I recently solved the AKQ game with pot P and one half-street of betting (i.e. P1 is forced to check, P2 can bet or check, and if P2 bets P1 can call or fold) but with an unequal distribution of A, K, and Q, which are dealt with probabilities x, y, and z (x+y+z=1). The optimal strategy is:

• P2 bets the Q with frequency x/z(P+1)
• P1 calls with the K with frequency (yP-x)/y(P+1)

In the original game, x=y=z, so the frequencies simplify to 1/(P+1) and (P-1)/(P+1), as demonstrated in MOP (yet note that the bluffing ratio α=1/(P+1), which is somewhat idolized in MOP, may take on a different value when the assumption of equal distribution is removed).

I find it interesting that the optimal betting and calling frequencies can be significantly altered according to the distribution of hands. Since the AKQ game is studied with the purpose of applying its concepts to poker (specifically river scenarios), this begs the question, what is a player's distribution of value, bluff-catcher, and bluff hands at the river? Of course, this depends on the player's strategy. But perhaps some general statements can be safely made. For example, what is the ordering of x, y, and z for strong strategies in general? (My guess is y>x>z.) With such statements, inferences may then be made about the optimal frequencies.

x is the frequency of Aces? If x is big and z is tiny, P2 bets his unlucky Q almost always, and is almost always facing an A? How does this work out? Do you have some restrictions on y?

Can you explain why all three must add up to 1? I dont see how y interacts with x and z.

Correct. No restrictions. Counterintuitive, perhaps.

Here are the indifference equations (c is the probability that P1 calls with the K, and b is the probability that P2 bets the Q):

<Y bets Q> = (-1)x+(-1)yc+Py(1-c)
<Y checks Q> = 0

<X calls w/ K> = (-1)x+zb
<X folds K> = -Pzb

Setting the first and last 2 equal to each other and solving for b and c results in the solutions I gave in the OP.

They are all the probabilities of a player getting dealt each card. Since they are the only possibilities (you are either dealt an A with probability x, K with probability y, and Q with probability z), they must add up to 1.

Did some more simple math and discovered something interesting:

• c decreases as x increases and increases as y or z increases
• b decreases as z increases and increases as y or x increases

In other words, you should call with a smaller percentage (of bluff-catchers) the more value hands are in your opponent's range and with a greater percentage the more bluff-catchers are in your range or the more bluffs are in your opponent's range, and you should bluff with a smaller percentage (of bluffs) the more bluffs are in your range and with a greater percentage the more value hands or bluff-catchers are in your opponent's range.

One counterintuitive result is that when x is increased and y is decreased while z is held constant, b increases. In other words, as the number of value hands relative to bluff-catchers in your opponent's range increases, you should bluff more often.

Wait, does this mean that both player ranges are {A,K,Q}? I thought p1 only has a K... Imo if p1 has {K} and p2 has {A,Q} then only the frequencies of A and Q can vary.

When P1 has A or Q the optimal strategy is not mixed (always call w/ A, always fold Q, regardless of frequencies), so it's only worth discussing the time he has K. But the variance of the K also matters because when P2 holds the Q it matters how often P1 holds the A or K.

say x=.9999999999999, y=0,z=1-x

"<Y bets Q> = (-1)x+(-1)yc+Py(1-c)" becomes -.9999999999999, which would mean Y should never bet to win the split*. Sounds reasonable

"P2 bets the Q with frequency x/z(P+1)" becomes 9999999999999/(P+1), which is pretty big for reasonable pot sizes. Larger than one, anyway.

Either I miss something here, or the equations do.

*Is this an infinite deck AKQ (i.e. no card removal, both players can get the same card)? If so, shouldn't there be a +Pz term there too, to mark the times X has a Q he's folding? Or just AhKhQh in the deck? If so, when Y gets the unlucky Q (and y=0), X will always have the A, and the equation still doesn't capture that... Y's value for betting the Q would be -1.

y can't equal 0. If there's no K in the deck, there's no mixed strategies in the optimal strategy, so my equations are useless.

say x=.9999999999999, y=.00000000000000000000000000000000000000000000000 0000000000000000001,
z=1-x-y
is still very negative, more than befor
isn't affected, is still larger than 1, which is a pretty high frequency.

Ah I see.

I doubt the equations themselves are mistaken. I think it's that x, y, and z must be bounded in relation to P. In other words, in some games with certain values of x, y, z, and P, the optimal strategy is not mixed but pure; in the example you gave, K and Q are not dealt often enough for bluffing (or calling either, I'm guessing) to ever be profitable (or even break even). Since the equation <Y bets Q> = (-1)x+(-1)yc+Py(1-c) is set equal to <Y checks Q> = 0, it must be possible to choose a value of 0<=c<=1 in the former (for the values of x, y, z, and P given) to actually satisfy the equation. Indeed, no such value of c exists for the values that you gave for x, y, and z (unless P is extremely large). But since your example (and pretty much all else like it) is so far from actual poker anyways, I wouldn't worry about it.

The fact that P2 bets the Q isn't affected by y leads me to think that the equations still fail. And the fact that first, no restrictions on y, now it can't be 0 or small. And the fact that x/z(P+1) becomes huge whenever x>z.

Edit: is this an infinite deck game?

It actually is affected by y. y=1-x-z, so y is inextricably linked to x and z, and since x and z affect the equation, so does y. Or if that doesn't convince you, I can rewrite x/z(P+1) as (1-y-z)/z(P+1), and now it should be clear that it is affected by y.

"Huge" is vague. If x=.4 and z=.2, then x/z(P+1)=2/(P+1), which I wouldn't consider huge for most P (which in the vast majority of limit poker situations is much greater than 1, the size of the bet).

Basically what you're saying is that it's counterintuitive and therefore you won't accept it. You need to find a problem with the mathematical derivation to demonstrate that the result is mistaken.

No, only one of each card (but dealt with unequal probabilities).

"If x=.4 and z=.2, then x/z(P+1)=2/(P+1)"

so if pot is1, Q bets with probability of 200%. You do realize that does not compute? Also, as this is a finite stack game, Q will face A half of the time (x=.4, y must be .4). Which mean it's called 100% of the time. Does not sound like a good bluff to me.

If P=1, 2/(P+1)=1, not 2. You could make P smaller than 1, but doing so 1. does not exemplify real poker, 2. is what I explained in my previous post that you can't reasonably do, and 3. even MOP assumes (for the very reasons I described in my previous post) that P>=1.

Not sure I understand your math. Yes, y=0.4. When you hold a Q, however, your opponent will hold the A 2/3 (0.4 out of the remaining 0.6) of the time and K 1/3 (0.2 out of the remaining 0.6) of the time. Where did the "half" you mentioned come from? And what's calling 100% of the time? The A? That's true in any version of the AKQ game because folding A is a dominated strategy. With the K, P1 calls (yP-x)/y(P+1)=(P-1)/(P+1) of the time, which is not 100%.

Again, you're trying to find an example to demonstrate that my result is wrong, but what you really need to do is find a problem with the mathematical derivation (which you have avoided completely until now).

"If P=1, 2/(P+1)=1, not 2." Ok I failed. Still, Q gets bet 100%. Change the probabilities a bit and you get over 100%. That's not a good sign, IMO.

"Not sure I understand your math." Me neither. Need to stop posting under the influence, it seems.

If x=y=.4 and z=.2, the Q will face A or K with same probability. The A will call always, the K will fold. This is fine for Q.

If x=.5, y=.4 and z=.1, Q will be called more than 50% of the time by the A. So it shouldn't bluff. But x/z(P+1) becomes 5/2 (for pot size of 1). So Q bluffs 250% of the time, and gets called 5/9 of the time.

"you're trying to find an example to demonstrate that my result is wrong"
That's often easier to do than the math, when things seem wrong. If you need to start setting arbitrary limitations to the model to remove disturbing results, without the derivation specifically pointing to the limitations, that's not usually a good sign. I find
"Q bluffs 250% of the time, and gets called 5/9 of the time" a disturbing result. Did I now break some limitation?

Too lazy to do the indifferences, but if you show the full math you did, I'm interested to take a look.

The limitation I specified, namely that x, y, and z must be bounded in relation to P because in some games the optimal strategy is not mixed but pure (do you understand what that means?), is not arbitrary; even MOP uses the same kind of limitation (would you also like to critique MOP as arbitrarily creating the limitation P<=1 because they know that otherwise their math can't stand up to counterexample such as yours?). I don't claim that my solution applies to the game you specified. The GTO strategy for your game is not a mixed strategy.

Let me show you an example of how I could do what you're doing with MOP's solution to the AKQ game. They say K is supposed to call with frequency (P-1)/(P+1). Well I'm gonna make P=0.5. So K calls with frequency -1/3. Do you find this result disturbing? Did I now break some arbitrary limitation? If you're going to continue to claim I'm wrong for the reasons you give, you're also going to have to claim that MOP is wrong for the same reasons since it's possible to give similar examples, as I just did. Good luck.

The indifference equations are the full math lol. Solving them gives the frequencies.

numbers beyond 0-1 simply mean that you cant make him indifferent.

MOP did bother giving the limitation P>1 for the AKQ solution, and showed why the indifferences break, i.e. no more bluffs happen at 1 and below. You gave us an equation, which had no limitation on the parameters. Then, there appeared some limitations, i.e y can't be 0. Then it appeared it can't be .0000000000000000000000001 either. Then it appears the parameters are bound in some way in relation to P. I hope you realize why this approach of presenting your work might bring some scepticism around.

Your equations might very well be right. How about just showing us one case how to actually safely use them, I seem to be bad in picking situations where they don't break.

Yeah that's pretty much what I said too.

I didn't present them the way you claim I did here. I didn't present it, then say "oh but it can't be 0," then later say "oh it can't be .000000000000000000001 either," then finally say "well the parameters are bound in some way in relation to P," all in response to counterexamples you gave that I otherwise had no satisfactory answer to.

Initially, I simply gave the equations for the mixed GTO strategies. Notice that this implicitly assumes that the equations only apply to situations in which mixed GTO strategies actually exist. Therefore, I didn't need to state this at the outset, and I only made this clarification explicitly once you began mistakenly applying the equations to situation in which mixed GTO strategies don't exist.

Anyhow, if you see why your examples aren't really counterexamples, we're now on the same page now and can move on.

Sure. Let P=5, x=.2, y=.3, and z=.5. Then

• x/z(P+1)=1/15
• (yP-x)/y(P+1)=13/18

Sorry, missed the part on applicability on some cases only. Had I remembered the original AKQ solution by heart, maybe I could have made the implicit assumption.
But w/e.

Ok, so let's say player with starting range 2xA,3xK,5xQ bets. His betting range consists of 2 units of A and 5/15 units of Q. So 2xA, 1/3xQ.

When facing a bet, player calls with 2 units of A and 3*13/18 units of K, that's 2 1/6 of a K. He calls total 4 1/6 of 5 combos, that's 5/6 of the time.


If the original distribution was 2,2,2 (i.e. equal probabilities), the betting range consists of 2 units of A and 2/(5+1) units of Q. So 2xA, 1/3xQ.
When facing a bet, player calls with 2xA and (5-1)/(5+1) K, that's 4/6 of his kings, or 8/6. So he calls 3 1/3 of 4 combos, that's 5/6 of the time.

Nothing breaks, but not sure we gained much.

When I wrote "No restrictions," I was referring to when mixed GTO strategy solutions actually exist, and when I wrote "Ah I see," I was communicating that now I understand what is causing you to disagree with me.

Just to be clear, is your point that since the amount of times which we call (5/6) is the same in both examples, what's the point of my equations?

I happen to be discussing this same objection in this thread. This is part of what I wrote over there:

The key number in your examples is not 5/6 but 13/18 and 2/3, which are slightly different.

Well, you found another way determining how many of your Queens you should bet. The old way was: "I have 2 combos of nuts, I want to bet 1/3 combo of air. Since my range includes 5 combos of Q, I bet them 1 time out of 15."

Similarly, you found a new way of determining how often to call when holding a K. The old way: "When facing a bet and able to beat a bluff, I need to call 5/6 (==25/30) of the time. My range holds 12 combos of A, 18 of K. Calling with A is 12/30, I need to call with 13 Kings to get to 25. That's 13/18 of all my kings."

By 2/3 you refer to the 4/6 of Kings in the original x==y==z example? Yes, it's different from 13/18. Caller has now less A and more K in his "can beat a bluff" range, so after exhausting his A's, he needs to now call with more of his Kings to keep bettor's Q indifferent (in proportion to his now bigger frequency of K, and thus also in absolute combos). Alpha is still good on both sides of the table.