Agree 100% 
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Thatâs almost true, except if P2 has a Queen, then P2 doesnât know if P1 has a King or an Ace.
Anyway, so far my strategy is just a candidate for the Nash equilibrium. Iâd happy if others can come up with better strategies.
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Why not raise small with everything as P1? Say 1/4 pot?
Itâs also funny here, how without later streets, the button/small blind is always out of position, with an information disadvantage to the big blind.
Edit: oh wait⊠what is the smallest legal raise here? 1/3 pot ($2 raise into the $4 pot)?
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The thought with raising as small as possible with everything as the button/sb was mostly to try and mitigate the information disadvantage. Unfortunately, the raises with 2/3 of our hands seem destined to be -EV. It is weird how strapped the button feels, even though it is the big blind with the forced full bet. But without folding any of your hands pre-flop, you are surrendering the one clear advantage that you actually have: that your not forced to commit money to the pot with your worst hands.
Hereâs another starting idea for P1:
- A min raise 100% of the time
- K min raise 2/3 of the time; fold balance
- Q min raise 1/3 of the time; fold balance
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The minimal legal initial raise out of the SB is to $4, i.e., put $3 extra into the initial $3 pot. (I think this would be called â1/2 potâ as itâs a $2 raise over the $4 pot after completing the SB.)
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I kind of like the idea of limping K when first to act, at least at some frequency. This would be tricky to pull off, because you would also have to defend sometimes.
In order to do this, one would have to limp an A at some frequency with the intention of re-raising if raised. I would expect K and Q to raise some of the time, so this might be worth doing.
How often to limp, how often to defend K against various sizings, and so on would be hard to get right, so maybe this is just another one of my dumb ideas, but I would try to work in a limp now and then.
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I just canât see the value of time in a 2 person 3 card game. If itâs a simple exercise fine , if not then I donât see it as a game.
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I think it is practice in finding an equilibrium strategy. It is pretty hard to find perfect play in something as simple as this, but many of the factors you need to evaluate are the same. The hope would be that by going through a lot of exercises like this, youâre able to come a bit closer to identifying equilibrium strategies in more complex, realistic situations.
Unfortunately, I find it rather comic how unable I am to quickly identify an equilibrium strategy even in something as incredibly simplified as this.
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Interesting idea. I notice that the Q raise frequency leads to a 3-to-1 value to bluff ratio when P2 has a K which exactly equals P2âs pot odds to call with a K, so P2 is indifferent between calling and folding. I donât think that bluffing with a Q is viable for P2 since P1âs range is too strong in this spot (60% A, 40% K). Thus, I think the following is the best that P2 can do:
- A 100% call (or raise, doesnât matter)
- K 100% fold
- Q 100% fold
The minimum defense frequency against P1âs Q bluff is 50%, which is already covered by P2âs Aâs, so no need to bluff catch with Kâs.
Now, itâs easy to calculate that P2âs EV with this strategy is 2, which exactly recovers the big blind. And accordingly, P1âs EV is 1 (so itâs unlikely to be optimal for P1 who has the advantage of paying less blind).
Edit: Thinking a bit more about this and despite what I said above, a Q bluff might be viable for P2 still. For example, in the situation where P1 min raises to $4 (pot $6), one could consider the following strategy for P2:
- A 100% min raise to $6
- K 100% fold
- Q 20% min raise to $6, 80% fold
After the raise to $6, P2âs value-to-bluff ratio is 5-to-1, there is $10 in the pot, and P1 needs to pay $2 to call, so 5-to-1 pot odds. I havenât calculated the EV yet, but Iâd guess it should be slightly larger than 2 for P2.
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I donât know anything about a Nash equilibrium or GTO strategy but as I see it based on the criteria set in your OP,
â⊠a Nash equilibrium is a set of strategies for all players so that given the strategies of all other players, no individual player has an incentive to deviate from their strategy. In the context of poker, such a set of Nash equilibrium strategies are often referred to as âGame Theory Optimalâ or GTO. âŠâ
Your conclusion seems correct i.e., always fold a Queen for a + E/V. As long as no one deviates from their strategy.
But, I donât think this is a game of chance since the ânutsâ (Aces) should occur 1/3 of the time.
To make this format a game of chance, I agree with @Chasetheriver 's suggestion of â⊠Would a deck with two of each card be a more viable game?â
But, if you wish to continue using your format of 2 players, 3 cards and no streets etc., change it where each player canât see their own card but only the card of their opponent which obviously canât be done in an on-line format. Except, everyone would always be deviating from a set strategy.
Now, this would change it to a game of chance.
As far as my strategy for this game would be is to always bet the pot in SB and min raise in BB. But, max raise if Ace in BB. My additional bet(s) or a fold would depend on the otherâs response.
Uncertain if this will result in a + E/V but, I would now have fun playing this type of game.
I was wrong again. P2âs EV with this strategy is also 2, exactly the same as with the simpler call-or-fold strategy. So bluffing is viable here but doesnât seem to add any value for P2.
I found this puzzling at first, but then it occurred to me that this is likely due to the fact that the ratio of P1âs value hands in this spot (60% Aces) exactly equals the minimum defense frequency (also 60%), so P1 doesnât have to bluff catch with any of their Kâs. I suspect that there is a general principle behind, something like this:
Adding bluffs to a betting or raising range add value if and only if the ratio of Villainâs value hands is less than the minimum defense frequency against the bluffs.
If this principle is true, it also explains why min raising with a Q does not add value for P1 (compared to just raising with Aces). In this spot, P2 has 50% value hands (Aces) and the minimum defense frequency is also 50%.
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Those seeking a more rigorous solution might consider reading, âThe Search for GTO: Determining Optimal Poker Strategy Using Linear Programming,â Stuart Young, The College of Wooster, 2017
An excerpt from the paperâs abstractâŠ
âThis project applies techniques from game theory and linear programming
to find the optimal strategies of two variants of poker. A set of optimal poker
strategies describe a Nash equilibrium, where no player can improve their
outcome by changing their own strategy, given the strategies of their
opponent(s). We first consider Kuhn Poker as a simple application of our
methodologyâŠâ
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Kuhnâs variant looks quite close to what we had been discussing, with just 3 cards in the deck (J, Q and K). A quick read of that section alone⊠lol, Iâm left thinking about people complaining when someone takes too long to bet, call or fold.
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Yeah, itâs a lot more complicated than one might think.
I havenât done the mathing, but this doesnât seem optimal to me. If we play each card value 100 times, this would have P2 folding 180 times, which is 60%.
Facing this fold frequency from the BB (P2), wouldnât it be correct for P1 to min raise every time? Wouldnât doing so win 3 SBs 60% of the time, while only losing 2 SBs 40% of the time, assuming P1 will never defend against P2âs raises?
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I only claim that the two P2 strategies from the post you quoted from are optimal (maximally exploitative) against the specific P1 strategy proposed by @Yorunoame (quoted in my post). They are not unexploitable themselves as you correctly pointed out. This actually shows that these strategies do NOT form a Nash equilibrium.
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Yeah, that was my point. It would be much easier if we were looking for Waldo or Carmen San Diego. 
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So then WHY is this such a difficult problem?
In a zero-sum game, a Nash solution would seem to imply a 0 EV state for both players. I could be wrong about this, but I donât see any other possible outcome.
This is easy to demonstrate in a simpler game such as rock, paper, scissors, where the GTO solution is to throw rock, paper or scissors at random, with no patterns at all. If both players use this strategy, they would expect to win 1/3, tie 1/3, and lose 1/3, for a net result of exactly 0 EV.
If this principal also applies to the problem at hand, the Nash solution, over the TOTALITY of all P1 and P2 lines, should be 0 EV too.
Since the ace lines for both P1 and P2 should always be +EV, wouldnât this suggest that the totality of king and queen lines across P1 and P2 will always be -EV?
At first, I thought this might point to an elegant solution where the K lines, which have 50% equity, would end up at 0 EV, leaving the - EV burden on the Q lines, but this canât be the case because itâs hard to realize this equity when the queens bluff.
I now suspect that the K and Q lines for P1 AND for P2 will all turn out to be -EV.
The toy game discussed here is much less symmetric than rock, paper, scissors though. P1 only has to pay $1 blind. P1 can always choose to pay an additional $1 (or more) if this is profitable, while P2 is forced to pay $2 from the start. This suggests that P1 has an advantage over P2 in this game.
Even if both players had to pay the same amount, the Kuhn poker solution shows that the EV (including paid blinds/antes) is generally not 0 for both players.
NB: Of course, if P1 and P2 take turns and play SB/Button and BB equally often, then the total EV (including paid blinds/antes) of optimal play will be zero for both.
One also has to be careful about what one exactly means by âEVâ. I think it is useful to distinguish between the following:
- action EV (given a specific hand): e.g. the EV of raising with an A from the SB is the expected profit of that action, i.e., the expected reward minus the cost of the action. As such, the EV of folding is always 0.
- averaged action EV (given a specific hand): e.g., if a player decides to take one action 60% of the time and another action 40% of the time with the same hand, then the averaged action EV is the weighted average of the corresponding two action EVs.
- spot EV: e.g., at the start of the game, P1 has either an A, a K, or a Q, each with probability 1/3. So the spot EV is the average of the averaged action EVs.
Note that anything that a player has already paid into the pot before deciding on an action is a âsunk costâ and doesnât enter into the EV calculation.
With these definitions and understanding that P1 has an advantage in the game at hand, one would expect that P1âs optimal strategy has an initial spot EV of slightly more than $1 (i.e., an expected profit after subtracting the $1 SB) and P2âs optimal strategy has an initial spot EV of slightly less than $2 (i.e., not completely recovering the $2 BB.)
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I thought it was a given that P1 and P2 would take turns being SB/BB.
Is this not the case?
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