### Solution

The brainteaser involved studying the following game:

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:

- 1/3 * 100 -- we bet and they fold
- 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
- 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.

### Takeaways

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.
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