TITLE: Markov inequality on tossing a fair die n times
QUESTION [5 upvotes]: I am tossing a fair die $n$ times. What is the the probability that the sum of the numbers is at most $2n$.
I am doing this in the following way:
I found $\mathbb E[X] = 3.5n$
So 
$$\Pr[X\leq 2n] = 1 - P[X > 2n]$$
According to Markov inequality:
$$P[X>2n] \leq E[X]/2n = 3.5n/2n = 3.5/2 =1.75$$
So $P[X \leq 2n] > 1 - 1.75 = -0.75$.
I am not sure where I am going wrong? How a probability can be negative?

REPLY [6 votes]: There is nothing wrong with using Markov's inequality here. It is only that (since the calculated upper bound is greater than one) the inequality does not forbid $P[X\le 2n]$ from taking certain values per se, and is thus ineffective in this situation.

REPLY [2 votes]: Observe that in order for $n$ rolls to sum to no more than $2n$, each die roll can only be $1$ or $2$. Thus
$$P[X\le 2n]=(1/3)^n.$$
No Markov inequality required!
There is nothing wrong with your inequality by the way. Using the Markov inequality, you simply noted that the probability is bounded below by $-0.75$, which is true, technically!
EDIT: It has been observed that my initial attempt at this was flawed. While restricting the die rolls to all $1$'s and $2$'s is sufficient, it is not necessary, so the most we can say is that 
$$P[X\le 2n]\ge (1/3)^n.$$
I am trying to compute this completely and will update again soon.