Monday, November 17, 2014

Improving your turn play with GTO -- Check raising vs leading on the turn

Today I'm going to get into a practical example of how to use GTO theory and GTORangeBuilder to directly improve your play in a common real world situation that many players struggle with.  Usually I tend to keep my blog posts more purely focused on theory, but I've gotten a number of requests for a more practical post that more directly mirrors what I work with my students on during coaching sessions.

For the actual analysis I decided a video demo would be the most instructive, see below.  You can browse the solution that is discussed in the video here:

Sunday, November 16, 2014

GTO is so much more than unexploitable

One of the most common misconceptions that people tend to have regarding GTO poker play comes from the idea that somehow the key element of a GTO strategy is its "unexploitability" or "balance" and the belief that any unexploitable strategy is inherently GTO.

The conditions required for a strategy to be GTO are much stronger than simple unexploitability (although of course any GTO strategy must be unexploitable), and in a practical sense, the elements of GTO play that are generally going to be the most valuable to try and use in real world poker games are the elements that have nothing to do with unexploitability.  By focusing on unexploitability people minimize and miss what is actually the a huge part of the value of understanding GTO play.

Today I'm going to take a look at why people have come to often confuse the idea of unexploitability with GTO and go through the core definitions and a simple example that illustrates the key difference between a GTO strategy and one that is only unexploitable.  This will also serve as a nice lead in to my next post which will present a practical example of how to analyze and improve your 6-max OOP turn play in raised pots by better understanding GTO.

Toy Games and GTO

GTO play often gets confused with unexploitable play due to the fact that in the very simplest toy games the two are equivalent.  People learn the solution to the toy games, without fully understanding the definition of GTO and assume that they now know what GTO means.

Games like Rock Paper Scissors, or the Clairvoyance game from the mathematics of poker only have a single "reasonable" unexploitable strategy, which happens to also be fully GTO, which means that people who are new to game theory are prone to mistakenly assume that GTO and unexplotable are equivalent.

Furthermore, in these toy games, you can solve for that GTO solution using only indifference conditions (which only can be used to identify unexploitability) and thus the mechanism for finding the solution reinforces the idea that unexploitability is all there is to GTO.  In fact, most arguments I've heard against GTO play stem almost entirely from generalizing from the Clairvoyance game to all of poker without any thought to the idea that a game that is trillions of times bigger might be fundamentally different.

This happens because when solving toy games we usually automatically discard strategies that might be unexploitable but intuitively are obviously dumb.  However, this completely breaks down in extremely tough games because the "obviously dumb" decisions are no longer at all obvious, and in fact identifying and avoiding these "obviously dumb" leaks in large games is where a huge amount of the value of studying GTO play comes from.


Lets go back to the core definition of a GTO strategy, which is a strategy for a player that is part of a nash equilibrium strategy set.  In a 2 player game a strategy pair is a nash equilibrium if, "if no player can do better by unilaterally changing his or her strategy" (source wikipedia).

How does this actually tie into the concept of exploitability and in a technical sense, what does exploitability actually mean?  The idea of exploitability is relatively intuitive, if your strategy is exploitable, it means that if your opponent know your strategy they would be able to use that information to alter their own strategy in a way that would increase their EV against you.  Formalizing the above, gives us an accurate definition of exploitability, but we need to define one more concept first.

A "best response" (sometimes called a "maximally exploitative strategy", or "counter strategy") to a given opponent strategy is a strategy that maximizes our EV against that opponent strategy, assume that his strategy is completely fixed.

Exploitability can now be defined (and measured) as follows.  Call G our GTO strategy, and S our opponents strategy.  Call B the best response to S.  S is exploitable if our EV when we play B against S is higher than our EV when we play G against S and that EV difference is the magnitude of the exploitability.

Intuitively, this should make perfect sense, our opponents strategy is only exploitable if we can alter our own strategy to exploit him and increase our EV, and the amount of EV we can gain when we maximally exploit him is an accurate measure of the magnitude of his exploitability.

Any GTO strategy must be unexploitable to satisfy the definition of a nash equilibrium, but in complex games there are usually infinitely many inferior unexploitable strategies that are not GTO.

A GTO strategy, is a strategy that is using every possible strategic option and every synergistic interaction between various hands in our range to maximize our EV while also still being unexploitable.  In most real world cases, understanding which of our strategic options are strong and how to correctly leverage that strength against our opponent in ways they cannot prevent is what makes GTO play powerful.

An Example -- GTO Brainteaser #6

Armed with our definitions we can now look at a very simple example of a unexploitable, non-GTO strategy.  Keep in mind that even this example is a relatively simple toy game and that the real game of poker generally has infinitely many unexploitable strategies that pass up EV and are not GTO for far more complex reasons.

We're going to revisit the model game from GTO Brainteaser #6.  The setup is as follows:

There are 2 players on the turn with 150 chip effective stacks, and the pot has 100 chips.  The Hero has a range that contains 50% nuts and 50% air and he is out of position.  The Villain has a range that contains 100% medium strength hands that beat the Hero's air hands and lose to his nut hands.

For simplicity, assume that the river card will never improve either players hand.

The hero has 2 options, he can shove the turn or he can bet 50 chips.  If the hero bets 50 chips on the turn he can then follow up on the river with a 100 chip shove.

As I show here it turns out that GTO play for the hero is to bet 50 chips on the turn and then 100 chips on the river with precisely constructed ranges that contain the right relative frequency of nuts and air, and to check/fold with the rest of his air.  Following this strategy gives the villain an EV of 11.11 chips.

GTO play for the villain is to call a 50 chip turn bet or a 100 chip river bet 2/3rds of the time and to call a turn shove 40% of the time.

Now consider the non-optimal strategy S where the hero shoves the turn 100% of the time with the nuts and 60% of the time with his air, and check/fold sthe rest of his air.  It is easy to check that the EV of this strategy is 20 chips for the villain.  So we are giving our opponent almost double the EV by playing this weaker strategy S where we jam the turn.

Clearly S is not GTO, we could unilaterally increase our EV by switching to the GTO strategy, because S is fundamentally not wielding our range and our stack to optimally prevent our opponent from realizing his equity.

However, S is completely unexploitable.  Because our turn shoving range is "balanced" to be 3/8ths bluffs our opponent is exactly indifferent between calling and folding to a shove so he cannot increase his EV by switching from his GTO strategy to a maximally exploitative strategy.

Someone looking for an "unexploitable" strategy might be happy with the strategy S, but in this case, S misses the entire practically valuable lesson that we can learn from the model, which is that by betting half pot twice we actually utilize our polarized range much more effectively than we do by shoving it, to the extent that we cut our opponents EV approximately in half.  The entire point of the example and its power is completely lost if we focus on unexploitability rather than on EV maximization and the true definition of GTO.

In fact our EV if we play an exploitable strategy where we bet 50 chips and then jam the river for 100, but with slightly incorrect value bet to bluff ratios is much higher than it is when we play the unexploitable strategy S, even if our opponent perfectly exploits us.

Similarly, when analyzing complex real world situations, focusing on unexploitable play is generally going to completely miss out on valuable lessons that a thorough study of GTO play has to offer.

Special thanks

This post was actually largely inspired by an email from a GTORangeBuilder user who was confused that using CardRunners EV's "unexploitable shove" option didn't give him a GTO strategy like GTORangeBuilder does.  Of course the feature does exactly what it says, it gives a range that is the best range for us to shove if our opponent perfectly counters our shove, which in no way suggests that the shoving range is actually GTO.

Wednesday, November 12, 2014

GTO Brainteaser #9 -- Multistreet Theory: Range Building with Draws

This week we are going to examine a multi-street game, starting from the turn where one players range consists of the nuts and draws while his opponents range consists of a medium strength hand that never improves to the nuts.   I'll be going over this game in depth and providing a solution in part of my next CardRunners video which should come out in a few weeks.

Not that this is almost identical to the game that we examined in GTO Brainteaser #6 except that in this case, the hero's turn bluffs have the potential to improve on the river.  In both games the player with the merged range has about 50% equity (in this game he has 53.8%).

The setup is as follows:

  1. There is a $100 pot, and we are on the turn with $150 effective stacks.  
  2. The board is AsTs9c2d
  3. The hero is in position and his range is 1/3 nuts, 1/3 straight draws and 1/3rd flush draws, specifically, the hero holds either 7s3s, 8h7h, AcAd.
  4. The villains always holds a medium strength hand, KcKd
  5. Both players are allowed to either check, bet 50% pot, or shove on both the turn and the river.
  6. Your opponent plays GTO
The villain offers to always check the turn to you if you pay him $3 should you accept his offer?  What does GTO play look like in this game for both players?  Is this game higher EV or lower EV for the villain than the nuts/air game from GTO Brainteaser #6?

Would the hero be better or worse off if his range was 1/3 nuts 2/3 flush draws?  How would optimal play change?

Thursday, November 6, 2014

GTORB Turn Launched -- Multistreet theory video coming soon

The latest version of GTORB can now solve turn scenarios and is available for purchase!  For a sneak peak of what a solution looks like see this post.

I've been putting most of my time towards the turn solving code lately so I haven't had time to do as much blogging / video creation, but now that the turn solver is released I'll be releasing a series of blog posts as well as a CardRunner's video on multistreet GTO theory with example turn solutions over the coming month, stay tuned!

Monday, October 20, 2014

GTO Brainteaser #8 Bonus Solution -- Optimal Betsize Calculations

I got a request for a numeric answer to the bonus of GTO Brainteaser # 8 which involves solving for an optimal bet size in a two street bluffing game.  The solution to the non-bonus question is here and is worth reading first.  I haven't actually done a post on deriving optimal betsizing in multistreet play before so I thought it would be useful to demonstrate the mathematics involved.  I'm not going to restate the game structure again here so please check out the original problem statement if you are not familiar with the original game.

The basic technique for calculating optimal bet sizing is as follows.

  1. Rather than using a fixed betsize in your calculations, make the bet size a variable and solve for GTO strategies as a function of that variable
  2. Compute the EV of the game when both players play the GTO strategies as a function of the bet size variable
  3. Maximize the EV of the person making the bet with respect to the bet size variable.  That is your optimal bet size.
While this technique is quite simple conceptually, the actual algebra involved can be hairy so I usually just make wolfram alpha do it.  So lets get started.

To calculate the optimal betsize we will make a few assumptions that are reasonably easy to verify and that I have shown in other posts / videos.

  1. The hero should always bet the nuts on the turn.  This allows him to "compound the nuts" over multiple streets which as I showed here and in more depth in my CardRunners videos is always +EV compared to betting on a single street with a polarized range.
  2. On the river it will always be most profitable to shove with the nuts with our polarized range.  This is quite simple to prove and I showed it in my first CardRunner's video.
  3. The hero's EV with his air on the river will be 0 unless z is so large that it is optimal for the villain to always fold the river.  However, clearly if the villain were always folding in a spot where the hero might hold air on the river, he would never call a turn bet, so any time a turn bet is called, the hero's EV with air must be 0 on the river
Combining observations 1 and 2 we can parameterize our bet sizing strategy with a single number x, the number of chips that we plan to bet on the turn.

If we bet x chips on the turn, then we know we will jam the river and bet the rest of our chips.  Given a starting turn pot of 100 chips if we bet x chips on the turn, the river pot will be 100 + 2x when we are called.

Our river jam will thus be a bet of  (150 - x) into a pot of 100 + 2x.  This means that we will be making a 

b = (150-x) /  (100 + 2x) percentage pot bet on the river.

Now lets call the frequency of the villain calling the turn c.  Observation 4 tell us that since the hero's river EV is 0 with air, his EV for bluffing the turn is very simple to calculate.

EV[turn bluff] = (1-c) * 100 - c * x

Clearly the villain always folding or always calling the turn is highly exploitable so we know his turn calling strategy is mixed and we can apply indifference conditions to see that 

c = 100 / (100 + x)

What about the villains calling frequency on non 3d/2c rivers?  This will of course just be determined by indifference conditions that depend on the pot / bluff size.  Call the villains river calling frequency rc.

EV[river bluff] = (1 - rc) * (100 + 2x) - (150 - x) * rc

Indifference conditions imply that

rc = (2x + 100) / (x + 250)

Now we can write the optimal turn bluffing frequency as a function of the turn bet size as well by looking at the villains EV for calling.  I calculated the villains EV of calling when the bet size was 50 chips in my previous post but I will duplicate the calculation here, assuming that the hero is bluffing with frequency with his air and always betting the nuts.  This means that (1-z) = a / (1 + a) of his betting range is air and z = 1 / (1 + a) of his betting range is the nuts.

Since the hero's EV with air on all rivers is 0, when he bluffs and we call we win the 100 chip pot plus his turn bet size in EV.  When he holds the nuts on 3c/2d runouts our EV is 0 on the river and on other runouts our EV when our opponent holds the nuts .

EV[call] = (100 + x) * (1-z) + z * (2/44 * -x + 42 / 44 * (-x - rc * (150 -x)))

If we apply indifference conditions to say that the EV of a call must be 0, this a relationship between z and x that we can solve for z.

Wolfram Alpha is much better at algebra than me so I just computed that relationship here.

Now the EV of the game for the villain is just how often the hero checks the turn, which is just (1-a) / 2, because by indifference conditions, when the hero bets the villain is indifferent between calling and folding and thus his EV is 0.

Since z = 1 / (1 + a), a = (1/z) - 1, so (1 - a)/2 = (2 - 1/z) / 2

So the EV of the game is (2 - 1/z) / 2 for the villain and we know z as a function of x.  Thus the optimal bet size for the hero is the value of x that minimizes his opponents EV, (2 - 1/z) / 2, where z is between 1/2 and 1 (because our betting range is at least 1/2 nuts and at most 100% nuts.  Since this is clearly decreasing in z, we just need to minimize z.

Again I calculated this using wolfram alpha here.  The result is that the optimal bet size is 52.69 chips.  This intuitively makes sense, as we would expect to bet slightly larger on the turn with some of the river runouts killing our action than we would without that risk.

The EV of this game is ~87.89 so the EV gain by changing betsize in this case is tiny, about 0.02 chips.

Thursday, October 16, 2014

GTO Brainteaser #8 Solution -- Multistreet Theory vs Practice

In this post I'm going to discuss the solution to GTO Brainteaser 8, check it out here if you missed it.  I'm also going to provide some introduction to multistreet theory and some simple examples of understanding the impact of runouts and the flow of information across streets.  A browseable GTORB version of the turn solution is also presented near the bottom for those of you interested in a sneak peak at the GTORB turn solution interface (it still needs some polish).


The brainteaser involved studying the following game:

  • You are on the turn and the board is  AsAhKsKh
  • The hero a hand range of AcAd and 3c2d
  • The villain a hand range of KcKd
  • The pot is 100 chips and stacks are 150 chips
  • The hero can either bet 50 chips on the turn and 100 on the river or he can shove for 150 on the turn.

  • The goal was to determine how and why this game was different from the nuts vs air multi-street game that we looked at in GTO Brainteaser #6 and from the multi-street polarized vs merged range theory in the mathematics of poker.

    The basic answer to this question is quite simple, the real world game with an actual deck is worse for the hero than the model game from GTO Brainteaser 6, because when a 3c or 2d hits on the river it reveals to the villain that the hero must hold the nuts.  The villain is able to convert this information into money by folding 100% on either of these rivers.  Perfectly polarized ranges are always the strongest possible ranges, so for the hero, having his range depolarized on the some river runouts decreases his EV.

    Note that the villain also gains information when an Ac or Ad comes on the river, but this information is not valuable because our EV when we hold 3c2d is 0 anyways.  A GTO opponent calls enough to make us indifferent between bluffing and folding, so if we are forced to always fold that doesn't actually decrease our EV.  In this case the villain still gains information but has no way to convert it into money.

    We can calculate the exact EV decrease quite simply.  1/2 of the time we hold AcAd and if we recall the solution to GTO Brainteaser #6, our EV with AcAd in this spot without river runouts giving away information is 16/9th of the pot.  Now when we hold AcAd, since we know our opponent holds KcKd, there are 44 river cards that might come and 2 of them reduce our EV to 1.5 pots in the case where our opponent calls our bet.

    Clearly, the villain must still make us indifferent between betting and checking a Q, on the turn which means that he must call 2/3rds of the time to make the EV of betting 0.  Thus the hero EV with an Ace in the new game can by calculated by adding up:

    1. 1/3 * 100 -- we bet and they fold
    2. 2/3 * (42/44 * R)  -- hero bets, villain calls and a non-3c/2d river comes where R is the hero EV on that river
    3. 2/3 * (2/44 * 150) -- hero bets, villain calls, and a 3c or 2d river comes and villain just folds
    On the unblocked rivers, as in Brainteaser #6, the villain must call our bet of 100 chips 2/3 of the time so our ev with an A on the river is R = 150 * 1/3 + 250 * 2/3 = 650 /3.

    Thus our EV with an A is 175.76 chips or 1.7576 pots.  Our EV loss with an Aces is 16/9 * 100 - 175.76 = 2.02 chips.

    Since we hold the nuts half of the time, our overall EV loss is 1.01 chips.  So the EV of the actual game for the hero is about 87.87.  As you can see in the solution browser below, GTORB computes the EV as 87.83 which is within the given margin of error of 0.05 chips.

    How does this EV loss effect optimal play?  Intuitively this is actually pretty simple.  Of course we still always should bet the nuts, and we should bet enough of our air that our opponent is indifferent between calling and folding to our turn bet.  In this game, where our opponent's EV when we hold the nuts is higher, we need a higher nuts to air ratio to maintain his indifference which means that we must bluff the turn less frequently.

    Mathematically figuring out the optimal  bluffing frequency is a bit complex as we need to make sure to properly weight the probability of various river cards coming, using all of the villains information about his opponents range and his own hand.

    The villains EV for calling the turn and then playing GTO on the river is 150 when his opponent holds air (On average he wins the entire river pot of 200, but 50 of those chips were his own).  When his opponent holds the nuts, on 3c or 2d runouts his EV is -50 because he called 50 and the turn and always folds the river.  On other all other runouts he calls the 50 on the turn plus an additional 100 on the river 2/3rds of the time for a total EV of -350/3 (-116.66).

    If the hero is betting x% of his air and all of his nuts then when he bets he holds air x/(1+x) of the time and the nuts 1/(1+x) of the time and given that he holds the nuts, 3c or 2d come 2/44ths of the time.

    And the villains EV for calling a turn bet of 50 is

    150 * x / (1 + x) - 1/(1+x) * (100 * 2/44 + 350/3 * (1-2/44)).

    Setting that equal to 0 and solving for x gives that the hero should bet 25/33 or 75.76% of his air on the turn and 43.1% of his turn betting range should be air which matches the GTORB solution precisely.  The villain still calls 2/3rds of both the turn and the river bet as that is all that is required to make the hero indifferent between bluffing his air and checking it.  This is our exact mathematical equilibrium solution, you can browse the approximate GTORB solution below.

    I also solved the optimal bet sizing bonus question in a separate post here for those who are interested.


    This example may seem trivial, but as it turns out, the existence of river runouts shift the range distributions of players and transfer information.  Being in a position to put as much money into the pot as possible when you have an informational edge over your opponent or when your equity distribution is polarized  and as little as possible when it is merged is very powerful.

    I'll demo a much more powerful example of this in the next brainteaser where we will see an example of how equity transitions and river information can make protecting your hand via turn bets that never fold out better hands and are never called by worse hands, still be GTO, even if you were required to pay your opponent his turn hand equity when he folds worse.

    Note on epsilon equilibrium:  One quick note on the GTORB solution which is a 0.05 chip (5/10,000ths of the pot) epsilon equilibrium. Due to the approximation techniques used, the GTORB strategy actually has the hero checking the nuts on the turn a tiny fraction of a % of the time.  This has almost no impact on the game EV or solution accuracy but it does mean that if you examine a river after both players check you will see the hero bet with a very low frequency.  This is because the approximate solution has him holding the nuts with a tiny probability.  These rounding errors are why the solution has a nash distance of 0.05 chips which in this case means an opponent who played perfectly could exploit the approximate GTO strategy for 0.05 chips out of the 100 chip pot.

    Thursday, October 9, 2014

    GTO Brainteaser #8 -- Solving the Turn, Theory vs Practice

    The internal alpha version GTORB is now capable of solving turn scenarios for GTO turn and river play so this brainteaser is going to focus on multi-street theory.  In the solution (probably a week from today) I'll post the first fully browse-able GTORB turn solution to the model game below for those of you who are excited play around with a GTO turn strategy.  Note that the version of GTORB that can solve the turn won't be released commercially for a month or two as there are some performance / scalability issues that I need to solve before it is ready for mass use.  It will likely cost extra.

    The problem

    In GTO Brainteaser #6 I looked at a model scenario where the hero had a range of 50% nuts, 50% air while the villain had a range of 100% medium strength hands.  There was a 100 chip pot, 150 chip stacks and two streets of betting.  The hero could either bet 50 chips on the turn and then have the option to bet 100 chips on the river or he could shove the turn for 150 chips, and the question was which option is higher EV and what are GTO strategies for both players in this game.

    The key simplification that made this scenario quite different from real world poker is that it was assumed that no river card was actually dealt, there were just two rounds of betting.

    For those who are curious you can check out the full solution to brainteaser #6 here.  It turns out that it is optimal for the hero to bet 50 chips on the turn with all of his nut hands and 7/9ths of his air and then to bet 100 chips on the river when he is called with all of his nut hands and 3/7ths of his air hands.  The villain calls each of these bets 2/3rds of the time and folds 1/3rd.  The hero wins 8/9ths of the 100 chip pot in EV in this game.  Furthermore, it turns out that betting 50 chips on the turn and 100 chips on the river is the exact optimal bet sizing for the hero to maximize his EV, all other bet sizes are lower EV.

    Lets now look at a very similar game.  Imagine the following (completely made up) scenario.

    1. You are on the turn and the board is  AsAhKsKh
    2. The hero a hand range of AcAd and 3c2d
    3. The villain a hand range of KcKd
    4. The pot is 100 chips and stacks are 150 chips
    5. The hero can either bet 50 chips on the turn and 100 on the river or he can shove for 150 on the turn.

    Clearly no matter what river card comes, the relative strengths of the hands in both players ranges will not change so in that respect this game seems identical to the model game from brainteaser #6.  AcAd will beat KcKd on every possible river and 3c2d will lose to KcKd on every possible river.

    However, it turns out that GTO play in this game is different from GTO play in GTO Brainteaser #6.  Why?

    1. What is the EV of the game for the hero, is it higher or lower?  
    2. What are the optimal strategies in this game and what is the hero's EV when both players play optimally?  

    Bonus:  Is betting half pot on the turn and the river still optimal or is there a higher EV bet size?