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
$$(1-q)\sum_{k=0}^\infty[(1-p)(1-q)]^k p = \frac{p(1-q)}{p+q-pq}.$$
Hint:
By throwing $3$ dice let $p$ denote the probability that all faces will be the same and let $q$ denote the probability that all faces are distinct.
Then $p=\frac66\frac16\frac16=\frac1{36}$ and $q=\frac66\frac56\frac46=\frac{20}{36}$.
The probability of winning will take the shape:$$p+(1-p-q)\times\cdots+q\times\cdots=\frac1{36}\times\cdots+\frac{15}{36}\times\cdots+\frac{20}{36}\times\cdots$$
Can you fill in $\cdots$ yourself?
$$\frac1{36}+\frac{15}{36}\left[\frac16+\frac56\frac16\right]+\frac{20}{36}\left[\frac1{36}+\frac{15}{36}\frac16+\frac{20}{36}\frac1{36}\right]$$
Best Answer
As others have remarked, this can be solved exactly, but the solution would be rather tedious and uninspiring. You can find the probability for $A$ to win by introducing as variables the probabilities for $A$ to win in each of $22$ states determined by the last roll and whether it increased, and solving the system of linear equations in these variables that's determined by the transition probabilities. The initial state could be taken to be $12$ (increased or not).
You can find lots of examples of such calculations on this site; instead of going through all the details of that, I'd like to do something perhaps more interesting and illuminating and estimate the desired probability on the basis of a simple approximation. The various winning events are not independent; for instance, given that $A$ didn't win on the last roll, it's less likely that she will win on this roll, since she can't have had a $7$ in both last rolls. But since the probability of winning in a given roll is rather low for both players, let's neglect this effect and see how good the resulting estimate is.
So let's model the process by one in which on each roll there's a probability $p_A$ of $A$ winning, a probability $p_B$ of $B$ winning and a probability $1-p_A-p_B$ of the game continuing, and let's use the marginal probabilities of winning in a general round for $p_A$ and $p_B$, neglecting the fact that $A$ can win on the second roll and $B$ can't. Then the probability $q_A$ that $A$ wins the game satisfies the recurrence $q_A=p_A+(1-p_A-p_B)q_A$, so $q_A=p_A/(p_A+p_B)$. We have $p_A=1/36$, and here's an evaluation of $p_B$ in Sage:
Thus the probability $q_A$ that $A$ wins is estimated to be $\frac1{36}/(\frac1{36}+\frac{895}{7776})=\frac{216}{1111}\approx0.1944$. This compares surprisingly well with the result of a simulation:
The good agreement seems somewhat spurious; taking into account that $A$ has a chance to win on the second roll and $B$ doesn't "improves" the estimate to $\frac1{36}+(1-\frac1{36})\frac{216}{1111}=\frac{8671}{39996}\approx0.2168$.
On second thought, it didn't seem quite so tedious to set up the linear system and solve it:
So the desired probability is
$$ \frac{106460465616}{556534555787}\approx0.1913\;, $$
in agreement with the simulation result. The large numerator and denominator suggest that there's unlikely to be a simpler solution that would have been worthy of a "riddle book".