Consider the following game: You throw a die 1000 times, and win 1 dollar for every you get a 6. You don't lose money for the other rolls. Apply both Chebyshev's and Chernoff's to bound the probability that we are $X$ or further above/below the expected value for $X = 50, 100, 150$. Simplify each answer using a calculator.
I tried solving for $X = 50$ first by using the Law of Large Numbers but I don't think I'm correct.
So $A_n$ is the average number of 6s after 50 throws.
$E[A_n] = 1000 * (1/6) = 166.667$
Since $Var(A_n) = Var(X)/n$ and $X$ has Binomial distribution (or is it Bernoulli?) we can use $Var(X) = p(1-p)n$.
$Var(A_n) = [p(1-p)n]/n = 5/36$.
The observed value of $A_n$ is $1/20$ therefore it is $1/6 - 1/20 = 7/60$ away from the expected value.
$$P(|A_n - 1/6| \geq 1/20) = Var(A_n)/(7/60)^2 = (5/36)/(7/60)^2 = 10.2$$
This answer is way off what I expected and I'm also not sure how to differentiate between above and below the observed rolls.