Part 2: Hat problem.

Let $a$ be the number of $\color{royalblue}{blue}$ hats a certain person can see and $b$ , $c$ be the number of $\color{#D61F06}{red}$ , $\color{#333333}{black}$ hats accordingly.

One strategy that maximizes the chance of winning is the following:

If $a \equiv b \equiv c \pmod 3$ , guess $\color{royalblue}{blue}$ .

If $a-1 \equiv b+1 \equiv c \pmod 3$ , guess $\color{#D61F06}{red}$ .

If $a-1 \equiv b \equiv c+1 \pmod 3$ , guess $\color{#333333}{black}$ .

First, note that every person with the same hat color gives the same answer , as they see the same amount of $\color{royalblue}{blue}$ , $\color{#D61F06}{red}$ and $\color{#333333}{black}$ hats.

Furthermore, $2$ people who wear different hat colors see the same amount of hats with the third color (the color which neither of them wears) and a different amount of hats of the other $2$ colors ( $\pm 1$ hat and therefore different values $\pmod 3$ ). Thus, any $2$ people wearing different colors guess different colors , because any $2$ different combinations of $a$ , $b$ and $c$ within the same "guessing group" differ in $a$ , $b$ and $c$ .

Lastly, whenever the group of people wearing $\color{royalblue}{blue}$ hats guesses $\color{royalblue}{blue}$ , everyone guesses their own hat color correctly. A person wearing $\color{royalblue}{blue}$ only guesses $\color{royalblue}{blue}$ if he sees $a \equiv b \equiv c \pmod 3$ . A person who wears $\color{#D61F06}{red}$ sees one $\color{royalblue}{blue}$ hat more and one $\color{#D61F06}{red}$ hat less than a person wearing $\color{royalblue}{blue}$ and therefore sees $a-1 \equiv b+1 \equiv c \pmod 3$ and guesses $\color{#D61F06}{red}$ . This means that a person wearing $\color{#333333}{black}$ guesses $\color{#333333}{black}$ , because it is the only color that is left. (Note that this strategy does not "favor" any color. You can make the same argument with $\color{#D61F06}{red}$ or $\color{#333333}{black}$ hats instead of $\color{royalblue}{blue}$ .)

Since the probability that someone guesses their own hat color correctly has to be $\frac{1}{3}$ , because a person cannot gain any information about his own hat color , and the only two possible outcomes are that everyone guesses their own hat color correctly or incorrectly at the same time, the chance of winning is $\frac{1}{3}$ . This also guarantees that there can not be a better strategy.

This yields a solution of $1+3=\boxed{4}$ .

Just sharing some thoughts here. Based on part one of the question, we may assume the following as a strategy: we have colours "blue" and "non-blue". If you see an even number of blue hats, guess blue. Otherwise flip a coin to guess between black and red.

With the described strategy the winning probability is

$\frac{1}{2}(\frac{1}{2^9}+\frac{1}{2^7}+\dots+\frac{1}{2}) \approx \frac{1}{3}$

However, we might be able to improve the method, based on the colour of the hats that appears the most. For example, if there are $8$ red hats, we better change the rule to: "If you see an even number of red hats, guess red. Otherwise flip a coin to guess between black and blue.". This way, you can increase the chance of winning. So, The strategy can be a bit more sophisticated. Something like:

1- If a colour appears ore than 6 times, choose that colour as the dominant colour. then, the rule would be: If you see an even number of dominant hats, guess dominant. Otherwise flip a coin to guess between the non-dominant colours."

2- If no colour appears more than 6 times, then blue is the dominant colour. If you see an even number of blue hats, guess blue. Otherwise flip a coin to guess between black and red.

Please comment your opinions.

Cheers.