Hi,
reading quantum poker, was hoping someone could clarify this equation for me
In discussing a situation where you have a G.S+F.D vs top pair
"You figure they will not call a potsized bet on the river if you make your flush, but they will call if you make your straight. You have 9 outs to win the pot, plus 3 outs to win the pot plus a pot sized bet. you estimate your equity at around 27% and need to figure out if you can call a pot sized bet. Your pot odds are 1/(1+2), 33%, not enough to call."

that i understand, here's the part i can't wrap my head around,

"but 3/44 times you will collect another potsized bet, which will be 3 times larger than the bet on this street. So 3/44*3 can be added to your pot odds calculation giving you about 31% equity needed to call"

I understand there is some weighting going on here, but,.. why are you doing 3/44*3

basically, 6% of the time you're making 3x as much..?
doesn't 3/44*3 come out to 18-20%? how does 3/44*3 give you the extra 4% equity thats getting you from 27% to 31%

thanks in advance !

I agree that the way they worded things is confusing.

Here is what I think they are saying. Call the current pot 1 so it "cancels" throughout.

Case 1: Only Direct Odds Matter

Let your winning percentage be given as E1 and your change in chip stack be given by C1.

Then you should call a pot-sized turn bet if C1 > 0, where

C1 = E1*(2) + (1-E1)*(-1)

Clearly this is equivalent to saying that you should call if E1>1/3.

When your actual winning pct E1 is 12/44, C1 is (12/44)(2) + (32/44)(-1) = -8/44 [being negative means Hero should Fold]

Case 2: Implied Odds Matter

Here Villain will call a pot-sized river bet if Hero's gutshot straight hits, but not if Hero's flush hits on river.

Let your winning pct again be given as E1 and your change in chip stack in this case be given by C2.

Then you should call a pot-sized turn bet if C2 > 0, where

C2 = E1[(9/12)*2 + (3/12)*5] + (1-E1)*(-1)

= E1*(2) + E1(3/12)*3 + (1-E1)*(-1)

When E1 is 12/44 this becomes

= (12/44)*(2) + (3/44)*(3) + (32/44)*(-1)

= 1/44 [being positive means that Hero should Call]

So you can see that there is an "extra" (3/44)*(3) term in Hero's "equity" equation.

The terminology is definitely confusing. It appears they are using two different (but related) definitions of "equity" and lumping everything under the "pot-odds" rubric.

But their point is valid. The additional "equity" the implied odds gives Hero is just enough to change his decision from Fold to Call.

I know I have hand-waved a few things, but I hope this helps.

+1

I often see people using the terms equity, pot odds and EV interchangeably. I think this is wrong:

Equity – the chance you win (+tie/2) the hand or your share of the pot based on your winning chances

Pot Odds – a ratio equal to ( pot you may win) / (amount you have to invest)

EV = the expected profit/loss if you make the required investment – depends on pot size, card and fold equity , investment

hmm my formal math training is low, so im going to have to ask for some clarification. forgive me for any technical mistakes

Let your winning percentage be given as E1 and your change in chip stack be given by C1.
Then you should call a pot-sized turn bet if C1 > 0, where

this is just saying i should call if its going to gain me >0 chips?

When your actual winning pct E1 is 12/44, C1 is (12/44)(2) + (32/44)(-1) = -8/44

so E1 is my odds of winning, why is C1 (how much i gain or lose?) my odds multiplied by two? where is the 2 coming from

the second half makes sense, it's just averaging the first half ([how much i win* the probability i win] - [how much i lose*probability i lose]), right?

once i understand this, the second part may become clearer for me.
thanks again

also, this is all very new and exciting for me. As long as the posts actually pertain to poker theory, will i get in trouble for posting 1-2 Q's per day here?

The "2" in that equation reflects the amount of chips you will add to your chip stack if you call and win (compared to your chip stack at this moment).

I assumed that the existing (before Villain makes a turn bet) pot is 1. I did all the EV calculations in terms of multiples of the current pot size to make things simpler. Hopefully that is not adding to the confusion.

The second pot-sized amount that you will add to your current chip stack comes from Villain's pot-sized turn bet. In Case 1 we are assuming that you are deciding whether to call (or fold) a pot-sized turn bet from Villain with no possibility of future bets on the river.

If you call and win you will win what is currently in the pot plus Villain's pot-sized bet. This would add 2*(amount of the current pot) to your current chip stack.

Calling and losing, of course, costs you 1 pot-sized bet (which is where the -1 comes from).

Edit to add: Yes, calling if C1>0 reflects the fact that you can fold to Villain's bet and your chip stack would remain unchanged from its current amount. That is, folding to the bet is 0, the baseline option.

ooh, that makes sense.
So basically im just doing an equity calculation of:
[my odds to win*amt i win]-[odds i lose*amt i lose] and then combining that with
[odds i win with a straight*amount i win]

Yes.