You're correct! If our villian has a perfectly balanced betting range (and we can't beat any of this value bets), then calling with our bluffcatchers is neutral EV (aka worth 0% pot), which leaves him with 100% pot.

Of course, in almost all situations villain either needs to check a number of hands or be terribly unbalanced, and he is doing much worse than capturing the full pot with his checking range.

The important thing to note is that villain can't do anything to improve upon the strategy you outlined (unless he has more hands that should be value bet or a number of small disclaimers like this). Regardless of how often he wants to bluff, he never captures more than 1x pot.

We also can't improve our expectation of 0% pot against the range he's betting, regardless of what we do.

The easiest way to look at this to start from villain's shoes.

Ask yourself how often does his pot sized bluff have to work to break even. In MOP they call this value a (actually alpha).

a= risk/(risk+reward)

So in this scenario:

a=1P/(1P+1P)
a=P/2P
a=1/2

Then, the amount of our range that we must defend is (1-a ) because we have to defend 100% of our range (1), minus how ever much he needs to break even (a).

When the value is pot, a= (1-a)=0.5

If you start by solving for "a" given the %pot (which is easier than jumping straight to what % of ur range u have to defend) using a=risk/(risk+reward), then getting from there to your own % is easy. --> Just subtract that value from 1.

People always confuse this math (which is separate but related to) "what does villain's range composition need to look like in order for my bluffcatcher's to break even".

That's really just an EV equation.

In this equation below, x denotes how often we win against his pot size bet, 2 is how many pot size bets we win when we do win (his 1PSB+ the 1PSB in the pot). (1-x) expresses how often we lose (strictly speaking it's "all the time minus the times we win"), and -1 represents how much we lose (our 1PSB that we called with).

You can then just solve for x after you substitute a value in for EV (in this case, we want to break even so we set EV--> 0.) if you're dealing with ICM or w/e you could set EV > 0.

EV= x(2)+ (1-x)(-1)

0=2x-1+x
-3x=-1
x=1/3

(1-x)=(1-1/3)
(1-x)=2/3

Remember we let x denote how often we needed to win and (1-x) denote how often we can lose.

Their values are:
x=1/3
(1-x)=2/3

So we can win only 1/3 of the time to break even.

Amazing.
Thank you very much RIO pros for your great responses. I'm so glad I made my post.
Happy grinding!
Later.