I'm used to hearing that the unexploitable calling frequency on the river should be the proportion of your range that implied by the pot odds offered by the bet. i.e., a pot-sized bet offers 2-1 odds, so its unexploitable to call with top 2/3rds of your range; a half-pot-sized bet offers 3-1, so unexploitable calling range is the top 3/4s of your range, etc.

The Chen & Ankenman Mathematics of Poker book uses the simple (0,1) model game to show how value betting, bluffing and calling frequencies vary with pot size (chapter 11). I found this a really cool way of getting insight into the fundamentals of balanced ranges.

But I noticed that the calling ranges that are produced in the solution of this game are tighter than the above ranges. i.e. when facing a pot-sized bet, the calling range is the top 4/9s of your range, not 2/3rds; and when facing a half pot-sized bet, the game's calling range is the top 60% of your range, not 75%.

Does anyone have any intuition / explanation for why this is the case??

I haven't looked at it in a long time, but is there raising in addition to calling?

No, can only call or fold.

Defense vs a psb should be 50%, not 4/9, but I don't know the specifics of the scenario you're referencing. There are multiple errors in the book actually and I believe a thread somewhere on the forums documenting them, maybe in the books section of this site.

In 0,1 game and in real poker only bluff that has higest equity otr should be indifferent.
If all bluffs have 0% equity then you get MDF, but if you bluff hand that has some SDV opponent should defend MDF of the range that beats all the bluffs. That's why you call less in 0,1 game.

Also here is the thread I think:

https://forumserver.twoplustwo.com/1...atics+of+poker

To piggy back off this, the equilibrium will end up being such that the EV of bet = EV of check

if the EV of check > 0, then the situation haizem described would have you folding a bit more than alpha (bet/(bet+pot)) such that the EV of bet is also > 0 and the two end up being equivalent (the two being EV of check and EV of bet with said hand).

thanks, will take a look. i had noticed one or two typos in some formulas!

This is maybe a dumb question, but as a pot-sized bet is offering odds of 2-1 to call, isn't the orthodox defence 2/3rds not 50%?

First, its not a dumb question. Second, the answer is no.

In a perfectly polarized situation such that there are no blocker effects and all of Hero's calls beat the opponents bluffs, but do not beat any value, you would not fold at a frequency greater than (bet/(bet+pot)). (bet/(bet+pot)) is often referred to as alpha and Minimum Defense Frequency (MDF) is referred to as 1 - alpha.

For a pot sized bet, this is equal to: 1/(1+1) or 1/2. If you fold more than this frequency, then villain can profitably bluff any and all combos of bluffs.

In the situation where the bluff from villain has EV of check > 0, then the equilibrium would be such that the folding frequency equates to the EV of bet = EV of check... this frequency would be greater than bet/(bet+pot) in order to have EV > 0.


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Also, from the other perspective, the bluffing frequency of the better will be equal to the pot odds given to his opponent in order to achieve indifference/equilibrium. In the pot sized bet situation, you need equity equal to or greater than pot odds to have neutral EV. Pot odds would be equal to 1/(1+1+1) = 1/3 = 33%, so the overall bluffing frequency of the villain should be equal to this value. If he exceeds it, then you would have equity > pot odds and would always call. If he bluffed less than this frequency, then you would always fold. It is only in the perfectly polarized situation and when the bluffing frequency is equal to pot odds that you would be indifferent, and the EV of all calls would be 0, but you would only fold at a frequency equal to alpha in order to make opponent indifferent to bluffing all combos.

Thank you, that's super useful.

Also, just to tie this up with my original post, the solution to the game given in MoT is a calling frequency that is more complicated than 1 - alpha. It is: 2x(1-alpha)/[(2-alpha)*(alpha+1)]

I'm sure they go through the derivation themselves, but it should somehow be equal to the EV of the check.

So, if the EV of the bluff is > 0 and let us just say it is equal to x. EV of the bluff is equal to folding frequency*pot

you'd set EV of bluff = EV of check = folding freq*pot = x and just solve for that required frequency.