Suppose that a balanced die is rolled repeatedly until the same number appears on the two successive rolls, and let $X$ denote the number of rolls that are required. Determine the value of $Pr(X=x)$, for $x=2,3,\dots$.
I guess the answer is $\big(\frac{5}{6}\big)^{x-2}\big(\frac{1}{6}\big)^2$. Am I right?
Best Answer
The case $X = x$ occurs precisely when:
$\bullet$ The $x-2$ rolls $2,3,\ldots,x-1$ are different from the previous roll.
$\bullet$ The $x$-th roll is the same as the previous roll.
So the probability of this happening is $\left(\frac{5}{6}\right)^{x-2} \cdot \frac{1}{6}$