Now let’s start with the guesses A and C make: For a winning strategy, we intuit that these as mentioned earlier, cannot be entirely random but rather functions of the colors b and d that they see on B’s and D’s heads. Similarly B and D will make guesses based on the a and c, they see as well as knowing the functions A and C will be using. What could these functions be? Turns out that sums and differences work well here as pointed out by Sandi Sandman who helped me when I was stuck.

So let’s say that A and B guess: 

\[
f(b,d) = (b+d)  \\ \\
g(b,d) = (b-d) 
\]

This is just saying that A takes the numbers corresponding to B’s and D’s hats (remember R=0, Y=1, B=2), adds them up and if their sum happens to be greater than 2, uses the remainder on dividing by three i.e. picking. the correct number from our clock with three points (red text above). C, does the same with the difference of B’s and D’s colors.

As an example, let’s assume the colors on B D are 0 1  i.e.  Red, Yellow. This means A will guess \((0 + 1) mod 3 = 1 \equiv\) Yellow (Y) and C will guess \((0-1) mod 3  = 2 \equiv \)Blue (B).

Now A and C could themselves have any one of 9 combinations on their heads. Their guesses based on what they see on B and D (0,1) will set them all free in 5 of those 9 instances. These 5 instances in our example are: (1,0)  (1,1),  (1,2),  (0,2),  (2,2).  But there are 4 other instances where their guesses do the group no good. These are when A and C have hats (0,0),  (0,1),  (2,0),  (2,1).  But B and D still have their guesses available. We now have to figure out a way for them to use their guesses to cover the remaining situations such that they have at least one correct guess. 

Now remember that B and D can actually see the hat colors on A and C (but not on their own heads of course!). So they have some information on what they need to cover. Let’s assume they see (0,1) on A and C, an instance where A’s and C’s guesses previously discussed are wasted.

So here’s where we are at so far:

A and C see (0,1) on B and D and per agreed upon strategy guess (1,2). This guess would have gotten them free in 5 out of 9 possible combinations on their head but unfortunately not in this instance when they have (0,1). With me?

B and D have  to work on the worst case assumption that the hat colors they themselves have and the functions they know A and C will use for their guesses will result in both A AND C making incorrect guesses. At least one of them (B or D) will have to guess correctly in those four remaining possibilities.

It follows then that if B and D assume that A’s and C’s guesses were wrong, then A and C were either off by 1 place higher or 1 lower in their guess using \(b\) and \(d\) (since there are only 3 choices!):

\begin{align*}
(b+d) 
& \begin{cases}
& =(a+1) \qquad \textit{call this case } m_1 \\ 
& \quad OR \\ 
& =(a-1)  \qquad \textit{call this case } m_2 \\ 
\end{cases} \\ 
& \textit{since } (b+d)  \textit{ didn’t provide A with a correct guess}
\end{align*} 

Using the same reasoning:

\begin{align*}
(b-d) 
& \begin{cases}
&= (c+1)  \qquad \textit{call this case } n_1 \\ 
& \quad OR \\ 
&= (c-1) \qquad \textit{call this case } n_2  \\ 
\end{cases} \\ 
& \textit{since } (b-d)  \textit{ didn’t provide C with a correct guess}
\end{align*}

B and D now have some information on their own hats in the worst case situation. We’re getting closer to a workable strategy now! On seeing b+d and b – d if you’re now thinking of recovering b and d from an addition and subtraction operation, you’re on the right track! These are mod 3 operations though and so a little more complicated, but the usual intuition of getting \(b\) and \(d\) holds. 

Adding \((b+d)mod3 \) and \((b-d)mod3 \) gives us something only dependent on \(b\), eliminating \(d\) for all b and d! Let’s call it \(\hat b\). You can check that \(b\ = (2\hat b) \ mod3 \).

Similarly,  \( (2((b+d)mod3 – (b-d)mod3))mod3 \) gives us \(d\).

So now B and D can deduce that B’s hat \(b\) and D’s hat \(d\) are one of the following four combinations that cover the cases when A and C both get their guesses wrong:

\begin{align*}
& b = 2(m_1 + n_1) \quad and \qquad \color{blue}{ d = 2(m_1 – n_1)}  \quad OR \\
&\color{red}{ b = 2(m_1 + n_2)} \quad and \qquad  d = 2(m_1 – n_2) \quad OR \\
&\color{red}{ b =2( m_2 + n_1)} \quad and \qquad d = 2(m_2 – n_1) \quad OR \\
& b =2( m_2 + n_2) \quad and \qquad \color{blue}{ d = 2(m_2 – n_2)}  \\
\end{align*}

But these are four \(b,d\) combinations you say. How can we pick so that at least one of the \(b\) or \(d\) is a correct guess? The key is seeing the second and third above will give the same result for \(b\)  and the first and fourth give the same result for \(d\).  This is because by using \(m_1, \ m_2, \  n_1, \  n_2\)  the choices can be rewritten as: 

\begin{align*}
& b = 2((a+1) + (c+1)) \quad and \qquad \color{blue}{ d = 2((a+1) – (c +1))}  \quad OR \\
&\color{red}{ b = 2((a+1) + (c-1))} \quad and \qquad  d = 2((a+1) – (c-1)) \quad OR \\
&\color{red}{ b =2( (a-1) + (c+1))} \quad and \qquad d = 2((a-1) – (c+1)) \quad OR \\
& b =2( (a-1) + (c-1)) \quad and \qquad \color{blue}{ d = 2((a-1) – (c-1))}  \\
\end{align*}

(you can verify this — remember these are all mod 3 operations)

What this now implies is that if B and D pick the red and blue options respectively:

\begin{align*}
b &= 2(a+c)  \\
d &= 2(a-c)
\end{align*}

they are guaranteed to get one of their guesses correct. Remember this is only true when A’s and C’s guesses based on their strategy, both end up being wrong. But this is precisely the scenario we set out to cover since otherwise one or both of A and C will guess correctly and set the group free.

Let’s finish up our example where ACBD have on (0,1,0,1). A and C guess (1,2) – a lost cause! B and D per the above formula guess 2,1. D guesses correctly and they yelp for joy as they are set free by the warden who is true to his word.  

So there you have it! This solves the problem.

Summary: 

Here is the strategy of guesses  then:

\begin{align*} 
& \bbox[yellow, 10px]{A: (b+d) \ mod \ 3} \\
& \bbox[yellow, 10px]{C: (b-d) \ mod \ 3 }\\
& \bbox[yellow, 10px]{B: 2(a+c) \ mod \ 3} \\
& \bbox[yellow, 10px]{D: 2(a-c) \ mod \ 3 }
\end{align*}

This is all too easy to verify in a spreadsheet or with a few lines of code.