Ram throws a fair six-sided die until six appears for the first time. Independently Shyam rolls the die until six appears for the first time, let $\alpha , \beta $ be the number of throws required by Ram and Shyam respectively to obtain six as a result and 'P' be the probability that $|\alpha – \beta|\le1$, then show that P is equal to $\frac{8}{33}$
My approach is as follow.$\alpha , \beta $ are natural number such that if Ram gets a six in 5th throw that Shyam needs to get a six in either 4th, 5th or 6th throw so that the condition $|\alpha – \beta|\le1$ is satisfied.
Hence the success in nth term is $P_n=1-(\frac{5}{6})^n$
Hence the success in n+1th term is $P_{n+1}=1-(\frac{5}{6})^{n+1}$
Hence based on the condition the required probability is $(P_n)^2+2P_n.P_{n+1}$ but not getting the answer
Best Answer
Probability that R. gets six in exactly $n$th roll and S. gets six in exactly $(n+1)$th roll:
$$P_n = \frac{5^{2n-1}}{6^{2n+1}} = \frac{1}{30}\cdot\left(\frac{25}{36}\right)^n$$
Probability that they both get 6 in exactly $n$th roll: $$P'_n = \frac{5^{2n-2}}{6^{2n}} = \frac{1}{25}\cdot \left(\frac{25}{36}\right)^n$$
Hence the total probability is $$\sum_{n=1}^\infty (2P_n + P'_n) = \left(\frac 2{30} + \frac{1}{25}\right)\cdot \frac{25}{11}=\frac{8}{33}$$