Problem:
Suppose that a committee with $n$ members needs to vote on whether to accept a proposition.
Each member in the committee can cast a yes/no vote ($q_i\in\{1,0\}$ for $i\in \{1,2,...,n \}$), and each member's vote has a different weight ($w_i\in[0,1]$ for $i\in \{1,2,...,n \}$).
The committee rules stipulate that the proposition is to be accepted if the weighted average of votes is more than 50%, namely, if $$\bar{q}(n)=\sum_{i=1}^{n} w_i q_i>0.5$$
Suppose that (i) voting is sequential, (ii) each vote is a random draw from a Bernoulli distribution with unknown mean $p$, and (iii) the order of voting is independent of voting weights.
I would like to define a 'dynamic quorum rule', such that a decision (accept vs reject) is made with only a subset of the committee's votes (e.g. the first $k$ votes), and yet be statistically confident that the decision made is the same as the one that would have been made if all members had voted.
What I currently have:
This problem is very similar to one that I posted earlier. The only difference is that I am now interested in a weighted solution. That means that for $w_i=w$, the bayesian solution proposed in the related question also applies here.
Exact treatment (intractable?)
You're asking for an answer similar to the one accepted in your related question, but that approach seems unlikely to be tractable here; however, the development is sketched below in case someone else might have an insight.
Let $$S_{j,k}=\sum_{i=j}^k w_i q_i$$ where $1\le j\le k\le n$ and $q_1,...,q_n$ are i.i.d. Bernoulli($p$) random variables. The sequential decision procedure is as before, mutatis mutandis, using the following posterior probabilities:
$$\begin{align}p_k &= P\{S_{1,n}<acc\mid (q_i)_{i=1..k}\}\\ &= \sum_{t\ <\ acc}P\{S_{1,n}\!=\!t\mid S_{1,k}\!=\!s\}. \end{align}$$
Using a Beta$(\alpha,\beta)$ prior for $p$, as before, now leads to the following form for these posterior probabilities:
$$\begin{align}&P\{S_{1,n}=t\mid S_{1,k}=s\}\\ &=\int_0^1 P\{S_{1,n}\!=\!t\mid S_{1,k}\!=\!s, p\!=\!p'\}\,\frac{P\{S_{1,k}\!=\!s\mid p\!=\!p'\}f(p')}{\int_0^1 P\{S_{1,k}\!=\!s\mid p\!=\!p''\}f(p'')dp''}\,dp'\\ \\ &=\frac{\int_0^1 P\{S_{k+1,n}=t-s\mid p\!=\!p'\}\,P\{S_{1,k}\!=\!s\mid p\!=\!p'\}f(p')\,dp'}{\int_0^1 P\{S_{1,k}\!=\!s\mid p\!=\!p''\}f(p'')\, dp''}\\ \\ &= \frac{\int_0^1\ \ \sum_{A\subseteq\{w_{k+1},...,w_n\}\\ \sigma A=t-s}p^{\lambda A}(1-p)^{n-k-\lambda A}\ \ \sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}p^{\lambda A'}(1-p)^{k-\lambda A'}\ \ f(p)\,dp}{\int_0^1\ \ \sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}p^{\lambda A'}(1-p)^{k-\lambda A'}\ \ f(p)\,dp}\\ \\ &= \frac{\sum_{A\subseteq\{w_{k+1},...,w_n\}\\ \sigma A=t-s}\sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}\int_0^1 p^{\alpha+\lambda A+\lambda A'}(1-p)^{\beta+n-\lambda A-\lambda A'}\,dp}{\sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}\int_0^1 p^{\alpha+\lambda A'}(1-p)^{\beta+k-\lambda A'}\,dp}\\ \\ &= \frac{\sum_{A\subseteq\{w_{k+1},...,w_n\}\\ \sigma A=t-s}\sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}\mathrm{B}(\alpha+\lambda A+\lambda A',\,\beta+n-\lambda A-\lambda A')}{\sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}\mathrm{B}(\alpha+\lambda A',\beta+k-\lambda A')}\tag{*} \end{align}$$
where, as before, $\mathrm{B}(\alpha,\beta)$ denotes the standard complete Beta function. In the above, we have used the combinatorial fact that, for $1\le j\le k\le n$,
$$P\{S_{j,k}=s\mid p\} = \sum_{A\subseteq\{w_j,...,w_k\}\\ \sigma A=s}p^{\lambda A}(1-p)^{k-j+1-\lambda A} = \sum_{A\in \mathcal{P}(\{w_j,...,w_k\})\\ \sigma A=s}p^{\lambda A}(1-p)^{k-j+1-\lambda A} $$ where $\sigma A$ denotes the sum of the elements in $A$, $\lambda A$ denotes the number of elements in $A$, and $\mathcal{P}(\{w_j,...,w_k\})$ denotes the power set of $\{w_j,...,w_k\}$.
It can be seen that in the case of equal weights (e.g. $w_i=1$), $(*)$ reduces to the previous result -- the outer summation in the numerator provides the factor $\binom{n-k}{t-s}$ and each of the other two summations (the inner one in the numerator and one in the denominator) provide a factor of $\binom{k}{s}$ that cancel one another, while in the summations $\lambda A = t-s$ and $\lambda A'=s$. However, in the case of arbitrary weights the combinatorics make it difficult to proceed any further.