Flush Draw Problem

[crazycrazy]

Situation:
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pot on flop $10, HU, $20 behind each
me on FD he has TP
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he checks, i bet $3 (3/16 < 0.2 i have odds for turn), no folding eq from him
if he raises me all in (17/50) i have odds for the flush
if he raises me some other amount then i can reraise AI and have odds. (i think i didnt count that for all)

BUT i'm still commiting 20 to 50 pot all-in on flop and thats not correct for my draw so wheres problem --COMMITING $20 ON FLOP IS WRONG ODDS BUT THE STEPS IN WHICH I COMMITED IT ALL HAVE ODDS ($3,$17 reraise,etc).

> if i exclude FE, is it different from calling and betting with FD ????

i'm missing something out on something here... hope someone know what it is

[PapalRage]

if you are sure that you have no fold equity, there is no point to bet after he checks. The reason we often bet with strong draws is because we can win when they fold or if we hit. In your example if he bet 3 dollars into you, than you would have odds to call. However, there is no reason for you to bet the 3 dollars if he checks without fold equity.

[crazycrazy]

if you are sure that you have no fold equity, there is no point to bet after he checks. The reason we often bet with strong draws is because we can win when they fold or if we hit. In your example if he bet 3 dollars into you, than you would have odds to call. However, there is no reason for you to bet the 3 dollars if he checks without fold equity.

How about building a pot ?

[BeeJall]

if you are sure that you have no fold equity, there is no point to bet after he checks. The reason we often bet with strong draws is because we can win when they fold or if we hit. In your example if he bet 3 dollars into you, than you would have odds to call. However, there is no reason for you to bet the 3 dollars if he checks without fold equity.

How about building a pot ?

Why would you want to build the pot with a draw assuming no fold equity?

[crazycrazy]

Because of profit when i hit it. I have good implied odds still even if i'm behind in the hand at flop.

[BeeJall]

But if your opponent has top pair, he's probably going to be willing to get it in on the flop, what are you going to do if you're raised? If you end up calling his all in with just a draw then you've done exactly what he wants. Where's the implied odds there?

[crazycrazy]

But if your opponent has top pair, he's probably going to be willing to get it in on the flop, what are you going to do if you're raised? If you end up calling his all in with just a draw then you've done exactly what he wants. Where's the implied odds there?

That is explained in the initial post and thats the question. i have odds to call the all-in.

[BankItDrew]

crazy, you've got it all wrong. always put the least amount of chips in the middle when you are behind and have no fold equity.

[Pants_101]

Situation:
BUT i'm still commiting 20 to 50 pot all-in on flop and thats not correct for my draw so wheres problem --COMMITING $20 ON FLOP IS WRONG ODDS BUT THE STEPS IN WHICH I COMMITED IT ALL HAVE ODDS ($3,$17 reraise,etc).

Ok so If I follow you correctly you feel you have discovered a paradox. On the flop you describe if you were to always push all in and he was always to call you would expect to lose money overall because you need to win 40% of the time and you will only win 35% of the time. BUT if you bet and then he raises all in you can call and expect to break even. You put all your money in both times but once you are losing and once you are breaking even. I think the point is in the second case you break even for how much money you have after you have bet. The money in the pot isn't yours anymore so you are getting better odds on your bet. So the OVERALL EV of a hand is different from any one individual EV decision during that hand. Does that help? Another way to look at it in this case is the stacks aren't deep enough to chase the flush as you can't win enough when you hit to offset the amounts you lose when you miss, even though you can make a +ev call for all your chips if you mistakenly bet and get raised. The initial bet though is a mistake. I think...

[crazycrazy]

Pants:
u understood what i mean correctly and it hink u made a good point that i have actually positive expectation on the reraise i call. i mean if i fold right away i loose more (more negative EV) then calling it when i end up wiht more money. the $3 bet into flush draw is mistake just from some limited mathematical stand point if u add things into equation like how much u can win if it hits - cos u get him more pot commited - then i think its not mistake at all,depending on opponent.

[sunnyday]

seems you got an interesting point here crazy.

it is true that the 1st bet when considered a semi-bluff is +ev in the case it is either called or opposition folds. it is +ev even for 0 fold probability.

but... opposition may also raise! now how to factor that into the ev equation of a semibluff? standard way of calculating ev for a semibluff only takes into account equity for fold- and call-cases.

i am at a loss here myself, but will do some thinking on it since it is indeed a very interesting point. seems you need to expand the standard semibluff equation (fold equity + call equity) by a third term for raise equity. i.e.: fold equity + call equity + raise equity. this should prove quite difficult to formulate because you would have to factor in many different possible raise sizes with respective raise probabilities. actually i am guessing you would need quite complex mathematics to write something like this down. my guess is you would have to form an equation describing the relation between all possible raise sizes and corresponding probabilities. then do some integration on that. whatever? maybe one of the math wizzes around this forum could help with that one?

2nd bet of course is clearly +ev.

cheers

[gingerwizard]

So the OVERALL EV of a hand is different from any one individual EV decision during that hand.

Sorry pants but this is not correct. Expectation is linear so that OVERALL EV of a hand is the sum of the expectations or each particular decision in that hand. That is mathematical fact (and has it's own proof if you want to see it), so the real question is what is incorrect in OP's reasoning that leads to this paradox?

Answer: Perceived EV is not actual EV. You perceive that your bet of $3 is +EV but of course it may not be (and in this case is likely not)

You see your decision to bet $3 commits you to the pot, i.e.. commits you to putting the rest of your $20 in. So to work out EV of the bet you need to work out the EV of what the bet implies:

x is % he folds
y is % he calls
z is % he pushes all in.
EQ is pot equity % on turn.

EV = 10x + y(13EQ - 16(1-EQ) ) +z(30EQ - 20(1-EQ))

Here EQ is equity of flush draw.
You see however that the decision to bet $3 has impacted on the rest of the hand, and if that decision is -EV then it was of course bad.

To sum up, you probably shouldnt be betting unless x you perceive x to be high and z to be low.

[badgers]

Expectation is linear so that OVERALL EV of a hand is the sum of the expectations or each particular decision in that hand.

YES!!! Great post gingerwizard, thanks!

Your EV calc assumes no implied odds, right? (may just be my poor maths/poker understanding)

[crazycrazy]

i come to conclusion its impossible to calculate exact EV, cos it depends on opponent playing style, what he does when flush comes on turn and more things. what is wrong is commiting all ur chips on flop in this situation but if u have odds to call some reraise from him then u should do it cos that investement actually makes u money while u have to take the loss of the rest of the pot -- u simply flopped worse then he did and ur underdog in the hand... but u can turn this around by making good play on him on later streets, or he making mistakes by giving u good odds...
ok enough odds...poker is about estimating where ur in a hand in a first place....

[crazycrazy]

lol badtgers i just noticed ur location: 20 tabling ur mad

[badgers]

i come to conclusion its impossible to calculate exact EV, cos poker is a game of incomplete information

And I only 20 tabled a few times but it was hilarious! Really should change that...

[gingerwizard]

i come to conclusion its impossible to calculate exact EV,

In most cases yes. The best players are those that get the closest.

There is certainly no paradox though. To answer your other question implied odds come into your estimation of equity on the turn.

[Pants_101]

So the OVERALL EV of a hand is different from any one individual EV decision during that hand.

Sorry pants but this is not correct. Expectation is linear so that OVERALL EV of a hand is the sum of the expectations or each particular decision in that hand.

Yes I accept this of course. I didn't perhaps express myself very well
all I was trying to say was that if the opp plays this hand 1000 times betting then calling the all in then his call will always be +ev yet he will still be steadily losing money because in this case the initial bet is -ev to a greater degree than the call is +ev. So overall the hand is -ev even though one of the individual decisions is +ev. Or is it the fact that the stacks aren't deep enough rather than the initial bet? Or are they related? Anyway if that's still wrong then gah! I should leave this stuff to the experts