If what you mean is that during this 10 day window, the onset of rain (like a monsoon "breaking") has a probability of 0.5 on any day, and it will go on raining till the end thereafter,
E(x) = $\sum_{k=1}^{10} 0.5^k\cdot(11-k) = \frac{9217}{1024}$, ≈ 9.001
This is not an integer, but as has been pointed out, that is not a problem !
Denoting the current day as index $i$ (so that yesterday is $i-1$ and the day before yesterday is $i-2$), the previous state comprises two elements, $State_{n-1}=\{S_{i-1},S_{i-2}\}$, where $S_k$ is the state of the weather for day $k$. Thus the current state is $State_{n}=\{S_{i},S_{i-1}\}$
Lettting $R$ be the occurrence of rain on a particular day, so that $\overline{R}$ is when no rain occurs for that day, there are $4$ possible values of the previous state $State_{n-1}$
$$\{\overline{R},\overline{R}\}, \{\overline{R},R\},\{R,\overline{R}\},\{R,R\}$$
For each of these $4$ states, there are only two possible next states:-
$$\begin{align}
State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},\overline{R}\}
\\
State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},R\}
\\
State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,\overline{R}\}
\\
State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,R\}
\end{align}$$
With the information given in the question, and based on the constraints of what the next state is based on the current state, you can figure out the structure of the transition matrix.
For example, the information Specifically, suppose that if it has rained for the past two days, then it will rain tomorrow with probability $0.7$, corresponds to
$$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=0.7\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=1-0.7=0.3$$
and if it rained today but not yesterday, then it will rain tomorrow with probability 0.5 corresponds to
$$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},R\})=0.5\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},R\})=1-0.5=0.5$$
Best Answer
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,
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.