Suppose I have observations $x_k\stackrel{iid}{\sim}Bern(\theta)$ $k=1,...,2n$ distributed over an interval, say $(0,1]$. $n$ is known, but $\theta$ is not. The likelihood is then $$ \mathcal{L}(\theta|x_k's)= \prod_{x_k=\{0,1\},k=1,...,2n}\theta^{x_k}(1-\theta)^{1-x_k}=\theta^{\sum_k x_k}(1-\theta)^{2n-\sum_k x_k} $$ I have two partitioning schemes:

1. Partition interval $(0,1]$ into equal-length segments, $(0,1/2]$ and $(1,2,1]$ with potentially unequal number of $x_k$'s in each of the two equal-length segments.

2. Partition interval $(0,1]$ into segments of potentially unequal length, but $n$ $x_k$'s in each segment.

Let's say $X_1=\sum_{\text{segment 1}}x_k$ and $X_2=\sum_{\text{segment 2}}x_k$ are two random variables. Given my data generating process, is there any reason to think that my parameter $\theta$ should be different for $X_1$ and $X_2$ when I choose one partition scheme over the other?

In other words, for partition scheme 2, I expect $X_1$ and $X_2$ to follow a binomial distribution with size $n$, but can the parameter be the same because they came from $iid$ Bernoulli trials? For partition scheme 1, my gut tells me that the parameters won't be equal.