The first equation is % of pot to bet. So if you have 33% air, you should bet .333/1-2*.333 = 100% of pot. If you have 40% air, you will bet .4/1-.8 = 200% of pot. etc

I realize that wasn't clear at all, my bad on that one.

By the way, the last equation is indeed linear and very close to the function 2*(1-%Air) as stack sizes get large. This is because we will obviously bet all of our nuts (i.e. with 1-%Air) and when stack sizes get large, we get to close to having a 1:1 value to bluff ration.

I still think there is some confusion over what the strategy does, so I will work your examples and hopefully it will make more sense.

Example 1: Air% is 33%. We use graph 1 to find that we should bet 100% of the pot. And we are done since we know we will always bet 100% of our range when %Air is less than 50%. (Quick sidenote: one can easily verify that the EV of this strategy is the size of the pot regardless of our opponents action -- which means it must be an equilibrium strategy for us)

Example 2: %Air is 66%. Since %Air is >50%, we know we will NOT bet 100% of our range. So we need to determine what percentage of bluffs we can use. Therefore, we use Graph 2 and find as you did that the bluff frequency is approximately .4975.

Finally we use Graph 3 to find our total betting frequency (and therefore, the percentage of the pot we win!). As you correctly stated, we will bet very close to 2/3 of our range (since once again, we are close to the limiting situation where we bet 2*(1-%Air).

However, the bluffing frequency is the % of our betting range (not our entire range) that is a bluff, so in this case it will be ~2/3*.4975 and our value frequency will be ~2/3*(.5025). We are betting about 2/3 of our total range as we calculated above.

As I showed algebraically earlier, this strategy makes our opponent indifferent (i.e. 0 EV call) with his entire range of bluff catchers. We are also betting all of our value bets and as many bluffs as we can without letting our opponent have a +EV call. (This should be fairly obvious, because even if we bluff very slightly more often, our opponent would no longer be indifferent and in fact would call their entire range).

And as I've also shown, when our opponent is indifferent between calling and folding vs our bets, our EV is the size of the pot*total betting frequency. So this strategy which clearly maximizes our betting frequency in indifference must also maximize our EV.

Again, I strongly challenge you to propose another strategy and calculate its EV. Then compare it to the EV of this strategy.

Lot's of good posts, Spreek.

If Rob can do this then this conversation can near its close.

The question is bet size. Spreek has answered the question. The size is all in, whenever air is greater than 33 percent, which is nearly always, and that all river bets with stacks of 10000 and pot of 100 will be bluffs, 49.75 percent of all river bets.

Spreek has a system, and says he would play this way 10000BB deep stacks.

That is fine and good. And it is only fair that I give a answer for what I think is a GTO way to calculate bet size.

I will include EV, but I’m not sure I can improve on Spreek’s system where 100 percent of every river pot he ever bets belongs to him because reasons.

Just to be clear, the sizing is NOT all-in unless % of air is very close to 50% or larger. It is definitely not the case that we go all-in when we have 33.33% air.

As I mentioned in my previous post, the formula for bet size is based on % of pot. So if you have 33.33% air, one will bet .333/(1-2*.333) = 100% of pot. If you have 40% air, one will bet (.4)/(1-.4*2) = 2x pot. As you get close to 50% air, your bet sizing blows up to infinity.

And I do want you to recognize why our betting frequency*pot = EV. So let me try to explain it more clearly. Obviously, if we have the nuts we are going to bet. So whenever we check, we always have air and therefore always lose. Therefore, our EV given check is 0.

By law of total probability: EV = EV|Check*P(Check) + EV|Bet*P(Bet), and so we have that total EV = EV|Bet*P(Bet)

Assuming we use my formula, our opponents will have an exactly 0EV call. Of course, folding is also 0EV. Raising is of course suicidal for them in this scenario for obvious reasons. So whatever they do, their EV = EV|Call*P(Call) + EV|Fold*P(Fold) = 0*something + 0*something = $0

Since the $100 in the pot will not disappear (i.e. Our EV + Their EV = +$100 no matter what), the logical conclusion is that our EV|Bet is the entire pot size prior to us betting.

So total EV = EV|Bet*P(Bet) = Pot Size*P(Bet). Does this make sense to you? If not, please explain where I'm losing you.

I understand, completely. I your system, you will be placing a bet 75 percent of hands, and when you check, you lose. Of your bets, 49.75 percent are bluffs, 50.25 percent are nuts. Bet size is located on the bet size graph, which needs the label for percent stacks to be corrected to percent pot.

So, 75 percent of hands, you win the pot, and 25 percent of hands you win zero.

This is an overall EV of 75 percent of pot.

Now, if I am in fact lost, correct me before I propose a GTO approximation of river bet sizing.

In the case that we are using an all-in sizing: The % of hands we bet is determined by the % air in our range (i.e. from graph 3). I believe in the example we were doing, we had 66.66% air and so we bet about 2/3 of hands (i.e close to 2*value% = 2*1/3 = 2/3), not 75%.

(Of course, if we have 62.5% air, then we will indeed bet close to 75% of hands -- since 37.5% of our total range is value and we want to have close to a 1:1 ratio for value to bluff).

Everything else seems correct.

According to linear graph of the final function, you will place a bet 75 percent of total hands, since every hand with air percentage below 50 is bet, and bet frequency descends linear from 100 percent down to zero as air percentage goes to 100. And you have demonstrated that when you place a bet, the EV is pot, and when you check you lose.

Just want to make absolutely sure we are in agreement!

Assuming that %air > 50%, you will place bet with a percentage of your range based on graph 3 (which is close to the function 2-2*Air for large stack sizes)

But the percentage of our total range that we bet is obviously not always 75%, it depends on %air. So as I was saying, the statement is true only if you consider %air = ~62.5%. Of course it's fine if you want to specifically consider that case -- but your statement confuses me since we haven't been considering that case.

just to be clear, the function we evaluate in graph3 is literally the % of all hands in our range that we bet. So it is not necessary to do any further calculations after that to find our bet %.

Your entire air percentage range is represented in the graph.
The graph function is defined by stack size, pot size, and air percentage.
Thus, the graph holds true for all values of air involving stacks=10000 and pot=100.

When you have air equal to or less than 49.75 percent, you bet 100 percent of hands....

When you have more air>49.75 percent you bet less often...

When your air is 62.5, you bet with frequency 74.63 percent.

When your air is 66.66 percent, you bet with frequency 66.35 percent.

When your air is 75 percent you bet with frequency 49.75 percent.

When your air is 95 percent you bet with frequency 9.95 percent.

.....all the way to zero with air = 100 percent.


At first glance, this is very very close to 75 percent total bets placed on the river. This is the area under the line, and inside the 1x1 box.

The exact value happens to be .74875, which is the total percent of the time you place a bet on the river, across your entire air range, 0<air<100.

Ok I see what you are doing now. I don't think this is a good approach though.

First of all, it makes a dubious assumption that our %air is uniformly distributed on [0,1]. Second, obviously it is necessary and sufficient to show that there is a strictly superior strategy for some fixed value of %air. I'd really prefer that you try to do that for simplicity sake.

These assumptions are limitations of the nuts/air toy game. If I am to demonstrate a more robust GTO approximation that allows for balanced GTO participants on the river, you can’t call foul if my solution is +EV across said balanced ranges, by saying “my air is more/less dense than that.”

I mean that's not necessarily an assumption of nuts/air. But whatever, that's not even the main point.

A much much easier task for you to accomplish is to name a single value of %air for which you can describe a strategy with a higher EV than the one I named.

As I said above, you obviously have to show this anyway for the integral (or expected value over the uniform distribution) to be greater than mine.

And you have to calculate the EV at EVERY point anyway, so this is much much easier. Plus, we have no chance of confusing each other over calculus or analysis minutia.

Then I will proceed. However, a GTO approximate solution will not be in terms of nuts/air, but more applicable to real poker in terms of value and bluffs.

Certainly, a GTO approximate solution will have a ratio of value to bluffs, and a specific bet size.

Ugh, please just give your solution to the nuts/air game we are talking about or admit I'm right about it.

It's quite telling that you are always trying to add more complexity while still claiming I'm wrong about simple examples.

Nuts/Air is your answer to the question:

Pot is 100, stacks are 10000, hero IP and OOP checks. Both players are GTO. What is the bet size for hero?

This is not a simple question....

Since nuts/air is a well documented toy game, you are the one constraining complexity, to fit within your system of deciding bet size.

Nuts/air apparently wasn't well documented enough for you to understand it. I'm still not convinced you do, but I hope I was at least somewhat informative.

IMO it is absolutely pointless to discuss more complicated situations before one understands the basics. How is one supposed to discuss how to play a very complicated range when you don't know how to play the most simple one?

How are you even supposed to discuss what a GTO solution if someone refuses to accept basic ideas like indifference, the relationship between bet sizing and polarity and value:bluff ratios? If for unknown reasons someone refuses to accept the single best software tool we have for calculating nash equilbria ?

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No one can answer this question, because nobody solved heads-up no limit holdem. All GTO-Solvers are doing only partial calculation with limited depth analysis. Researchers at Carnegie Mellon University have poker AI player, which plays better than top human players. So they can give you the closest value – But all other can only guess.

On the river, if the pot is 100 and effective stacks are 10000, OOP player checks and IP player decides to bet, what is the size of the bet?

Both players are perfect GTO, both players know the optimal solution.

No other delimiters are placed on the question.



With specific regard only to the nuts/air model proposed by Spreek, some assumptions are made:

1) When hero has air percentage below 50 percent, all his bets with the nuts win, on average, with EV the size of the pot. Both GTO players know this to be true, and any calls are zero EV. Thus hero EV in this region is the size of the pot.

2) Hero has to decide how often to bluff when his air percentage is above 50 percent. In this region, hero will win the pot when villain folds, and will lose the bet when villain calls. All of this averages out to hero having EV=Hero Bet frequency in this region.

3) In fact, for all values of air, the EV=Bet frequency across the entirety of hero range.

After double-checking several available resources on this model, and reviewing Spreek's explanation of this model, I think the assumptions above hold true.

So, first some observations about this model. Bet size selection is a function of the percent air in range, and varies drastically with small movements within the nuts region, and then shoots to all-in whenever hero places a bet from the air region. At 30 percent air, we bet 75. At 33 percent air we bet 100, 40 percent air we bet 200 and from 40 to 50 percent we jump to 10000. So, any casual GTO observer will know that hero is betting from his nuts range when he bets less than 10000. Whatever the range composition of hero is, villain knows this also.

There are no secrets in GTO, only combinations of mixed strategies and pure strategies.

Now, I do not hold this model to be optimal, but a dominant strategy would be:

IF air percentage is less than 50, bet full range with bet size = the pot.

IF air percentage is greater than 50, bet frequency=(2-Air) with bet size = the pot.

The number 2 is the value of the betting frequency function with the pure nuts range (air=zero) which is achieved when we grant the players Doctor Evil level of infinite stacks. In fact, if we calculate this number for the more pedestrian stacks of 10000, it is 1.99. So, we start evaluating hero betting decision with his nut hands, and work our way down to pure air. If theories that solve for infinite bet sizes are considered valid, then we can also give the players lolfinity stack sizes. Let's get all that EV!

This strategy does not require bluffing 10000 into 100, and does not require the strategy of betting smaller when our range has less than 1/2 air.

This conjectured +EV strategy has the following advantages.

1) Bet size is the same, when hero bets from either region of his nuts/air.

2) Every two bets from the Air<50 region will be matched with a bluff from the air>50 region. This keeps villain indifferent. If hero has a non-uniform distribution of air in his range, villain will know this and adjust accordingly.

3) This is maximum EV, including the EV of infinite stacks, and higher than the previous strategy for all bets placed in the air>50 region, by just a tad, not enough to really matter. This tiny +EV along with the other more serious obvious strengths with regard to exchange of information make this strategy the more optimal of the two, about as optimal as a nuts/air toy game of poker can be.

4) This strategy is more resistant to villains who may make errors with overcalls. We do not expect any errors, but when two strategies are conjectured optimal, accounting for MORE LIKELY mistakes is a plus. Here, if villain makes an overcall, our exposure to this is limited to 100 instead of 10000.

5) Possible objections to the newly proposed strategy will include the question of bet size being much less than 10000. Among GTO players, there will never be a call made that is -EV. Also, among non-GTO players who may make erroneous overcalls, our bet size disguises where in our range our bet was placed from. And as stated previously, our bets are from the air<50 region twice as often as the air>50 region.


Verification of the improvements in EV are:



Air Percentage = 60 percent

Strategy 1: Bet 10000 with +EV of 79.60

Strategy 2: Bet 100 with +EV of 80.00



Air percentage = 50 percent

Strategy 1: Bet 10000 with +EV of 99.50

Strategy 2: Bet 100 with +EV of 100



Air percentage = 20 percent

Strategy 1: Bet 33.33 with +EV of 100

Strategy 2: Bet 100 with +EV of 100



So, one question. Where does this number of Bet=Pot come from? It is not allowed to just imagine a number and think it sounds right and put it into a strategy. Well, it turns out, this number is very old, and probably represents one of the first calculated bet sizes in the history of poker theory. This is the solution to the first toy poker game in the book 'Theory of Games and Economic Behavior' by John von Neumann and Oskar Morgenstern, (1944 Princeton University Press). The reason it works so well in this nuts/air game is that in the book, OOP was also forced to check, and could only call. Neat!



To solve a toy poker game with variable bet sizing, as well as more complex extensive form models, there are several steps. The steps go in this order....

Define all actions that both players may take, with any frequency above zero.

Define the relationships of these actions.

Create ranges that include these actions.

Solve for the boundaries of these sections within range, via indifference.

Double check the optimality of the conjectured strategy in piece-wise linear graphic form.

Choose between multiple admissible optimal strategies via possible player mistakes, if any.

Solve for the expression of the value of the game, in terms of bet size(s).

If multiple bets are allowed, start with the last bet placed, and resolve this equation in terms of the first bet placed.

Set the derivative of the resulting bet equation equal to zero and solve the resulting equation for B.

Use this bet size to solve for the other bet sizes, if any.

Now insert these bet sizes into the indifference equations and find the resulting optimal indifference frequencies.

Thus, the game is solved and the last number you calculate is bet frequency.



This is why you cannot learn about bet sizing from a tool that requires you to enter bet size.

Also, when you pick an EV you want, calculate the frequencies that yield this EV, and THEN solve for bet size, you get lolinfinty bet size.



-Rob

Robert, your posts in this thread are borderline ridiculous.

The nuts-air toygame is not that difficult to understand. You clearly have some background/experience with math. There's no reason you can't easily understand the math of the nuts-air game.

Yes, you are correct that the nuts-air game is not an accurate representation of actual poker. It's a simplified version with, as you said, constrained rules. But this fact (nuts-air not being real poker) does not mean it's impossible study and calculate solutions for the nuts-air game. Those solution have been correctly presented to you in this thread and you keep rejecting them or dismissing them as nonsensical.

I have seen the bet size in nuts/air calculated as infinity. This is incorrect for the reasons in my post. Sorry if this is ridiculous.

It does trend towards infinity. Your understanding on this is wrong. Also your post is riddled with inaccuracies. One of which,

Strategy 2 of betting pot (100) only yields +EV of 75 and not 100. Much less than strategy 1.

EV = Bet Frequency*Pot ONLY if villain has a 0EV call.

In your proposed solution this is not the case at all.

If you have 60% air, you have 40% value. if you bet 80% of your range, you are betting 50% value. Your opponent has 50% equity with their entire range of bluff catchers. They call whole range and win 50% of the pot. Your EV is 50% of the pot. The EV of my strategy is much higher.

If you bet pot with a balanced range (i.e. with 2:1 value to bluff ratio), then you are only betting 60% of the time. In this case, your EV is really BetFreq*Pot, but again, my strategy has higher BetFreq so it is superior.

And dude, %air is a fixed value. both players know it once they get to river. It's not a random variable, so it makes no sense to try to balance your range by treating it as one.

This statement is pure nonsense. Your opponent knows your percentage air. it's not a random variable. If you use a mixed strategy, they know the expectation of your % air. you can't balance like this because they know your full strategy.

i.e. They will use the following strategy: Ask you your %air. You tell them its an average of 60% (or whatever). They ask you which hands you are betting. You tell them 80% of which on average of half of them are value and half of them are bluffs.

They say lol, call their whole range, and win half the pot.

Ahhh, now I see it. My strategy does not bet enough to force villain to fold when Air>50 percent. Then I will have to bluff less often when the exact fixed and known air percentage on a particular river is above 50, if I want to fix the size of the bet to pot. Does my system still win the whole pot when Air<50? I don’t see a GTO bluff catch calling in that area.

if air is less than 33% yeah.

if air is lets say 40%, then you are betting pot with your whole range. your opponent has 40% equity and they need >33.33% to call a pot sized bet, so they still have a +EV call with their entire range of bluff catchers.