Alternate moves, first one to roll a 6 wins: what's the probability of winning?

Solution 1:

Player A can win straightaway with probability $\frac{1}{6}$ and with probability $\frac{5}{6}$ we are in exactly the same situation from player B's perspective. Let's name $p$ the probability of winning (in any number of moves) for the player who is about to roll the die. Then:

$$p = \frac{1}{6} \cdot 1 + \frac{5}{6}(1-p) = \frac{6-5p}{6} \iff 6p = 6-5p \iff p= \frac{6}{11}$$

Solution 2:

Another way: $A$ wins the game if either $A$ wins the turn, or both $A$ and $B$ loose the turn and $A$ wins the game from there.

$$p_A= \tfrac 16 + (\tfrac 56)^2p_A$$

Thus $p_A=6/11$, $p_B= 5/11$

in this case, the games dont seem to have a end point (don't go default).

It is almost certain that that the game will eventually end in a victory for one player or the other, rather than continue indefinitely.   The only question is "when?" will the game finish, not "if?"