Compute Minimum Mean for At-Most Two Clusters?

Question

Let X be a set of 1000 points, say [0,1] x [0,1]. Let a,b be another two uniform points from the same square. We construct more than two clusters centered at a and b.

Answer 1

Number the points with $1,2,\dots,100$ and let $X_i$ take value $1$ if point $i$ is an element of $C(a)$ and value $0$ if point $i$ is an element of $C(b)$.

Observe that for point $X_1$ the probability of not being an element of $C(a)\cup C(b)$ is zero so that: $$P(X_1\in C(a))+P(X_1\in C(b))=1\tag1$$ On base of symmetry we also conclude that:$$P(X_1\in C(a))=P(X_1\in C(b))\tag2$$Then combinining $(1)$ and $(2)$ we find:$$P(X_1\in C(a))=\frac12=P(X_1\in C(b))$$

Then $|C(a)|=\sum_{i=1}^{100}X_i$ and with linearity of expectation and symmetry we find:$$\mathbb E|C(a)|=100P(X_1\in C(a))=100\times\frac12=50$$

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addendum:

Evidently $|C(a)|$ has binomial distribution with parameters $n=100$ and $p=\frac12$, and if random variable $Z$ has binomial distribution with parameters $100$ and $\frac12$ then: $$\begin{aligned}\mathbb{E}\min\left(Z,100-Z\right) & =2^{-100}\sum_{k=0}^{100}\binom{100}{k}\min\left(k,100-k\right)\\ & =2^{-100}\left(\sum_{k=0}^{49}\binom{100}{k}k+\binom{100}{50}50+\sum_{k=51}^{100}\binom{100}{k}\left(100-k\right)\right)\\ & =2^{-100}\left(2\sum_{k=0}^{50}\binom{100}{k}k-\binom{100}{50}50\right) \end{aligned} $$Focusing on the first term we find:

$$\sum_{k=0}^{50}\binom{100}{k}k=100\sum_{k=1}^{50}\binom{99}{k-1}=100\sum_{k=0}^{49}\binom{99}{k}=50\sum_{k=0}^{99}\binom{99}{k}=50\cdot2^{99}$$

So finally we find: $$\mathbb{E}\min\left(Z,100-Z\right)=2^{-100}\left(2\cdot50\cdot2^{99}-\binom{100}{50}50\right)=50-2^{-100}\binom{100}{50}50$$

Check me on mistakes though.