A and B are shooting at a shooting range, but they only have one six-chambered revolver with one bullet. Therefore, they agreed to take turns randomly spinning the cylinder and shooting. A starts. Find the probability that the shot will occur when the revolver is with A.
The shot occurs on an odd attempt (when the revolver is at A), and no shot occurs on the other attempts. Therefore, the desired probability is the sum $1 / 6 + (5 / 6)^{2} \cdot 1 / 6 + (5 / 6)^{4} \cdot 1 / 6 + \ldots = 6 / 11$.

## Answer

$6 / 11$.