Hi, I'm a relative beginner on poker...

Want to give bad implied odds for my opponents, so here is an example:

Suppose I know exactly what he has, which is an OESD on the turn, let's round his equity from 0,18 to 0,20. So the math is the following:

Call/(Call+Bet+Pot+X)=0,20. So in this case, the X (implied odds) is 2
1/(1+1+1+2)=0,20

If his stack is less than 2 he cannot call profitable OTT, since he must win 2 in average in order to be breakeven. So let's say his remaining stack is 2 and he will always win another 2 if he hits. If he miss he never bluffs and always lose without spending more money.

So, his EV is:
EV = W%*W$ - L%-L$
EV = 0,20*(1+1+1+2) - 0,80*1
EV = 1 - 0,8
EV = 0,20 $

Is this correct? Shouldn't be EV = 0?



Now the EV for the better:
EV = 0,80*(1+1+1)-(0,20*3) < is it correct? Losing 3 = 1 OTT and 2 OTR
EV = 2,40 - 0,6
EV = 1,80 $

Where I'm going wrong? Couldn't find a solution looking on Google...
Both EV can't be positive, right?

Edit: Title should be EV on implied odds for Aggressor and Caller

The decision point is bettor making the turn bet when pot = 1.

Your EV equation for caller includes his winning the $1 call he made on the turn and that is wrong. When you eliminate that, his EV =0 if his stack is 2 on the river and he always wins if he hits.

Your EV equation for the bettor includes his bet. It should be

EV = 0.8*(1+1) – 0.2*3 = 1.

The sum of EV’s = 0 + 1 = 1, the pot prior to the turn bet, which is what is required.

A more complete implied odds model would include villain not always winning if he hits (reverse implied odds) and hero not always calling if villain hits and bets.

Thank you! Simple and easy to understand!

But didn't completely understand the EV for bettor. Let's say if pot before betting OT is 55, than his "EV = 0" must be EV = 55? Anything bellow EV = 55 is a negative EV?
Like in the previously example, instead of the caller having $2 left he has $3 left, now the bettor EV is 0,80. It's already a negative EV?

I have too much of a headache to look at your equations right now, but it's frequently the case that it is profitable for one player to make a bet and also profitable for the other to call it. This is because there is money in the pot already.
Although in the vacuum of a single hand, only one player wins all the money at showdown, both players have equity, and a chance to get back more from the pot than they put in at the decision point.

e.g. Someone could bet the turn with 95% equity in a massive pot, but if the other player only has to put in 5% of the final pot at that particular point in the hand, they still break even in the long run. Thus it becomes a +EV bet, and a +EV call.
More commonly, someone shoves with a pot-sized bet on the flop and a made hand, and the opponent has a draw with about 33% equity. He calls and breaks even in the long run, while the other player has an EV of "pot". If the guy on the draw has 35% equity when faced with a pot-sized bet, he'll make a small profit on his call, and the shover will have an EV of less than pot.

For the bettor EV, if villain doesn’t hit on the river, he check-folds and hero wins the turn pot plus the turn call = 1+1= 2. This happens 80% of the time. If villain hits on the river, he bets 2, you call and V wins so you lose 1+2 = 3. This happens 20% of the time.

Again, you included a player’s current investment as part of his winnings if he wins.

I understand now. It would be ok to use a -Pot in the EV equation?
W%*W$ - L%*L$ - Pot? To make it easier to see exactly how much EV I have (or I'm giving) when I make the bet?

e.g. Played this hand few days ago
MP 0,20 97 ($6,97)
CO Fold
Hero Call 88 ($10)
SB Fold
BB Call

Flop KT8
BB check, MP bet 0,47, Hero raises to 1,55, BB fold, MP call

Turn 3 (3,64)
MP check, Hero bets 2,75, MP calls (leaving 2,47 behind)

He has 30% pot odds and 18,18% equity. His implied odds are 5,99 (2,75/(2,75+2,75+3,64+X)=0,1818

So, his EV, having 2,47 left, is: EV = -0,64 $, correct? (0,1818*(2,75+3,64+2,47))-(0,8182*2,75)
EV for the bettor is EV = 4,28$

If his remaining stack was exactly 5,99, his EV would be exactly 0.

If his remaining stack was exactly 5,99, the EV for the bettor would be exactly the pot, like statmanhal showed:

EV = (0,8182*(2,75+3,64))-(0,1818*(2,75+5,99))
EV = 3,64$ = The pot OTT before hero making the bet.

So this is the equilibrium point, right?

But can I add the -Pot on the Bettor EV equation?
EV for his actual remaining stack of 2,47
EV = (0,8182*(2,75+3,64))-(0,1818*(2,75+2,47))
EV = 4,28
If I subtract the pot before making the bet
EV = 4,28 - 3,64
EV = 0,64.

So the EV for the bettor is +0,64 and EV for the caller is -0,64. The EV numbers are always equal, just that one is positive and other is negative, unless both are exactly zero. That's beautiful lol.

But my EV after I bet on the turn is only 0,64 cents, thought it would be higher than this (I know that EV is higher if included pot before betting)... Still, a better result than showing OTT and never getting called, since by betting and getting called I'm winning +0,64

Now I want to go deeper on it, will try to add reverse implied odds to it.

I think that reads are much more important with marginal draws than what's going on in here, because strictly speaking these marginal draws have ev near zero, else they wouldn't be marginal. Reads swing marginal decisions; vs overaggressive players on the flop you have two more streets of implied odds which increases the implied odds of hitting on the turn, but if you miss the turn you'll be forced to fold a lot; overly tight players(that have significant folding ranges on each street; they get poker but they miss preflop and postflop ev by definition, else they wouldn't be considered tight) allow you to hit many more draws for free, and even when they bet it's usually a smaller fraction of the pot on average. This is ev gained in the form of free cards and cheap draw outs(I'd rather pay 1/3 pot to draw out vs 3/4 pot, of course), which actually allows you to expand the bottom of your range on a previous street with this read; If there were two different strategies(one that plays wild postflop and one that plays very tight) that use the same starting range, then I would have a wider range vs the very tight player going to the flop. It's about realizing draw value by hitting hands, which happens much more often vs the tight player, while you're forced to fold more vs overagressive players.

This is all much more important to me with marginal call/fold draws on the flop and turn than implied odds, particularly with draws that are vulnerable to redraws. If the draw isn't vulnerable to redraws, then it's probably not all that marginal except maybe facing a large turn bet. The vulnerable draws are more likely to be marginal(hint: vs strong opponents you're not really worried about making a small mistake here by calling, unless you're bad at estimating the margin; if you assume that unknowns are solid tags from the start, these marginal draws are not where the money is to be made vs unknowns, else you wouldn't assume they're a strong opponent(perhaps you play homegames with friends, or micro stakes online, or small stakes live)).

so, no reads: it doesn't matter as long as you're good at estimating a half decent call/fold margin.

with reads, the number of streets left on which you can win big bets matters, as does the average price to improve.