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Modifying the puzzle: Cooperative hat-guessing, no mistakes, 3+ colors

Puzzling Asked by klm123 on September 4, 2021

Following Guess your hat color, but you don't have to and it’s interpretation via Covering codes I tried to create a puzzle with more hat variativity, i.g. 3 types of hats.

4 hats are put on 4 logicians, each hat color is selected randomly: Red, Green or Blue.
As usual, every logician doesn’t see the hat on his own head, but sees the rest. They cannot communicate in any way possible.
Each logician at the same moment must answer the question – "what color is the hat on your head?". And there are only 3 possible answers they can say: "Red", "Green", "Blue" and "I don’t know".
If at least one color is named incorrectly logicians fail and die. If no one named a correct color they die just the same. Otherwise (if at least one answer is correct) – logicians survive.
As usual, they have time to discuss a strategy before the hats are put on their heads.
What’s the strategy, which gives the highest probability to survive?

I picked number of logicians $N=4$ and number of colors $M=3$, because these are the numbers a generalized covering code exists for ($N=(3^2-1)/2$, see wiki). There by the puzzle solution:

That’s all good, but the probability is much less than expected. The upper estimate of survival probability is $P_{survival} le N/(N+M-1) = 2/3$. Here is why:

This number was achievable in the similar cases for $M=2$ (when $N=2^k-1$). But now I have no idea how to achieve it. Thereby two questions:

Is there a solution for the mentioned puzzle ($N=4$, $M=3$) with probability $P_{survival} > 4/9$?

Is there a combination of $Nge 2$ and $Mge 3$ where $P_{survival} = N/(N+M-1)$ is achiavable?


Edit:
@tehtmi answer proves that $P_{limit} = N/(N+M-1)$ is not achievable. I’ve rewarded this proof with a bounty. Now I want to reward the best strategy with a bounty.

@Reinier’s strategy gives (if I’ve not messed up the calculations)
$P=16/27 approx 59.3%$ for $N=4,M=3$,
$P=55/81 approx 67.9%$ for $N=5,M=3$,
$P=17/32 approx 53.1%$ for $N=4,M=4$,
$P=75/128 approx 58.6%$ for $N=5,M=4$
Is there a better strategy for any of those cases?

2 Answers

Answering whether the $N/(N + M - 1)$ survival probability can be met:

Answered by tehtmi on September 4, 2021

Partial Answer

I am only going the answer the first question here, the answer is

The case $N = 2$, $M = 3$

The case $N = 3$, $M = 3$

The case $N = 4$, $M = 3$

Answered by Reinier on September 4, 2021

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