Calling Turn Raises due to Pot-Odds

hi all, i've been thinking about the following:

i dont want to make it too exact, but just consider this situation: we are on the turn, the pot is not small, hero bets and somebody raises, everyone else folds. Hero sees that he is behind but concludes that he has enough outs to call the raise.

the thing is this: on the river the pot will then be >10 BB and what should hero do if he doesn't hit? strictly speaking, he would have to fold, but do you really do it for a single bet?

if i don't do it, do i then count this bonus bet (or maybe 75% of it since villain could check) in my pot odds on the turn?

how do you do it? do you fold on the river every time? i have the feeling i'm setting up my calculation on the turn incorrectly since i call more than one BB on average.

wie macht ihr das dann? foldet ihr den river dann wirklich konsequent? ich habe das gefühl, dass ich dann meine rechnung am turn falsch gestalte, weil ich ja im schnitt mehr als eine BB calle.

thoughts?

Korn:

W = we win the SD
-W = we lose the SD
A = we are ahead on the turn
-A = we are behind on the turn

P is probability, | indicates conditional probability.

We have:

i) P(W) = P(A) * P(W|A) + P(-A) * P(W|-A)


The first part of the equation is the chance that we have the best hand on turn and that we still have the best hand at the showdown.

The second part is the chance that we are behind on the turn but hit an out and have the best hanbd on the SD.


But the first part of this equation is only valid if we assume that we call a bet on the river even if we don't hit an out, or if we assume that the opposition does not bet on the river.

So we must answer two questions:
A) Would we call a river bet even if we don't hit an out?
B) If not, what is the chance that our opponent bets again?

If we answer no to A) and our opponent bets the river again with probability P(B), we must modify equation 1 as follows:

ii) P'(W) = P(A) * P(W|A) * P(-B) + P(-A) * P(W|-A)


Now we can calculate P(W') depending on the answers to questions A) and B).

Whether we now call or not comes from:

Case 1) answer A) with yes

Call if

P(W) * expected pot at showdown > expected cost to showdown


Fall 2) answer A) with no

Call if

P'(W) * expected pot at showdown > expected cost to showdown


Roughly speaking, there are two factors as to whether we call on the turn: first, the chance that we are ahed and second the count of our outs. The interplay between these factors decides between a call and a fold.

I wouldn't mind seeing a few example calculations.

Gimli:

Example calculation:

I have J J on a board of 2 T K in UTG
I raised pre-flop. Only the MP called.
I bet the flop, he called.


Turn: 2 T K 9
I bet and MP raises.

Now there are ca. 7 BB in the pot.

MP could have:
- a finished straight
- 2 pair
- a set
- a strong draw (e.g. gutshot + flush + overcard)
- Top Pair with a good kicker

I have 2*J and 4*Q = 6 outs.
The outs must be discounted a little – as must the outs to a set.
I would have more outs against 2 pair, of course.

All in all, I assume that I am still ahead in 15% of cases and if I am ahead, then I will win 80% of the time.

=> P(A) * P(W|A) = 0.15*0.8 = 12%

As for my outs, I'll just assume 5 outs for now
=> P(-A) * P(W|-A) = 0.85 * 0.11 = ~ 9%

So just based on the outs a call is not profitable.

But I know that I would call a river bet if the pot is larger than 6.7 BB, which is also the case if I call down.

Assume that the opposition will be the river.

The cost for the SD is then 2 BB.
So I calculate:
12% * 10 BB + 9%*10BB = 2.1 BB > 2 BB

According to this, we have a close call...

Korn:

Here we could at least calculate under which assumptions a call is justified – and get some figures for it. This is much more useful than just saying: "In my experience, I would call here." It often happens that unconsidered experience leads to a wrong conclusion. A correct formula and an example can clear these things up.