D's chances of guessing correctly are either 100% (if B and C match) or 50% (if B and C mismatch). The chance that B and C match is 1/3, so D's overall chance of guessing correctly is 2/3.
C's chances of guessing correctly are 2/3 regardless of the initial setup.
Therefore D has the same chance as C.
Note: I'm assuming that all ${4 \choose 2} = 6$ initial set-ups are equiprobable.
Expanded Explanation:
There are 6 possible initial set-ups. Let's use lowercase for white and UPPERCASE for black. So initially we have one of these:
abCD
aBcD
aBCd
AbCd
ABcd
AbcD
Note that only the third and sixth possibilities (in bold) have B and C matching.
Now consider D's strategy: if B and C match, guess the opposite (100% chance to be right). If not, guess at random (50% chance to be right). So D is right with probability (1)1/3 + (.5)2/3 = 2/3.
Now consider C's strategy: just guess the opposite of B no matter what. From the table above (and from your question), C's chance is 2/3.
Note: D can achieve the 2/3 chance of winning by just using C's strategy (which is simpler): just guess the opposite of C. As you can see from the table above, this wins 2/3 of the time too since it implicitly incorporates the bold cases.
These kinds of problems are special because we're only looking for one solution, so that we can make educated guesses to stumble upon one.
In a magic square, 1 and the maximum number are usually paired. But in this puzzle, unlike a magic square, every number is included in exactly two sums. This gives us freedom we wouldn't normally have. We can pair 1 with 12 on the outside, but because of this freedom, we can put the next-most-extreme numbers together, to give us a place to start from:
$$\begin{array}{ccccccc}
& & & 1\\
? & & ? & & 2 & & ?\\
& ? & & & & 11\\
? & & ? & & ? & & 12\\
& & & ?
\end{array}$$
There is one undetermined number in a sum with 11 and a sum with 12. That can probably be made as small as possible: 3. (We could have done this symmetrically with 1 and 2 needing a big number, but probably not both at the same time.)
$$\begin{array}{ccccccc}
& & & 1\\
? & & ? & & 2 & & ?\\
& ? & & & & 11\\
? & & ? & & 3 & & 12\\
& & & ?
\end{array}$$
To complete $3+12$, we need a sum of $11$, which we could only get as $6+5$ or $7+4$. To complete $3+11$, we need $5+7$ or $4+8$. Therefore, we can't pick $7+4$ because it would wipe out both possibilities to complete the diagonal (this feels like kakuro). As such, the row needs a 6 and a 5. Which should go in the bottom left corner? Probably the middling value 6 since it's in a sum with 1 and a sum with 12:
$$\begin{array}{ccccccc}
& & & 1\\
? & & ? & & 2 & & ?\\
& ? & & & & 11\\
6 & & 5 & & 3 & & 12\\
& & & ?
\end{array}$$
This leaves $4+8$ for the diagonal on the right. But putting 4 in the top right would make the top row have no options since the numbers would have to be too big:
$$\begin{array}{ccccccc}
& & & 1\\
? & & ? & & 2 & & 8\\
& ? & & & & 11\\
6 & & 5 & & 3 & & 12\\
& & & 4
\end{array}$$
The only pair for the top row is 7,9 and the only pair for the upper diagonal is 9 and 10, so 9 must go in the intersection:
$$\begin{array}{ccccccc}
& & & 1\\
7 & & 9 & & 2 & & 8\\
& 10 & & & & 11\\
6 & & 5 & & 3 & & 12\\
& & & 4
\end{array}$$
As made clear in How to solve this system of linear equations, this is just one of many other solutions.
Best Answer
Let $a$ be the number of fish. Then you know that $a=3b+1$ for some non-negative integer $b$, since Ed saw that the remainder was one when he tried to divide the fish into thirds. He threw one back and took $b$ or himself, leaving $2b$. Similarly, $2b=3c+1$ and $2c=3d+1$, where $d$ is the number of fish Eric ended up taking at the end.
We know that $d$ must be odd, so equal to $2k+1$ for some non-negative $k$. Plugging that in we get $$2c=6k+4$$ or $$c=3k+2.$$ Therefore $$2b = 9k+7$$ and $k$ must also be odd, $k=2l+1$. Then $$2b = 18l+16$$ or $$b = 9l+8$$ and $$a = 27l +25.$$ The smallest value of $a$ occurs when $l=0$: Ed sees 25 fish, tosses one back, and takes 8 (leaving 16). Eddie tosses another back, taking five and leaving 10. Lastly Eric tosses one and takes three.