I like the following strategy, but I'm not sure it's optimal. However I think it's a good ice-breaker as being the first with an actual winning chance computed.
As my starting ideas were symmetrical and deterministic up to a point, yet the resulting strategy is somewhat asymmetrical, my hunch says its probably not optimal. But it's the best I have so far.
Also sorry, it's longer to read because I started writing something else and changed mid-way
The next Greedy strategy tries to win the middle cases (w=3,w=4) and group everyone errors in the losing cases:
Depending on z choose:
z=0 - Red*
z=1 - Blue
z=2 - Red
z=3 - abstain
z=4 - Blue
z=5 - Red
z=6 - Blue*
*-same chances if instead of color, abstain is chosen.
This greedy strategy has a 2*(7+35)=84 out of 128 winning cases = 65.625% It wins for B={1,3,4,6} instances.
Its not so good compared to 112 out of 128=87.5% for the perfect-Hamming strategy (if we knew the distribution of hats as well), but not far from that either.
Depending on z choose:
z=0 - Blue
z=1 - Red
z=2 - abstain
z=3 - Blue
z=4 - abstain
z=5 - Blue
z=6 - Red
This reverseGreedy strategy has 2*(1+21)+35=79 out of 128 winning cases. Its not as good as the Greedy strategy, but it is good if you think you're in a non-B instance {0,2,5,7}. As a side-effect, it also wins for a particular B-instance (w=3) i.e. when you think you're in a non-B instance but you're actually in a w=3 instance.
Step 0. Establish B={1,3,4,6} and assume teams form a circle (i.e. n=3 Hamming instance).
Assume A={000,111}.
Appoint a leader team as a starting point and a rotation direction: e.g. British, French, Italian.
Step 1. Compute the two bits of information you see from other teams based on set B.
So if you see a number that is in B mark it as a bit 1, else mark it as bit 0.
Step 2. Assuming the two possibilities (that you are either a B-instance or a non-B-instance) as bits for your team
you arrive at two unique 3-bit strings a (if you are a nonB-instance) and b (if you are a B-instance).
The fact that you established in step 0 a starting point and a rotation direction makes a and b unique.
Step 3. If neither a nor b belong to A - ABSTAIN, else go to step 4.
Step 4. Exactly one of a or b belongs to A:
If a is in A, then choose b (think that you are a B-instance), so play the individual Greedy strategy.
If b is in A, then choose a (think that you are a non-B instance), so play individual reverseGreedy strategy.
Note that this global strategy is independent of what you are told about your team, and only after arriving in step 4 and choosing an individual strategy you actually take the information about your team into account.
Chances of winning depend on the unknown global string w1w2w3 (British, French, Italian) where wi=1 iff the number of hats in team i is in B:
-000 - Everybody sees non-B instances, chooses b in step 4, tries to play Greedy and fails. Game is LOST.
-001, 010, 100 - One team (in a B instance) sees a=000 in A and plays the Greedy strategy which wins. Other two teams see 1-bit and 2-bit numbers not in A and abstain. Game is WON.
-110, 101, 011 One team (in a non-B instance) sees b=111 in A and plays reverseGreedy which wins. Other teams see 1-bit and 2-bit strings not in A and abstain. Game is WON.
-111 Everybody sees b=111 in A, so they choose to play reverseGreedy (for the wrong reasons). They usually fail. But the lack of symmetry in reverseGreedy gives you a rather unique global win if everyone is in a particular B-instance (w=3). Only in this case, all three teams play a winning reverseGreedy. In all other cases, at least one team fails and Game is LOST.
That lack of symmetry in reverseGreedy makes over-all winning chances harder to compute.
First, the individual probability of having a B-instance(1,3,4,6) is easy to compute (and was already computed) as it is the same with winning chances of the Greedy strategy 84/128. Conversely, 44/128 cases for a non-B instance.
Also, the chances of a w=3 instance are 35/128.
So, global chances (denote t=44/128):
Case 1) 000 means non-B instances for everyone = 44/128*44/128*44/128= t^3
Case 2) 001, 010, 100 means as single non-B instance = 3*44/128*44/128*84/128 = 3*t^2*(1-t)
Case 3) 110, 101, 011 means as single B instance = 3*44/128*84/128*84/128 = 3*t*(1-t)^2
Case 4) 111 means B instances for everyone = 84/128*84/128*84/128= (1-t)^3. Of which a particular case with global (35/128)^3 probability should be counted as a win (reverseGreedy side-effect).
So, global winning is 3*44/128*44/128*84/128 + 3*44/128*84/128*84/128+ (35/128)^3=(1486848+42875)/128^3=1529723/128^3.
Roughly 72.94%.
EDIT: Tried to make it more readable by two levels of spoilers.
Combining 3 teams' strategies in a greedy manner, each team playing the same local (inside-team) strategy is a little tricky.
We should use the fact that one knows the exact instance the other teams are in to obtain a global advantage.
Therefore each player should look at the global picture, then either choose to abstain or play a team strategy (e.g. Greedy strategy as above).
Assuming the third team's point of view for a moment i.e. imagine you see x and y and your own z<=w<=z+1.
First, if you play the Greedy strategy as described above you know it's a win for a team iff number of hats is actually 1,3,4,6.
So if you see that both x and y are not in the set B={1,3,4,6}, there's no point in abstaining since the others surely cannot get it right on their own. You might get lucky and be in a B-instance yourself, therefore winning the game. This would be Naive-Rule 1.
If both x and y are in set B, then you might be spoiling everyone's chances by playing since your set might not be a wining set. This would count as a Naive-Rule 2. However, assuming a symmetrical global strategy, if all see B-instances, they would all abstain and perhaps lose a chance of winning. Then perhaps choosing/appointing a team to take action if all else fails would be better but this does not actually lead to a simple Naive-Rule 3.
This starts to looks like a small version of the Hamming-problem. So, instead of choosing a strategy based on the Naive-rules above, you can actually choose a sort of Hamming n=3 strategy.
However, besides the individual Greedy strategy above, you will need a mirror strategy for when you're a non-B instance.
The following reverseGreedy strategy wins for non-B instances (i.e. when w=0,2,5,7)
Returning to the Hamming-style global strategy:
