Let p for every shot = probability of the chamber containing the bullet * probability that the bullet is still in the gun (hasn't fired yet)

P(1st shot) = 1/6 * 6/6 = 1/6 (1/6 chance that the bullet still contains the bullet * 100% chance the gun still contains the bullet)
P(2nd shot) = 1/5 * 5/6 = 1/6 (assuming first shot didn't fire, 1 in 5 chance of the bullet * 5/6 chance the gun still contains the bullet)
P(3rd shot) = 1/4 * 4/6 = 1/6
P(4th shot) = 1/3 * 3/6 = 1/6
P(5th shot) = 1/2 * 2/6 = 1/6
Now, there's only one bullet left and 100 % chance of it still being in the gun but 1 in 6 chance that the game would get to this point:
P(6th shot) = 1/1 * 1/6 = 1/6

Person 1 = 1/6 + 1/6 + 1/6 = 50%
Person 2 = 1/6 + 1/6 + 1/6 = 50%

Makes no differnce