Follow the reroll rule and translate a difference into an absolute value

An unbiased ordinary die is thrown. If its score is not 3, the result is that score. If it is 3, the die is thrown again and the result is the absolute difference of the two scores. Let R denote the result. Show that $P(R=0)=1/36$ and $P(R=1)=2/9$, and find the distribution of R.6 marks

Find

(i) the mean result,2 marks

(ii) the probability that the difference between $(R-1)^2$ and 3 is at most the mean from (i).4 marks