# Inequality only with mean

The is a set of 100 bins $$\rightarrow \{b_1,...b_{100}\}$$

Every bin as an owner $$\rightarrow \{p_1,...p_{100}\}$$

Every owner selects $$x$$ bins (but not their own) and throws a ball into them (e.g $$p_1$$ selects $$x$$ bins from the set $$\{b_2,...,b_{100}\}$$, $$p_2$$ selects from the set $$\{b_1,b_3,...b_{100}\}$$ and so on).

I now have that the probability of every bin to have more than $$x$$ balls is equal to $$\frac{1}{z}$$ and so the expected number of bins with more than $$x$$ balls is equal to $$100 \times \frac{1}{z}$$

Is there any inequality (similar to Markov's or Chebyshev's inequality), using this expected value and the way that the balls are distributed, that bounds the probability of the number of bins with $$x$$ balls being greater or equal than a specific number?