Question 135966
{{{p(d)=1+d/33}}}


{{{p(0)=1}}} because {{{p(red(0))=1+red(0)/33=1+0=1}}}


{{{p(33)=2}}} because {{{p(red(33))=1+red(33)/33=1+1=2}}}


So:
{{{p(red(110))=1+red(110)/33}}}, all you need to do is simplify