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:
sage: i,j,k = var ('i,j,k')
sage: p (n) = (6 - abs (n - 7)) / 36
sage: sum (p (i) * sum (p (j) * sum (p (k),k,j+1,12),j,i+1,11),i,2,10)
895/7776
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:
sage: count = 0;
sage: ntrials = 1000000;
sage: for n in range (0,ntrials):
... seven = false
... last = 12
... increased = false
... while true:
... roll = 2 + ZZ.random_element (6) + ZZ.random_element (6)
... if seven and roll == 7:
... count = count + 1
... break
... if increased and roll > last:
... break
... seven = roll == 7
... increased = roll > last
... last = roll
...
sage: q = count / ntrials
sage: print q.n (), "+-", sqrt (q * (1 - q) / ntrials).n ()
0.191416000000000 +- 0.000393415702462421
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:
sage: d=6
sage: nsums = 2 * d - 1
sage: nstates = 2 * nsums
sage: a = matrix (QQ,nstates)
sage: b = matrix (QQ,nstates,1)
sage: for prev in range (2,2*d+1):
... for up in range (0,2):
... index = prev - 2 + nsums * up
... for next in range (2,2*d+1):
... p = (d - abs (next - (d+1)))/(d*d)
... if prev == d + 1 and next == d + 1:
... b[index] = p
... else:
... if next > prev:
... nextup = 1
... else:
... nextup = 0
... if up * nextup == 0:
... nextindex = next - 2 + nsums * nextup
... a[index,nextindex] = -p
...
sage: for i in range (0,nstates):
... a[i,i] = a [i,i] + 1
...
sage: result = a.solve_right (b)[nsums-1]
sage: print result, "~", result.n ()
(106460465616/556534555787) ~ (0.191291743718327)
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".
You have multinomial distribution with $n$ trials, $k = 6$ outcomes and all $p_i = \frac{1}{6}$. So probability of getting two equal rolls is $\sum\limits_{x_1 + \ldots + x_6 = n} \left(\frac{n!}{x_1! \ldots x_6!}\cdot \frac{1}{6^n}\right)^2$. It's unlikely there is good closed form for it. For asymptotic see, for example, this answer on mathoverflow.
For example, if we have $n = 2$, we need to sum over all variants to partitioning $2$ into $6$ non-negative terms. There are $15$ of them with $2$ ones and $4$ zeroes: $1+1+0+0+0+0$, $1+0+1+0+0+0$, ..., $0+0+0+0+1+1$ and $6$ with $1$ two and $5$ zeroes: $2+0+0+0+0+0$, $0+2+0+0+0+0$, ...
First variant will give in our sum $15$ terms equal to $\left(\frac{2!}{1!1!0!0!0!0!} \cdot\frac{1}{36}\right)^2 = \frac{1}{18^2}$. Second will give $6$ terms equal to $\left(\frac{2!}{2!0!0!0!0!0!} \cdot \frac{1}{36^2}\right)^2 = \frac{1}{36^2}$. So the answer is $\frac{1}{18^2}\cdot 15 + \frac{1}{36^2}\cdot 6 \approx 0.05$.
Best Answer
The event that you have three consecutive numbers is the union of five disjoint events:
$$ \overline3456\cup123\overline4\cup234\overline5\cup\overline2345\cup2345\;, $$
where a number with or without a bar indicates that that number is absent or present, respectively. The first four of these have the same probability, so we only need to calculate $2$ different probabilities. By inclusion–exclusion, we have
$$ \mathsf P(\overline3456)=\sum_{j=0}^3\binom3j(-1)^j\left(\frac{5-j}6\right)^5=\frac{125}{1296} $$
and
$$ \mathsf P(2345)=\sum_{j=0}^4\binom4j(-1)^j\left(\frac{6-j}6\right)^5=\frac5{81}\;. $$
Thus the probability to get three consecutive numbers is
$$ 4\cdot\frac{125}{1296}+\frac5{81}=\frac{145}{324}\approx45\%\;. $$