provable evenly shuffle each element is swapped with **equal probability** to any positions ## proof: for the last element, there is no probability for other elements swap with it before, each position is 1/n; for the second last element, the probability for position [0, n-1] is 1/(n-1), and the probability for finally swapping at the last position is depending on the last element, i.e. 1/n; th
(&mut self, arr: &mut Vec<T>)