Let P(n) be the probability that the program terminates, starting from x=n. Clearly P(0)=1; the iteration step gives the eleventh-order recurrence$$P(n)=\frac12[P(n-1)+P(n+10)]$$(for all n>0). Assuming unbounded integer sizes, we also want [imath]P(n)\to0[/imath] as [imath]n\to+\infty[/imath].

The general solution to such a recurrence is generically of the form$$P(n)=\sum_k c_k (r_k)^n$$where [imath]r_k[/imath] are the roots of the characteristic polynomial, found by substituting [imath]P(n)=x^n[/imath]. [If the characteristic polynomial has repeated roots the form is slightly more complicated.] Here the characteristic polynomial is the eleventh-degree polynomial [imath]x^{11}-2x+1[/imath]. This has eleven distinct roots, so finding the coefficients seems difficult. But one of the roots is 1, and nine have magnitudes larger than 1, so there's really only one physical root, [imath]r\approx0.500245462[/imath], so$$P(n)=r^n\;\;.$$This makes the termination probability, starting from n=10, roughly 1/1019.