Here's a hint for the riddle. Vald had the back guy use the 3 people in front of him to the others' advantage. What if he utilized everyone in front of him in some manner? Also, an advanced mathematical background, while not essential, is extremely helpful in solving this dilemma.

You are correct. Now find this method.

Also, if no one gets it in a couple of days, I'll try to explain the answer as best as I can.

I wrote out a pretty elaborate answer. We'll see what happens.

Spoiler Alert:
Hopefully this makes sense, and hopefully I didn't make any errors.

No sage is capable of seeing the hat color of sage in the back, so he's pretty much screwed without luck.

The sages decide to assign values to the colors.

W=0
B=1
R=2

The back sage adds up all the values of the hats in front of him under a base 3 counting system (1, 2, 10, 11, 12, 20 etc.) But only keeps track of the ones digit, as that is all that is necessary. (This is essentially modular arithmetic, but I'll avoid using the terminology)

After he has added up the values, he says the color associated with the ones digit of his summation.

The following sage doesn't know the color of his hat, but he can do what the previous sage did with the 98 sages in front of him, and subtract that from the assigned value of the color he just heard to figure out his own color. For example, say he comes up with 1 and he previously heard "white", meaning 0. He then knows his hat must be red, or value 1 (0-1=2).

From this, any sage in line after this can subtract subsequent values of the previous guesses leading up to his place in line, and then subtract the sum of the values in front of him from this result to arrive at his own value, and in turn color. And, certainly the sages can keep subtracting the values of the guesses from the original as the other sages before them state them, as opposed to having to remember a long string of numbers.


Small sample size example:
BRRWB

First sage sees: RRWB. R+R+W+B = 2+2+0+1=2. He says "red"
Second sage sees: RWB. R+W+B = 2+0+1=0. (Previously obtained 2) 2-0=2. He says "red"
Third sage sees: WB. W+B = 0+1=1. 2-2-1=2. He says "red"
Fouth sage sees: B. B=1. 2-2-2-1=0. He says "white"
Fifth sages sees nothing 2-2-2-0-0=1. He says "black"

Extend this to 100 sages and it'll work just as well. At worst, 99 sages survive.