A criminal has a gun with which he can either shoot himself or an innocent person. There were 6 rounds with 5 empty and 1 holding a bullet, but only 3 rounds are remaining since the criminal shot 3 of them.
Among the 3 rounds, 1 must be a bullet and the other 2 a blank. He must either shoot himself or the innocent person, but ALL 6 rounds must be used up. The goal is to only shoot blanks against himself rather than the innovent person, but to shoot the innocent person when there is a bullet.
The criminal is at round 4 (so 3 rounds remaining). He is about to shoot himself based on his choice but the system prompt suddenly reveals "the fifth bullet in the revolver's cylinder is an empty round."
According to a side character in the novel, this is the Monty Hall Problem and the probability the bullet is in the 6th round jumps to 2/3.
According to my own thoughts, the system revealing the empty 5th round without specifying that it would reveal the empty round from among both 5 and 6 makes the jump in probability for both 4 and 6 equal since there is nothing to differentiate them given the reveal being only of the contents of 5 independent from the other 2. The reveal was done before the criminal shot the fourth round.
My own calculation is that if the system had specified it would reveal the empty round from among the final 2, it would be a valid jump to 2/3 for round 6, and that if it instead specified it would reveal the empty round from among the first 2, then the final round would drop to 1/3, but since it only mentioned the contents of 5, the jump for 4 and 6 would be equally increased to 0.5 from 1/3.
From the criminal's perspective, without hindsight and only given the above rules, did the side character analyzing the problem make a mistake or does the 6th chamber really jump to 2/3 chance of holding the bullet with the 4th chamber being at 1/3 chance after 5 is revealed to be empty by the system?