In this (https://www.youtube.com/watch?v=XlN2I4wLQ-w&t=447s) explanation of why you should balance your range. Doug shows that betting 100% value and 0% bluffs produces the same result as betting a balanced range of 66% value and 33% bluffs (for a pot sized bet) - the result is that you win the pot. However, he argues that your should choose the balanced option as in this case you get to bet more frequently, so you win the pot more often. So if you have 2 value bets and 1 bluff on the river. You could only bet the value and win the pot twice, or you could bet all three, be balanced, and win the pot three times.

However, when you crunch the number they don't seem to add up:

If you bet 3 times on the river, 2 value bets and 1 bluff, assuming villain calls 50% of the time (to remain balanced himself), you would win $200 from value bets, $100 from folds and $0 from bluffs (half the times it’s called the other half he folds and you win). Total $300

However, if you only bet your value hands. You bet 2 times, 2 value bets 0 bluffs. Assuming villain calls 50% of the times (to remain balanced himself), you win $200 from value bets, $100 from fold and $0 from bluffs (as they’re checked back). Total $300.

So in both cases, you win the same amount. In the first, you bet twice and won $300, in the second you bet three times, a higher frequency as Doug says, but you didn't win more, you still only won $300.

Am I missing something?

You are misunderstanding what he is saying.
Betting a balanced range wins the pot most often and **CANNOT BE COUNTERED**
Betting only for value wins the pot most often but can be countered.

He is comming from the assumption that if you only bet for value your opponent will fold 100% of the time.

I think his frame is not very useful in terms of looking at how hands in our betting range wins the pot on average, since obviously, value bets win more than the pot on average versus a balanced calling range, and bluffs win 0$.
It would be more intuitive if we framed it like this:
If you have a balanced betting range versus a balanced calling range, your valuebets make >pot$ and your bluffs make 0$.
If you only value bet, your opponet will adjust by always folding and therefor your valuebets only make pot$.

Those calculations imply that villain is calling the same frequency whether or not you are bluffing. If you're never bluffing, villain's best choice is to never call. You win $200 over the 3 hands. If you're bluffing once in every 3 hands, villain's best choice is to call 50% of the time, and you now win $300 over 3 hands.

Yh this is the exact conclusion I came to.

So would I be right in saying that there is no benefit to the balanced approach, besides the fact that it cannot be countered. So in a hypothetical situation where villain played perfect balanced and decided to never change (always just calling 50% to a pot sized bet), there would be no benefit to choosing one option over the other?

I'm not sure what he means by 'most often'. How can the balanced scenario and the value heavy scenario both win the post most often? Surely only one of these scenarios can win the pot MOST often. Or do they both win the pot an equal amount, and this happens to be greater than all other options?

Yes okay, so this is what I was thinking.

If villain starts to always fold, you win $300 in the balanced scenario and just $200 in the value heavy scenario. However, this assumes that villain adjusts

Doug gave the impression that the balanced scenario is superior to the value heavy scenario in general, regardless of villain's folding frequency (aka his balance). Because you get to bet more frequently and so win the pot ($100) more often?

Doug just uses the fact that villains will adjust to our imbalances as a given for all of his strategies, which is a relevant assumption when playing top high stakes regs HU for tens of thousands of hands

No, there would be no benefit, assuming villain doesn't change their approach. That's a big assumption though, and as soon as they do, you're losing 33% of what you would with a balanced approach.

I am surprised this has not been brought up yet:

Let’s make this a bit more abstract by using real numbers. So let’s say Villain’s range is (0,1) while Hero’s range is (-.4, 0) U (1, 1.6). Both ranges have measure 1, Hero is polarized.

If Hero bets all value for a pot-sized bet, he bets 60% of the time. Villain’s correct counter-strategy is to never call, so Hero wins the pot 60% of the time.

If Hero bets all value plus half his bluffs, he is now betting 80% of the time. But Villain’s best counter-strategy is still to never call—those calls will be losing against a 3:1 value:bluff ratio. Now Hero wins the pot 80% of the time.

Adding bluffs into a value-heavy range has allowed Hero to win the pot more often, *even though Villain’s counter-strategy has not changed*.

And I think the point here is that if Hero adds the optimal amount of bluffs he wins the pot 90% of the time, and this is the best Hero can do unless Villain continues to never call, in which case Hero can over-bluff, but that’s beside the point.

I don't quite understand what the numbers mean? What does (-.4, 0) U (1, 1.6) represent?

Oh sorry, it just is meant to represent ranges where the winner at showdown is the bigger number. So Hero can have a “hand” that is either between -.4 and 0, meaning Villain always wins a showdown, or between 1 and 1.6, meaning Hero always wins. It also implies Hero has the best hand exactly 60% of the time in this made-up scenario.

Ah okay, I see what you're saying now. So even though, heros bets increased from 60% to 80%, villain is still not getting the correct odds to call. So hero is winning the pot more often. But my question - what if instead of employing a counter strategy, villain just simply called at 50% frequency (the frequency he should call at to remain unexploitable against a pot sized bet). In this situation, it seems that hero earns the same amount of money, no matter if he is balanced or 100% value heavy.

Yes, everyone already responded to this