Ram throws a good six-sided die till six seems for the primary time. Independently Shyam rolls the die till six seems for the primary time, let $alpha , beta $ be the variety of throws required by Ram and Shyam respectively to acquire six consequently and ‘P’ be the chance that $|alpha – beta|le1$, then present that P is the same as $frac{8}{33}$

My method is as comply with.$alpha , beta $ are pure quantity such that if Ram will get a six in fifth throw that Shyam must get a six in both 4th, fifth or sixth throw in order that the situation $|alpha – beta|le1$ is happy.

Therefore the success in nth time period is $P_n=1-(frac{5}{6})^n$

Therefore the success in n+1th time period is $P_{n+1}=1-(frac{5}{6})^{n+1}$

Therefore primarily based on the situation the required chance is $(P_n)^2+2P_n.P_{n+1}$ however not getting the reply