Lets say we are heads up on the turn with AX flush draw with about 25% equity and 0 fold equity. The pot is $60 and we want to compare ev between $30 bet and check back.

I assume since we have 0 fold equity, then our ev equation is our equity times the total pot after bet and call minus our bet right?

ev(b30) = (0.25)(60+30)-30

And the ev of check is just our equity times the current pot?

ev(check) = (0.25)(60)

Are these right?

Should we also include the amount we might gain if we hit our draw?

Not right: EV = 0.25*(60+30) – (1-0.25)*30) = 0.25*(60 + 2*30) -30 = 0
This is only an approximation since it ignores the river action by basically assuming hand is checked down on the river. Also, it assumes villain will not raise. Okay for a first-cut evaluation.

Okay only if villain also checks turn and both check on river. If villain bets the turn then you have to consider your response to his bet size and you still have the river action to consider. This is also a first cut evaluation to be modified to the extent possible

This is what an implied odds analysis does. You bet or call the turn, even if EV < 0, with the chance you might win big if you do hit on the river and villain calls with a big stack that will result in a +EV. There are EV models to do this.

I was looking around on the internet and found the "Expected Value Calculation" series at pokervip. There I learned of the many variables that can go into an ev calculation. Beside the implied odds, there's also the fold equity, double barrelling tendencies, and the propensity to check/fold after a one&done cbet.

But these factors happen in varying degrees, for instance, villain will call $5 100% but fold to a $20 bet about 40% of the time, we will end up with too many different calculations. The approach then is to use averages because that is how much we will make in the long run.

As an example, lets say we have a flush draw with 18% chance of hitting, and villain will pay us off:

$100 0%
$80 25%
$60 40%
$50 60%
$40 75%
$20 80%
$10 100%

average = ($100×0) + ($80×0.25) + ($60×0.4) + ($50×0.6) + ($40×0.75) + ($20×0.8) + ($10×1) = $130

In this situation our long term implied odds =
$130 × 0.18 = $23.4

Though this may works for off table analysis, in real time we must choose the bet size that will net us the most ev, So in the end we will still have to calculate for all the differing variables. Thats a lot of variables to account for.

I've just started on this endeavor so im not sure what it entails. So far I found 1 constant which is the time we win the current pot and a bet when we hit our draw.

Ie. on the flop the pot is 1 and the bet is 1, our flush draw is 18%. thats 2(0.18).

Not sure if there are other constants that we can use, even if there are, thats a lot to think about in the heat of the moment. Is it even possible to think of all these thing while in play? Are there shortcuts we can use? Or is it only possible to formulate plans and strategies off table and use what we found to quickly infer what the most +ev play is in the moment?

Your implied odds equation does not cover all the factors. The question you asked was about a turn decision. Then, implied odds involves calling a -EV turn bet with a good draw like for a high flush and winning a big pot on the turn 1) if you hit else you fold , 2) bet/raise big 3) villain calls and 4) the hit wins.

Your equation only covers some of these factors and does not relate back to the turn decision.

For a fairly complete implied odds analysis, assuming you have a good draw on the turn, the factors are:

Turn pot before villain bet
Villain bet
Hero call amount
Hero chance for river hit
Hero bet amount if a hit on the river
Villain calls hero’s bet or folds
Hero wins with a villain call (a loss is termed reverse implied odds)

Then, Implied Odds (assuming villain calls river bet)
= Total Win Amount/Turn Investment = (Turn Pot after Villain Bet + River Bet)/Hero Turn Call

Below is an example of such an analysis written in Excel VBA, which determines the future bet you must make to break even.



I did a number of analyses with set, straight and flush draws and found that the conventional statement on needed implied odds of 15 to 1 is not unreasonable for a number of common situations.

statmanhal,
Just found your tumblr looking up implied odds models. Thanks for putting up your work.

Your welcome -- and thanks for giving me the opportunity to get 5000 posts