Bullet Spray

Solution

Answer: Yes

Consider a revolver with 6 chambers. Two bullets are placed in consecutive chambers. Label the chambers \[ 1,2,3,4,5,6 \] and suppose the bullets occupy chambers \(1\) and \(2\). The other chambers \(3,4,5,6\) are empty. The cylinder is spun randomly before the first shot, so the firing position is equally likely to be any chamber. We pull the trigger once and hear a click, meaning the chamber aligned with the firing pin was empty. Therefore the first chamber must have been one of \[ 3,4,5,6 \] Each of these four possibilities is equally likely. After a trigger pull, the cylinder advances by one chamber. The chamber for the second shot therefore depends on which empty chamber occurred first. If the first chamber was \(3\) \[ 3 \rightarrow 4 \] and chamber \(4\) is empty. If the first chamber was \(4\) \[ 4 \rightarrow 5 \] and chamber \(5\) is empty. If the first chamber was \(5\) \[ 5 \rightarrow 6 \] and chamber \(6\) is empty. If the first chamber was \(6\) \[ 6 \rightarrow 1 \] and chamber \(1\) contains a bullet. Thus the only way the second shot fires a bullet is when the first empty chamber was \(6\). Since the first chamber could have been any of the four empty chambers with equal probability, \[ P(\text{bullet on second shot without shuffle}) = \frac{1}{4} = 0.25. \] Now consider shuffling (spinning) the cylinder again before the second shot. After a full shuffle, the cylinder position is again random among all six chambers. Since two chambers contain bullets, \[ P(\text{bullet after shuffle}) = \frac{2}{6}. \] Simplifying, \[ \frac{2}{6} = \frac{1}{3} \approx 0.333. \] Comparing the probabilities, \[ \frac{1}{3} > \frac{1}{4}. \] Therefore spinning (shuffling) the cylinder increases the probability of firing a bullet, from \(25\%\) to about \(33.3\%\).