Problem
Let $s,e,k\in \mathbb{N}^*$ and $p\in[0,1]$.
We have the following experiment. We're performing the following sub-experiment $k$ times:
- we're drawing $s$ times a random biased coin with probability $p$ to get heads
- let $n$ be the number of heads
- the output of the sub-experiment is $ne$
The output of the whole experiment is the sum of outputs of the $k$ sub-experiments.
Question: what is the probability that the output is smaller than $\frac{2}{3}kesp$, that is, two thirds of the expected output?
Next you'll find my approach to formalize the problem, which does not seem to be of much help.
Formalization
Let $X$ the random variable for the whole experiment, and $Y_i$ the random variable for the $i$th sub-experiment.
We have $E[X] = \sum_{i=1}^k E[Y_i] = k \cdot e\cdot sp$.
The probability that the output is $n$ is $$P[X=n]=\sum_{n_1+\dots+n_{k}=n}\prod_{i=1}^kP[Y_i=n_i]$$ And we have:
- $P[Y_i=ne] = \binom{s}{n}p^n(1-p)^{s-n}$,
- $P[Y_i=m]=0$ for $m$ not a multiple of $e$.
We need to compute $$\sum_{n=0}^{b}P[X=n]$$ where $b := \lfloor\frac{2}{3}kesp\rfloor$.
But how to compute this numerically (exactly or approximately), or get a reasonably good upper bound? Values for $s$ and $k$ would be in the thousands (say $2^{13}$), and $e$ would be rather smaller, say $2^i$ with $0\leq i \leq 6$, and $p$ also small (say $p=10^{-5}$).