using fold eq calculator. for 100bb.. if villain is 3betting 10percent vs our open and we jam 22-88 for example,

we see a profit if villain is folding over 69 percent. Which he will be if he only continues with AK and JJ+ as this is only 3 percent of the 10 percent 3bet

Is my logic here right? If so is it therefore possible to do this every time against people who 3bet over 10 percent in certain spots

if you have about 33% equity when called and he folds about 2/3 of the time then you're roughly indifferent between jamming/folding

keep in mind a polar >10% 3b is super wide, looking something like JJ+, AQs+, A9s-A2s, KJs+, 76s, 65s, 54s, AKo, ATo, KJo+. You should also be able to widen your calling range because you won't be at such a large equity disadvantage and will also be able to realize more equity because spots won't be as polar post flop

in spots like btn vs sb they may 3b wider than 10%, but you also have to compare the ev of jamming to calling instead of folding

yeah thanks it was more for co v btn, sb v bb spots where oop we won't be able to realise as much equity. Thanks for the reply dude

fwiw people have started 4b jamming bvb this year if you check the highstakes nvg thread

I questioned how you can say this without any notion of stack sizes and pot and bet amounts. It turns out that with the given card and fold equities, the EV equation is (1/9)*(7 – B), where B is the jam in pot size units. Therefore the max B for +EV is 7, a pretty large bet range

lost me here

Quote:
Originally Posted by valuecutting View Post
if you have about 33% equity when called and he folds about 2/3 of the time then you're roughly indifferent between jamming/folding

The quoted statement claims indifference between jamming and folding for given card and fold equity with a jam. The indifference refers to EV and the EV of a fold Is 0. Therefore the EV of the jam has to also be 0. But EV has to include pot and bet sizes which were not given. So I wrote out the EV equation with a jam assuming pot = 1 and jam bet = B as

EV =2/3*1 +(1/3)*(1/3*(1+B) – 2/3*B),

which, if I didn’t mess up reduces to 1/9 * (7-B).

Set EV to 0 to find the indifference bet and it is B=7.

op specified 100bb deep and i assumed a standard ~10bb sizing with a lot of rounding everywhere to just get a rough estimate

so what does the number 7 denote in the actual hand

After some algebra, the EV equation for a jam (or bet) reduces to (7-B)/9, where B is the bet size. Now, for indifference between jamming and folding, you have to set the EV to 0

(7-B)/9 = 0

7-B = 0

B = 7

If the effective stack is 7, then the jam would result in an EV of 0. Any bet less than 7 has positive EV for the bettor. So, in pot size units, 7 is the maximum bet size for +EV.

Solvers. Hands like TT can do this.

When I first started studying poker I learned that 44 vs AK is a flip. I started going all in with any pair any time a raising war broke out preflop. I ran hot with this strategy for a while but then I started running into too many big pairs so I stopped doing it.

Any one else following this? not meaning to sound condescending, i'm just extremely lost... as in 7x pot jam?

gotta ride that variance train

Yes that is correct. 7 means 7*pot size in this case because the pot size= 1.

Basically that's the absolute maximum you could bet given the parameters used and still have betting = folding (0 EV). Any less than that size and the bettor's EV should be positive.

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