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

Let $X_1,X_2$ denote the first and possible second throws. $P(R=0)=P(X_1=3,X_2=3)=(1/6)(1/6)=1/36$. For result 1, $P(R=1)=P(X_1=1)+P(X_1=3,X_2=4)+P(X_1=3,X_2=2)=1/6+(1/6)(1/6)+(1/6)(1/6)=2/9$. The source’s last subscript is printed $X_3$; it must be $X_2$, since there is no third throw.

$P(R=2)=P(X_1=2)+P(X_1=3,X_2=5)+P(X_1=3,X_2=1)=1/6+(1/6)(1/6)+(1/6)(1/6)=2/9$. The final subscript here is also corrected from the source’s $X_3$ to $X_2$.

$P(R=3)=P(X_1=3,X_2=6)=(1/6)(1/6)=1/36$.

$P(R=4)=P(X_1=4)=1/6$; $P(R=5)=P(X_1=5)=1/6$; $P(R=6)=P(X_1=6)=1/6$.

$E(R)=0(1/36)+1(2/9)+2(2/9)+3(1/36)+4(1/6)+5(1/6)+6(1/6)=13/4$.

“Difference between” is interpreted as absolute difference, as in the source working. Therefore

$\begin{aligned}&P\!\left(|(R-1)^2-3|\le\frac{13}4\right)\\ & =P\!\left(-\frac{13}4\le(R-1)^2-3\le\frac{13}4\right).\end{aligned}$

Add 3 to get $P(-1/4\le(R-1)^2\le25/4)$. Since a square is nonnegative, this is $P((R-1)^2\le25/4)=P(|R-1|\le5/2)$.

This is $P(-5/2\le R-1\le5/2)=P(-3/2\le R\le7/2)$. Using the integer support, it becomes $P(0\le R\le3)=1/36+2/9+2/9+1/36=1/2$.