I have a question considering Chebyshev's inequality.
$$ P[|X-E[X]| > a] \leq \frac{Var[X]}{a^2} $$
Let's say we have 5 possible events, probability is uniformly distributed, so every event has $$ PR[event] = \frac{1}5$$
Now we want to know the upper bound of the probability of a specific event happening at least 10 times, if we have 50 trials. This would lead to $$E[X] = 10$$
If we then use Chebyshev's inequality:
$$PR[X > 10] = PR[X -10 > 0] = PR[|X-10|>0] \leq \frac{Var[X]}{a^2} $$
Which is a problem. But there surely must be a way of getting a upper bound for this case with Chebyshev's inequality right? Or did I miss something?
The Chebyshev's inequality is useful only when $a^2> Var[X]$
Otherwise you get some unuseful bounds such as $P(Something)<1000$!
P.S.
In the common proof of chebyshev you get this:
$a^2 P(|X-EX|\ge a)\le Var[X]$
So there is not any algebraic problem of dividing by zero