The executioner and the hats (3 colors)

1. All except the first are determined uniquely.
2. Person 9 knows $y_{10}$, sees $x_1,\dots,x_8$, and solves for $x_9$: $x_9\equiv-\big(y_{10}+x_1+\cdots+x_8\big)\pmod{3}$. He says it and he gets it right.
3. Coding: red=0, blue=1, green=2 (module 3).

Back to problem
Answer: 9 saved are guaranteed.
Coding: red=0, blue=1, green=2 (module 3).
Let $x_1,\dots,x_{10}$ be the real value of each hat.
Person 10 (first to speak), who sees $x_1,\dots,x_9$, says:

$$ y_{10}\equiv-(x_1+\cdots+x_9)\pmod 3. $$

It may fail, but it leaves the equation fixed:

$$ y_{10}+x_1+\cdots+x_9\equiv0\pmod3. $$

Person 9 knows $y_{10}$, sees $x_1,\dots,x_8$ and solves $x_9$:

$$ x_9\equiv-\big(y_{10}+x_1+\cdots+x_8\big)\pmod3. $$

He says it and he gets it right.
Then person 8 already knows $x_9$ (heard), sees $x_1,\dots,x_7$ and solves for $x_8$; and so on until 1.
All except the first are determined uniquely.
Guarantee: 9 saved always; The first one only has a probability of $1/3$ of being correct.