Interesting problem! These are simply the partial sums of rows of Pascal's triangle. So in this case, the answer is just the sum of the first six elements of the $n = 9$ row (divided by $2^9 = 512$):
$$
p_6 = \frac{1+9+36+84+126+126}{2^9} = \frac{382}{512} = \frac{191}{256}
$$
Here's how this comes about: There are only two possible final states: certain drought, and certain rain. For any $k, 0 \leq k \leq 10$, let $p_k$ be the probability that the final state will be certain rain, given that the initial probability of rain is $\frac{k}{10}$. (Here, initial only means "current" since the process is homogeneous in time.)
Then, there is a simple set of linear equations relating the $p_k$. Suppose $k = 1$ initially. That is, the current rain probability is $\frac{1}{10}$. Then with probability $\frac{1}{10}$, the next rain probability will be $\frac{2}{10}$, and with probability $\frac{9}{10}$, the next rain probability will be $0$ (and the final state is certain drought). We can represent this as follows:
$$
p_1 = \frac{1}{10} p_2 + \frac{9}{10} p_0
$$
where $p_0 = 0$, naturally.
Now, let us suppose that $k = 2$ initially. Then with probability $\frac{2}{10}$, the next rain probability will be $\frac{3}{10}$, and with probability $\frac{8}{10}$, the next rain probability will be $\frac{1}{10}$. We can represent this as follows:
$$
p_2 = \frac{2}{10} p_3 + \frac{8}{10} p_1
$$
Proceeding along these lines, we can write equations of the form
$$
p_k = \frac{k}{10} p_{k+1} + \frac{10-k}{10} p_{k-1} \qquad 1 \leq k \leq 9
$$
with boundary conditions $p_0 = 0, p_{10} = 1$.
Now, interestingly, because for any $k$ the coefficients of $p_{k-1}$ and $p_{k+1}$ sum to $1$, we can view each of the $p_k$ as a weighted mean of $p_{k-1}$ and $p_{k+1}$. That is to say,
- $p_0 = 0$
- $p_1$ is $\frac{1}{10}$ of the way from $p_0$ to $p_2$
- $p_2$ is $\frac{2}{10}$ of the way from $p_1$ to $p_3$
- $p_3$ is $\frac{3}{10}$ of the way from $p_2$ to $p_4$
and so on. This permits us to find $p_2$ in terms of $p_1$, $p_3$ in terms of $p_2$, etc. In particular, if we let $p_1 = q$, where $q$ is some quantity currently unknown, then
$$
p_2-p_1 = \frac{9}{1} (p_1-p_0) = \frac{9}{1}q
$$
$$
p_3-p_2 = \frac{8}{2} (p_2-p_1) = \frac{9 \times 8}{2 \times 1}q
$$
$$
p_4-p_3 = \frac{7}{3} (p_3-p_2) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} q
$$
and for any $k$,
$$
p_{k+1}-p_k = \binom{9}{k} q
$$
And since
$$
\sum_{k=0}^9 p_{k+1}-p_k = p_{10}-p_0 = 1
$$
it must therefore be the case that $q = \frac{1}{2^9}$, and
$$
p_k = \frac{1}{2^9} \sum_{i=0}^{k-1} \binom{9}{i}
$$
So your intuition that the binomial coefficients were involved was not far off; it's just that they represent not the actual probabilities themselves, but their first differences. An obvious generalization of this yields a similar expression when the probability of rain is $\frac{k}{n}$, with increments of size $\frac{1}{n}$:
$$
p_k = \frac{1}{2^{n-1}} \sum_{i=0}^{k-1} \binom{n-1}{i}
$$
Alas, this question suggests that no further simplification is likely for general $k$ and $n$. (Obviously, closed forms for some special cases may be obtained.)
ETA: The relatively simple form of the answer makes me wonder if there is a cleverer answer that relies on some analogy between selecting no more than $k$ of $n$ objects and the probability of certain rain starting with a rain probability of $\frac{k}{n}$. But I confess nothing quickly comes to mind.
