Hey all,
I'm reading through Expert hunlh by Will Tipton, and I'm stuck at the chapter about the indifference principle.
on page 122 there's an excercise: in the games you play, how frequently must the BB 5bet to make us indifferent between folding and 4b-folding after he 3bets.
Now this is how I approached it:
Asuming we're playing a husng with stacks of 1500, we're even in chips and the blinds are 15/30
We raise to 60
BB 3-bets to 150
And now we have the option to either 4-b to 360, or fold and give up the pot.
The EV of folding is 1440, so he should 5b the exact amount, so that our ev of 4betting is alfo 1440 right?
if we fold we lose 60 chips, if we 4b and he folds we win 210 chips. If we 4-bet and he 5bets, we lose 360 chips.
Is all this right? And if so, how do we put the final numbers together to calculate the frequency that he should be 5 betting with?
Thanks in advantage!
Yea so we're indiff between folding and 4-bet-folding if EV(fold) = EV(4betfold)
As you've said, EV(fold)=1440, and
EV(4betfold) = (chance Villain 5bets) (EV if he 5bets) + (chance Villain folds) (our EV if he folds) = X*(1140) + (1-X)*(1650)
where X is the chance Villain 5bets, and I've assume that if he doesn't 5bet he folds. (i.e. he never flat calls). To find the X that makes us indifferent, set those two EVs equal and solve for X.
Of course:
- our EV of 4betting changes if it is possible for Villain to flat our 4bet
- even if we're indifferent between folding a 4-bet-folding, flat calling the 3-bet could be better than both of those options
but both of those issues are beyond the scope of the current chapter.
Thanks for the Quick response and for clearing it up (:
I'm confused, once we open raise from the sb and the villain 3b us, I really think our EV from folding is 0, 1440 is the chips we end up with if we fold,
I mean the EV from folding vs a 3b is one thing and the expected chips we end with is something else, am I rite??
I do agree that for us to be indifferent between folding or 4betting vs a 3b we shguld be looking for EV (fold vs 3b) = EV (4b)
but once again I'm confused with the calculation of EV (4b), do we assume that he never flats our 4b??, do we always 4b fold??, if those 2 assumptions are correct,
then the math becomes easier, and we end up with this ecuation I believe: 0=EV (4b)= x * EV (4b folding ) + (1-x) * EV (him folding our 4b),
EV (4b folding) = -300
EV (him folding our 4b) = 150
So we get: -300x +150-150x= 0, and solving for x we get x = 1/3, so we're indifferent bewteen folding vs a 3b and 4b folding if he 5b 1/3 of the time.
Is this rite??
VHL
You're calculating your expected profit or change in stack size.
Will calculates the expected stack.
If you take your starting stack from the expected stack you get the expected profit.
So both methods will lead you to the same conclusions.
But for more complicated decision trees mapping out expected stacks at each payoff node is easier imo.
Will explains this approach very well in vol 1of his book.