The given solution is false because it should be 5/6a=5/6b. If we assume that b shot first then for a the chance to win the game is 5/6 * the chance that he wins the resulting 5 chamber game in which he gets to go first. But for b the chance to win is 5/6 * the chance that he wins the resulting game of 5 chambers where he goes last. The answer modified a to mean winning the resulting 5 chamber game but he left b to mean winning the original 6 chamber game. and then he made the formula 5/6a=b which is correct but a+b=1 is not. It is not true that the chance that a wins the 5 chamber game after the first shot has been made + the chance that b wins the original 6 chamber game before the first shot has been made is 1. In fact with his new a and b a+b<1 since a has only 5/6 to happen while the original 6 chamber game a had 1 chance to happen.
There have been multiple comments showing different good arguments as to why both players have equal probability of winning. Allow me to add yet another good argument:
Assuming that players take turns shooting themselves instead of shooting the other, the chance that the first player dies is 1/6 on his first turn, 1/4 on his second, and 1/2 on his third. His first turn has 1 chance of happening, his second has (5/6)*(4/5) chance of happening , and his third has (5/6)*(4/5)*(3/4)*(2/3) chance of happening. That is because for every turn to happen all previous turns need to not result in a death, and every turn that someone does not die one empty chamber is out of the game. Add all those probabilities to get the chance that the person shooting first dies: (1/6)+(5/6)*(4/5)*(1/4)+(5/6)*(4/5)*(3/4)*(2/3)*(1/2)=1/2. Thus he has 1/2 chance to die and 1/2 chance to win and so does his opponent dince there will always be a winner and a die'er.
The given solution is false because it should be 5/6a=5/6b. If we assume that b shot first then for a the chance to win the game is 5/6 * the chance that he wins the resulting 5 chamber game in which he gets to go first. But for b the chance to win is 5/6 * the chance that he wins the resulting game of 5 chambers where he goes last. The answer modified a to mean winning the resulting 5 chamber game but he left b to mean winning the original 6 chamber game. and then he made the formula 5/6a=b which is correct but a+b=1 is not. It is not true that the chance that a wins the 5 chamber game after the first shot has been made + the chance that b wins the original 6 chamber game before the first shot has been made is 1. In fact with his new a and b a+b<1 since a has only 5/6 to happen while the original 6 chamber game a had 1 chance to happen.
There have been multiple comments showing different good arguments as to why both players have equal probability of winning. Allow me to add yet another good argument:
Assuming that players take turns shooting themselves instead of shooting the other, the chance that the first player dies is 1/6 on his first turn, 1/4 on his second, and 1/2 on his third. His first turn has 1 chance of happening, his second has (5/6)*(4/5) chance of happening , and his third has (5/6)*(4/5)*(3/4)*(2/3) chance of happening. That is because for every turn to happen all previous turns need to not result in a death, and every turn that someone does not die one empty chamber is out of the game. Add all those probabilities to get the chance that the person shooting first dies: (1/6)+(5/6)*(4/5)*(1/4)+(5/6)*(4/5)*(3/4)*(2/3)*(1/2)=1/2. Thus he has 1/2 chance to die and 1/2 chance to win and so does his opponent dince there will always be a winner and a die'er.