Heads up. You have Axs nut flush draw on the flop. The flop is not paired and we also assume for our purposes that if we hit the flush we win the pot (yes not exactly true due to dirty full house outs/etc but we simplify). We also assume villain has us covered and can't reraise.

Pot Size on Flop: X

You shove (IP or OOP not relevant for our purposes) for what amounts to 2X pot. So if pot was $100, you just shoved in $200 for example.

Let's look at the case of villain calling off shove. Assign P as probability of our flush coming in and us winning the pot by river.

EV = P*(X+2X)-(1-P)*2X = 5PX-2X
Neutral EV case = 0 = 5PX-2X then P = 2/5 = 40%

Is this correct so far? So if we get called we would need 40% equity to breakeven?

If that is correct and we assume our actual P to be 33% the difference does not seem to be that much, only 7%. And so it seems that this particular shove will almost always be overall +EV if villain folds even a small % of the time such as 25% of the time.

Is the above correct? Is my thinking and conclusions correct?

EV = p(call) * equity when called * pot when called - shove amount + p(fold) * pot

Where p(call) is the probability that villain calls and p(fold) is the probability that villain folds.

So in your example:

If you need exactly a flush to win*** you have 1 - (38/47 * 37/46) =~ .35 probability of making your flush. So you would win ~.35 of a $500 pot =~ $175 when called. You risked $200, so that alone has a net of -$25. Now we need to calculate how often you are called. Let's say you are called 50% of the time. So 50% of the time you lose $25 which is -$12.50 and 50% of the time you win the $100 pot, which is $50. Adding the two together we get a net win of $37.50. We can find the equilibrium by doing a little algebra (not shown here), to see that if we are called 80% of the time we would break-even. Any less than 80% of the time and we are +EV.


*** It should be noted that it is pretty odd that you only need a flush draw to win and that it is a lock once you get it. Basically that means villain must have a straight when he/she calls. Villain never has top pair because you said that you only have the flush draw as outs (if his top pair was less than an A, you would have an A as an out...if it was an A, you would already have a pair of aces yourself). Villain also cannot have two pair or trips because that would give him outs to a full house which you said isn't possible. Thus, this calculation is only accurate if villain has a made straight on the flop and you have a flush draw with no straight possibilities.

In a more realistic case, you would simply calculate your equity against villain's calling range on the given board. Plug that in for your equity when called. If you know villain's range for going to the flop, you can easily calculate his fold % by subtracting the difference from what he will call a shove with from his preflop hand range.

edit: To note that I did this on the fly and did not double-check my work. BruceZ and others will no doubt catch any mistakes I may have made.

Do you find a mistake in the way I did it? Obviously your way is more comprehensive and I just considered the EV of getting called.

Your equation is correct for the condition imposed, namely that villain will call. Sherman expanded the condition to include a fold probability. He found that if your equity is 35% , then villain folding 20% of the time or more will be +EV. You conjectured that with 33% equity, villain may have to fold about 25% for breakeven. The two conclusions are quite consistent IMO.

No. I didn't find a mistake, but I had difficulty understanding your way (which is why I did it my way). In any case, statmanhal seems to have understood your way better and confirmed that it seems to provide a reasonably appropriate (if not perfectly correct) answer.

I like Sherman's way much better than mine, it's more comprehensive.

EV = p(call) * equity when called * (pot when called - shove amount) + p(fold) * pot

Now let's make this even more bad ass.

p1= probability of a fold
p2 = probability of a call = (1-p1)
p3 = probability of winning if called
p4 = probability of losing if called = (1-p3)
X= initial pot size
AX = shove amount, where A > 1

EV of calling = P3 * (X+AX) - (1-P3)*AX = P3X+ P3AX - AX + P3AX = P3X + 2*P3*AX - AX = X (P3 + 2*P3*A - A)
EV of folding = P1 * X

Neutral EV = 0 = P1*X + P2*X (P3 + 2*P3*A - A)
P1 * X = P2* X (P3 + 2*P3*A - A )
P1 = (1-P1) (P3+ 2*P3*A - A)
P1 = P3 + 2*P3*A - A - P1*P3 - 2*P3*A + P1 * A
P1 = P3 -A - P1*P3 + P1*A
We'll assign 0.33 to P3.

P1 = 0.33 - A - 0.33*P1 + P1*A
1.33*P1= 0.33 + A (P1-1)
A = (1.33*P1 - 0.33) / (P1-1)

Err nevermind, my algebra sucks. But I know I'm on the right track.

You didn't copy my formula correctly. You added a parenthesis where there is none. It should be this:

EV = p(call) * equity when called * pot when called - shove amount + p(fold) * pot

What you call p1 is my p(fold). What you call p2 is my p(call). What you call p3 is close to, but not the same as my "equity when called." Equity when called includes 50% of ties, whereas your p3 does not.

Now what is it that you are going for when trying to make it "more badass." That is, I am not sure what you mean by that and what you are trying to find out.

The biggest issue that can come up with these calculations is that changing one number (e.g., p(call) ) a little bit should change the other numbers (e.g., equity when called). That is, change in one number is not independent from change in the others. The dependence between p(call) and p(fold) is obvious. Less obvious is the fact that if an opponent calls with a wider / tighter range, your equity when called changes (in a direction that is unknown without knowing the specific hand ranges).

I'm trying to reduce the EV equation to only 2 parameters:

Probability of a fold.
Multiple of the overbet as in multiple of the pot size.

Then I will be able to plugin the two as estimates and see if it's +EV for a particular situation.

If pot before the shove is N and shove amount is M, how do you define "pot when called"?
Is it N+2M?

pot when called is the total size of the pot that will get pushed to someone when the hand is over. So yes. It is N + 2M assuming villain always covers M.

In EV analysis, the term "Pot" can be troublesome.

Take the simplest HU case of villain making an all-in bet and you call.

We have three possible pots:

Po = pot before villain bets
Po = pot after villain bets
P1 = pot after hero calls.

The EV equation for each will differ but if done correctly will give the same results. If
B is the bet and eq is hero's equity, then

EVo = (Po+3B)*eq -B
EV1 = (P1+2B)*eq -B
EV2 = (P2+B)*eq - B

These equations are derived using the fact that P1 = P0+B, P2 = P1 + B and
for an all-in bet,

EV = Pr(win)*Win Amount - Pr(lose)*Bet

Shouldn't EVo = (Po+2B)*eq-B ?

EV = p(call) * equity when called * pot when called - shove amount + p(fold) * pot

I will use your formula but will use my abbreviations.

p(fold) = F
p(call) = 1 - F
equity when called = E
pot = X
shove amount = A*X where A is a multiple
pot when called = X + 2*A*X

So we have: EV > 0

E*(1-F)*(X+2*A*X) + F*X - A*X > 0
X*E*(1-F)*(1+2A)+FX-AX > 0
X*(E*(1-F)*(1+2A)+F-A) > 0

Divide by X.

E*(1-F)*(1+2A)+F-A > 0

Assume E = 0.33 for flush draw simplified scenario.

(0.33-0.33F)(1+2A)+F-A = 0 we equate to 0 to solve for breakeven point.
0.33+0.66A-0.33F-0.66FA + F - A = 0
0.33-0.34A+0.34F-0.66FA = 0
0.33+0.34F-A(0.34+0.66F) = 0
A (0.34+0.66F) = 0.33+0.34F

A = (0.33+0.34F)/(0.34+0.66F)

Then we can assume some F and find out what A is breakeven. I
m pretty sure I f-ed up somewhere but at least now you guys know what I'm trying to do.

I had:

EVo = (Po+3B)*eq -B
EV1 = (P1+2B)*eq -B
EV2 = (P2+B)*eq - B

I messed up the pot notation. The above equations are consistent with the followng:

Po = pot before villain bets
P1 = pot after villain bets
P2 = pot after hero calls.