Your probabilities are correct: in a given round, player 1 has probability $p = 1/6$ of winning and player $2$ has probability $q = 5/36$ of winning. The expectation of each player (assuming you mean winnings) is $pd$ for player 1 and $qd$ for player 2, where $d$ is the amount of money in the pot.
If we now let the game continue until someone wins, a third possibility arises: each round, the probability of passing to the next round is either $1-p$ or $1-q$, depending on whose turn it is.
Let $A$ mean "player 1 wins", let $B$ mean "player 2 wins",
and let $\bar{A}$ mean "player 1 loses" and let $\bar{B}$ mean "player 2 loses" (on a round-by-round basis). From your clarification in your comment, it follows that if player 2 starts, the first round's possible outcomes are $B$ or $\bar{B}$, the second round's possible outcomes are $A$ or $\bar{A}$, and so on, alternately. Thus, the possible game sequences are $\{B, \bar{B}A, \bar{B}\bar{A}B, \bar{B}\bar{A}\bar{B}A, \bar{B}\bar{A}\bar{B}\bar{A}B, \dotsc\}$.
The probability of a sequence like $\bar{B}\bar{A}\bar{B}A$ is the product of the probabilities: $$(1-q)(1-p)(1-q)p.$$
Now, player 2 wins iff the game sequence was one of the following: $\{B, \bar{B}\bar{A}B, \bar{B}\bar{A}\bar{B}\bar{A}B, \dotsc\}$. Thus, the probability of player 2 winning is
$$\sum_{k=0}^\infty [(1-q)(1-p)]^kq = \frac{q}{1-(1-p)(1-q)} = \frac{q}{p + q - pq}$$
by the geometric series formula.
Player 1 wins iff the game sequence was one of the following: $\{\bar{B}A, \bar{B}\bar{A}\bar{B}A, \bar{B}\bar{A}\bar{B}\bar{A}\bar{B}A, \dotsc\}$.
Thus, the probability of player 1 winning is
$$(1-q)\sum_{k=0}^\infty[(1-p)(1-q)]^k p = \frac{p(1-q)}{p+q-pq}.$$
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:
$$p = \frac{1}{6} \cdot 1 + \frac{5}{6}(1-p) = \frac{6-5p}{6} \iff 6p = 6-5p \iff p= \frac{6}{11}$$
Best Answer
You can solve this recursively. Noting that after A and B both didn't throw double six the game basically restarts, you get
$p={1\over 36} + {35 \over 36}{35 \over 36}p$.
Thus $p = \frac{{1\over 36}}{1-{35 \over 36}{35 \over 36}}= {36 \over 71}.$