
12. CFCFC
13. FCCFC
14. CFFCC
15. FCFCC
16. FFCCC
17. CCFFF (Three players fold)
18. CFCFF
19. FCCFF
20. CFFCF
21. FCFCF
22. FFCCF
23. CFFFC
24. FCFFC
25. FFCFC
26. FFFCC
27. CFFFF (Four players fold)
28. FCFFF
29. FFCFF
30. FFFCF
31. FFFFC
32. FFFFF (All players fold)

Now imagine everywhere you see C's and Fs you need to not only come up with a probability of their occurrence but also the probability the game reaches that state (also these probabilities aren't independent of one another).

You could approximate by condensing the possibilities into their larger categories (i.e. all call, 1 fold, etc) and calculate that for approximations.

Also I might have messed up some of the f/c diagrams so there may be states that I missed.

so Global EV would be equal to :

P(all call)*EV(all call)
+P(One player folds)*EV(One player folds)
+P(Two players fold)*EV(Two players fold)
+P(Three players fold)*EV(Three players fold)
+P(Four players fold)*EV(Four players fold)
+P(All players fold)*EV(All players fold)

is that it?

Yup you got it. Now the EV parts will be broken down by everyone's equity in the pot....

Sent from my SM-G900R4 using Tapatalk

OK ty, am I right here :

P(One player folds) = 5*P(CCCCF)
P(Two players fold) = 10*P(CCCFF)
etc...

right?

I'm a little rusty and haven't atrempted to do a calculation of this size, but I don't think the relative frequency of an event happening in our list of possible outcomes has any bearing on the calculation.

I simply listed the game states to show you how many potential calculations you'd have to track if you wanted to be thorough.

The probability an event occurs is based on the relative frequency of each player calling or folding.

The easiest way to do this is to either just assign probabilities to the event happening (i.e. you could just say the probability of 1 player calling is 40%) or to assign each player the same calling frequency and calculate the probability of each event happening.

In another forum, someone asked a similar question for multiway all-in EV with hero shoving against 2 villains (more than 3 players is unlikely). I reproduce my answer below:

A general EV equation is

EV =Sum over all outcomes { Pr(Outcome Oi)*EV(Oi)}

The outcomes are hero shoves and beats both villains, hero beats A and loses to B, hero beats B and loses to A, hero loses to both. Note winning over an opponent has to consider a card showdown win or an opponent fold. That would make 9 detailed outcomes where hero is involved, for each outcome, oopponent can play and win, play and lose or fold. For each outcome there is a payoff based on bet, pot and stack sizes. Since you know your hand and villain calling ranges, you can use an equity calculator to estimate the probability of each outcome without folds and use that in conjunction with the payoffs to get EV(Oi). A tricky part is estimating joint folding probability outcome but assuming independence is usually adequate instead of combo counting.

As I wrote this, I thought perhaps I should replace the phrase “some effort” with “a great deal of effort”

this means the same as this :

P(all call)*EV(all call)
+P(One player folds)*EV(One player folds)
+P(Two players fold)*EV(Two players fold)
+P(Three players fold)*EV(Three players fold)
+P(Four players fold)*EV(Four players fold)
+P(All players fold)*EV(All players fold)

right?

Yes, if you are looking for an abstract formula.

But in the real world of poker, as just_grindin and statmanhal posted above, deriving each one of those terms is very difficult.

Stack sizes matter.
Position matters.
Who/how many callers matters in each villain's decision.
Etc.

To come up with an actual formula, you probably have to make so many simplifying assumptions that the formula is probably not very valuable.