acidboy

"You are correct that KJs makes more money 3betting versus a wider open range but that doesn't necessarily mean it's the best play to 3bet it."

Care to ellaborate this? Im not an expert, but the way I see it, if a play makes more money than other, is better.

Did you really need the calculator to do 1/2? :D

Cool :)

@ Case 3, ph33roX hand, you wrote: 

Q+ and 42-or-better straight draw, with double flush draw = (Q+,42+):(hh,cc)

Doesnt that expression mean Q+ or 42+ with hearts OR clubs draw? I mean you are defining hands which have 2 hearts or 2 clubs, instead having both, no? If you want to select the ones that have a double flush draw shouldnt it be like this: (Q+,42+):(hhcc) ?

I might be wrong but i wanna be sure :)

Great, thanks! :)

Thanks for the very fast answer!

Sounds great that they are planning to add that :) It would be def an interesting addon, would be great if we get that + the possibility of assigning a villains c/f % or fold % of each part (top mid and bottom) of his range if we combine it with HUD data and personal reads, to get a better idea of the EV of some of our bets, if that makes sense. 

Regarding the syntax, You wrote:

"Any boardpair/overpair + 2 backdoor FDs

2+:(cc,hh,dd)

That is, any hand, bottom pair or better, that also has two clubs, or hearts, or diamonds. "

This is not what i was looking for :P I think I explained it poorly (im spanish so sorry for the poor explanations). What i want is adding to the strong range the single non top board pairs which also have necesarily two backdoor flushdraws (only one BDFD is too weak imo), like:

4d5d7c8c

AcQcKd2d

etc

I guess it should be expressed with something like:

(2+!6+):(cchh),(2+!6+):(ccdd),(2+!6+):(hhdd),(2+!6+):(hhcc) ?

"Top pair + any nut BDFD

6:!77+:(cc,hh,dd)"

Here I meant all the top pairs and OVP with a nut backdoor fd, so i guess it should be something like

6+:(Acc,Ahh,Add) 

no?

Hi! Very good job, thanks for the hard work :)

I wanted to comment that, in the last hand (the Phil galfond vs analtmezer one) isnt the Top Range for Phil's preflop $3b15o on that board a bit too pesimistic? 

Flop is 4c2h6d, and you estimate that a top range for Phil there would be 53,22+,875+:2+

However, imo theres some more good holdings we could have on that board, like any boardpair/overpair + 2 backdoor FDs, top pair + nut BDFD, some two pairs...  or you think we shouldnt consider this top range?

So I wanted to check how adding that stuff would affect to our top range % (probably not much) etc, but then i realized that I dont know how to express this:

any boardpair/overpair + 2 backdoor FDs 

top pair + any nut BDFD

I have checked the vid @ pokerjuice website regarding the intelligent postflop syntax, but i still dont know how to express it.

Sorry for the noob question =P

This guy is clearly unable to beat the games hes playing in the video, maybe he can beat them in a couple of specific sites. On Stars he should move down to 2/4 if he wants to really help people, this is just insulting.

the reason KJ was doing better than AJ was because you specified AJ to be offsuit where as you just put KJ as any suits and therefore the KJ could have the fd aswell

You are right, forgot to take into account that I shouldn't take a look at my [0, 0.17], I should take a look at my bluffcatching range. Thanks. 

I see a problem with your solution, GT.

When you bet [0, 0.17]  of your range and get called by [0, 0.50], you are assuming OOP is valuebetting and winning always the pot. So, if you have 2/3 valuebets, 1/3 bluffs, EV equation for IP player when calling, assuming PSB, is:

EV=%OOPvaluebets * (Eq*Pot-Pot)+%OOPbluffs * (Pot)

So you are doing

 0.66(-2)+0.33(4)=0

But the thing is, when [0, 0.17]  bets against a calling range of [0, 0.50], hes gonna win around 2.5/3 of the times (so it has 83.33% equity). Thats because 2/3 of the times the calling range has hands between [0, 0.50], and 1/3 of the time they are going to show up with the same range. So, the EV equation for IP player should be in this example:

0.66(0.166*2-2)+0.33(4)=-1.11111+1.33333=0.22222.

This number obv should be 0.