<s>[INST] Throwing dice: Probability Total is Divisible by 3 
Suppose I throw a fair die $1995$ times. What is the probability that
  the total is divisible by 3?

I tried to attack this problem inductively, storing the total in a variable $t \mod 6$, and then showing that for any total $t$ there is a $1/3$ probability of the next throw of the dice creating a total $t+n$ divisible by three. However, I've had difficulty formalizing the proof and I'm wondering if there isn't a better way to do this.
 [/INST] Let $X_n$ be the $n$th number rolled by a fair die (modulo $3$) and let $S_n = X_1 + \ldots + X_n \pmod 3$ be the sum of the first $n$ numbers modulo $3$. We will prove by induction on $n \in \mathbb N$ that  for any $r \in \{0,1,2\}$, we have that $P(S_n = r) = 1/3$. The base case is clear; we will focus on the induction step.
Induction Hypothesis: Assume that the claim holds true for $n = k$.
It remains to prove that the claim holds true for $n=k+1$. Indeed, for any $r \in \{0,1,2\}$, observe that:
\begin{align*}
&P(S_{k+1} = r) \\
&= P(S_k = 0)P(X_{k+1} = r) + P(S_k = 1)P(X_{k+1} = r-1) + P(S_k = 2)P(X_{k+1} = r-2) \\
&= P(S_k = 0)\cdot\frac{1}{3} + P(S_k = 1)\cdot\frac{1}{3} + P(S_k = 2)\cdot\frac{1}{3} \\
&= \frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{3} \qquad\text{by the induction hypothesis}\\
&= 3\left(\frac{1}{3^2}\right)\\
&= \frac{1}{3}\\
\end{align*}
as desired. Note that $r-1$ and $r-2$ are taken modulo $3$.
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