Do we always have to giveup some of our value hands against a raise?
Assume the polar [0, 1] river toy game, where oop is Villain and ip is Hero.
Some variables:
p: pot size
b: bet size
r: raise size
Hero bluffs with probability: b/(2x b + p)
Villain has auto-profit if Hero folds more than r/(r + b + p)
Can we find some r such that Villain has auto-profit?
r/(r + b + p) < b/(2x b + p)
<=> r x (2x b + p) < b x (r + b + p)
<=> 2x br + rp < br + bb + bp
<=> br + rp < bb + bp
<=> r x (b + p) < b x (b + p)
<=> r < b => not possible to raise less than the initial bet size
In other words: Hero always has to fold some of his value hands. Is this statement right?
2 Replies
Yes.
Restating your math, the pot odds of the bet is always less than the alpha of the raise.
So if your initial bet is balanced, then there is no way to defend more then MDF facing the a raise, (assuming you always fold bluffs).
Then again, if OOP's polarity is off, hero could over-defend by 3-betting some bluffs rather than folding.
Restating your math, the pot odds of the bet is always less than the alpha of the raise.
So if your initial bet is balanced, then there is no way to defend more then MDF facing the a raise, (assuming you always fold bluffs).
These two statements seem to be mutually exclusive. Did you formulate them correctly?
mdf_raise = 1 - alpha_raise
-----------------------------------------------------------------
alpha_raise > pot_odds_bet =: prob_bluff_bet
1 - alpha_raise < 1 - prob_bluff_bet
-----------------------------------------------------------------
==> mdf_raise < 1 - prob_bluff_bet = prob_value_bet
Therefore, Hero has the potential to defend more than mdf_raise by calling all of his value hands. In other words: Hero has to fold all of his bluffs plus some of his value hands to meet gto conditions.
