You are right, of course.

It's a geometric sequence, and they would solve like this.. if S is the sum then

S = 1/6 + 1/6 * (5/6)^2 + 1/6 * (5/6)^4 + ...

S * (5/6)^2 = 1/6 * (5/6)^2 + 1/6 * (5/6)^4 + ...

-------------------

S - (5/6)^2 S = 1/6

as almost all the terms on the left side cancel out when the lower equation is subtracted from the upper.

11/36 S = 1/6

S = 36 / (6*11) = 6/11

***

Another way to do it, without referring to sums of infinite geometric sequences:

The first player loses if the 1st, 3rd, 5th etc. shot is the unlucky one.

The second player loses if the last shot is 2nd, 4th, 6th etc.

Every shot being the unlucky one comes with a probably of 5/6 of the one before; that is, one more "click" before boom. 5/6 can be taken as a common term, so the probability of player 2 losing is 5/6 is the probability of player 1 losing. Thus,

P(1st loses) + P(2nd loses) = 1

P(1st loses) + 5/6 * P(1st loses) = 1

from which P(1st loses) = 6/11