# Let $G$ be a finite group and define a probability $P$ on $G$ by $P(id)=1-\frac{\epsilon}{2}, P(s)=\frac{\epsilon}{2(|G|-1)},0\leq\epsilon\leq2$

Let $$G$$ be a finite group and define a probability $$P$$ on $$G$$ by $$P(id)=1-\frac{\epsilon}{2}, P(s)=\frac{\epsilon}{2(|G|-1)},0\leq\epsilon\leq2$$ Show that $$P^{*k}(id)=\frac{1}{|G|}+\frac{|G|-1}{|G|}\left(1-\frac{\epsilon}{2}\frac{|G|}{|G|-1}\right)^k$$ and $$P^{*k}(s)=\frac{1}{|G|}-\frac{1}{|G|}\left(1-\frac{\epsilon}{2}\frac{|G|}{|G|-1}\right)^k$$.

Using this, show that $$\|P^{*k}-U\|=\frac{|G|-1}{|G|}\left|1-\frac{\epsilon}{2}\frac{|G|}{|G|-1}\right|^k$$, and that $$\sum^*d_pTr(\hat{P}(\rho)^k(\hat{P}(\rho)^k)^*)=(|G|-1)\left(1-\frac{\epsilon}{2}\frac{|G|}{|G|-1}\right)^{2k}$$

Once showing the first two parts, I know the formula $$\|P^{*k}-U\|=\frac{1}{2}\sum_{s\in G}\left|P(s)-\frac{1}{|G|}\right|$$, however, that is also not very obvious to me.

any help is appreciated