Please excuse me for stating theory as fact once again. I have these ideas in my head and I need your help.

I was tinkering with numbers trying to find the common thread that links betsize, ranges, and bluffing frequency. Here's what I came up with:

In order to find the betsize that maximizes expectation, I think this series of equations works:

Assume: pot = 1

Count the combos of bluffcatchers.

Count the combos of nuts.

Divide bluffcatchers/nuts

The result = the final potsize when called

Final potsize = (pot+bet+bet)

Bet = the betsize that maximizes expectation

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Notice that (pot+bet+bet) is also a part of the bluffing frequency equation:

Bet/(pot+bet+bet) = bluffing frequency

From here we can explore how these equations interact by substituting bluffcatchers/nuts for (pot+bet+bet):

(Bet*nuts)/bluffcatchers = bluffing frequency

(Bluff frequency* bluffcatchers)/nuts = bet

(Bluff frequency*bluffcatchers)/bet = nuts

(Bet*nuts)/bluff frequency = Bluffcatchers

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I'm not sure where to go with all of this so in conclusion here are some emoticons:




[Mod Note] Added to OP at request of Bob148]

After toying with some more math, I've come to these preposterous conclusions:

If bluffcatchers/nuts > 5/4 then bet according to the formula given in the op.

If bluffcatchers/nuts < 5/4 then check.

I have a bad headache now so I'm not gonna post the proof but maybe later I will. It works out to never betting < 0.125 pots.

Now comes the fun part:

When ranges overlap, the equations become more complicated:

Bluffcatchers/(equity*combos of non junk non bluffcatchers) = final potsize

Equity of non junk non bluffcatchers needs some explaining:

Take the average equity of these hands vs the bettors betting range and multiply the result by the number of combos in the callers non junk non bluffcatchers range. Let's call this the nuttyness of the callers range.

I'm not entirely sure what you're trying to ascertain, but there's a quadratic equation in Janda's book (and MOP, I believe) that shows the "perfect betsize" for the amount of equity you have on the river.

0 = 1 – (2Y)(1+2X + X sqrd) where X is our bet size in PSBs and Y is how often we "effectively lose".

FWIW, in one of Janda's videos, he said that the "perfect bet size" to maximize EV when we win 90% of the time is 1.23*PSB.

Is that along the lines of what you're trying to calculate?

does that equation suppose a value range first and then determine betsize based on that? If so, I think that's wrong, but I'm clearly biased.

I'll elaborate on why I think it's wrong:

If y = how often we lose, then that means our value range is predetermined. I think this is incorrect because our value range should be dependent on our betsize. If our value range is determined before we determine our betsize then we will either miss profitable value bets or value bet too thin.

Instead I think we must first determine betsize and then determine which hands can profitably value bet with that sizing. Then we can add bluffs depending on the number of value combos we hold.

I now disagree with this part because it requires us to know the bettors betting range.

I explored a scenario with overlapping ranges:

Out of position has 75 combos of (.5) and 50 combos of (1).

In position has 2 combos of (1) and 1 combo of (0).

Bluffcatchers = 75

Nuttyness = (50*.5) = 25 effective nut combos.

Bluffcatchers/nuts = 75/25 = final pot

Final pot = 3 = (pot+bet+bet) = (1+1+1)

Bet = 1

We should bet pot in position with our whole range.

Out of position player calls 50 combos of (1) and 12.5 combos of (.5).

Here are the payouts:

(.5*2) + (.4*.5*.3) + (.1*3) = 1.36 pots of revenue.

Revenue - investment = profit

1.36 - 1 = 0.36 pots of profit for the in position player.

Fixed my math because a typo got my head spinning.

There’s gotta be some more to this. EV in the simplest lead bet case with pot =1 is

Ev=eq(1+2Bet/Pot)-Bet/Pot =(Bet/Pot)(2eq-1)+eq

Clearly for eq>0.5, EV increases with Bet/Pot.

Does Janda consider fold equity or something else? What is “perfect”?

This is where you lost me...

Earlier you say bluffcatchers/nuts = final pot size....I'm with you there.

So now you are substituting nuts with equity*combos of non junk non bluff catchers to come up with a different final pot size....are you reducing the number of nut combos in the bettors range in order to offset the nuttiness of the callers range ?

See above post #6. I disagree with the post you quoted because it required that we know the bettors betting range. However, I explored a scenario with overlapping ranges and I now think that "effective nut combos" = (combos/2) if the bettors value range is reduced to (nuts).

"So now you are substituting nuts with equity*combos of non junk non bluff catchers to come up with a different final pot size....are you reducing the number of nut combos in the bettors range in order to offset the nuttiness of the callers range ?"

Something like that, yes.

I explored another scenario with overlapping ranges.

Out of position has 2 combos of (1) and 5 combos of of (.8).

In position has 6 combos of (1) and 4 combos of (0).

I'm not sure but twice I've done the math for a 2x pot bet. Both times I got an ev > pot for the in position player. Either my math is bad or the out of position player should fold more than minimum defense frequency would suggest. Most likely the former.

Don't make it too complicated.

- Use Arty's formula to solve optimal betsizes, deduce the raise freq from equity if raising allowed.

- Tadaa! we got optimal betsizes for each valuehand

- Do basic math to see how much EV different hands lose from using nonoptimal betsize for the exact hand. Decide what betsize you want to use for your range.

Ok so Janda's formula gives an optimal bet size for each hand in the value range. What if you want to bet the entire value range rather than a specific hand....would you take the weighted win frequency for each value hand and subtract it from 1 ? This would give the weighted frequency that OOP player effectively wins, so a weighted average for the variable Y.


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For example the first scenario given by OP was

""
Out of position has 75 combos of (.5) and 50 combos of (1).
In position has 2 combos of (1) and 1 combo of (0)
""

What if the ranges were instead:

""
Out of position has 75 combos of (.5) and 50 combos of (1).
In position has 1 combos of (1), 1 combo of (.8) and 1 combo of (0)
""

The 1 combo of (1) will lose half the time it is up against the 50 combos of (1) - split pot ! It will win all the time it is up against the 75 combos of (.5). So this combo will lose (50*1/2)/(50+75) = 20% of the time. Let's call this Y1

The 1 combo of (.8) will lose 50/(50+75) = 40% of the time. Let's call this Y2.

When betting his value range the in position player will effectively lose (20% + 40%)/2 = 30%. We can call this Y for the range.

So plugging this into Janda's formula should give us an optimal bet size for the range.....is this correct ?

well just calculate how much each valuehand loses by using nonoptimal size and their portion of your range, should be quite simple.

Im editing the scenario a bit, not sure how to deal with splits lol, and cba figuring it on this second.

OOP: 75 combo of 0.50, 50 combos of 0.90, and always checks to IP.
IP: 1 combo of 1.00, 1 combo of 0.80, infinite combos of 0.00 (making it a bit simpler), we don't focus on the total range vs range GTO, just the vacuum of IP after OOP checks 100%.

Lets say stacksizes are 5x pot, no raising allowed.

Equities for the IP hands are 1.00 has 100% equity, 0.00 has 0% equity, and 0.80 has 60% equity.


Optimal betsizes for diff hands for IP

1.00 = 5x pot
0.80 = 12% pot.


So if we only want to use one size, it's pretty intuitive that best option is that we check 0.80 hands and just 5x pot 1.00 hands. If we bet bigger than 25% pot with 0.80, checking becomes higher EV and we just choose to check instead.

If we use 25% sizing with 1.00 hand, our EV is 1.2x pot, if we use the 5x pot, our EV is 1,833 pot.






Hmm, not sure about your formula, don't think you can solve it like that. Yeah, it's definitely not correct.

The amount of times .8 will lose depends on the betsize. You bet bigger, villain calls less of the .5 combos, and we lose more often. If we check or bet infinitely small, and villain calls all of his range, only then would we win 60% time.

If your ev is > pot, something is wrong with the out of position players calling range.

The EV's are for hands, not ranges.

Ok got it.

Forgot to multiply by value betting frequency.

Incomplete math was my mistake. Here are the payoffs:

Out of position calls 2 combos of (1) and 0.8 combos of (.8).

In position bets 2 pots with 4 combos of (0) and 6 combos of (1).

(2/7 * 5 * 0.5) + (0.8/7 * 5) + (4.2/7 * 3) = 7.6/7 pots of profit

Profit * value betting frequency = average profit

(7.6/7 * 3/5) = 4.56/7 pots of average profit, or just over 65% pot.

The river is the street on which you lose the most ev if you try to use only 1 bet size. A perfect solution to NL river play would likely involve a huge number of different bet sizes all tailored to specific hand strengths.

In practice we can get away with about 3 or 4 sizes but using only 1 would be a huge mistake.

What happens if we allow the opponent to raise?