We do not need to consider the other $6$ people and can only focus on the two Chinese and two Americans. In which case, the valid arrangment must be $CCAA$.

By the multiplication principle, there are $2\times 1$ ways to arrange these two Chinese, and $2\times 1$ ways to arrange these two Americans. Meanwhile, thre are $4\times 3\times 2\times 1$ ways to arrange these four people free from any restriction. Therefore the answer is $$\frac{(2\times 1)\times (2\times 1)}{4\times 3\times 2\times 1}=\boxed{\frac{1}{6}}$$