Random Integer Divisible by 3

Let $L$ be the number of digits needed to get a multiple of $3$ . Then $\mathbb{P}[L \ge n] \; = \; \big(\tfrac23\big)^{n-1} \hspace{2cm} n \ge 1$ Proof : For $L\ge n$ we must have not formed a multiple of $3$ by the time we have made the first $n-1$ digit choices. This means that the digit sum of the first $m$ digits cannot be a multiple of $3$ for any $1 \le m \le n-1$ . At any stage, no matter what the digit sum of the first $m-1$ digits is, there are $3$ out of the $9$ digits $1,2,3,4,5,6,7,8,9$ that could be chosen which would make the sum of the first $m$ digits a multiple of $3$ . Thus the probability that the sum of the first $m$ digits is not a multiple of $3$ is $\tfrac23$ , and this probability is independent of the first $m-1$ digit choices. Thus it is clear that the desired probability is $\big(\tfrac23\big)^{n-1}$ .

Thus the expected value of $L$ is $\mathbb{E}[L] \; = \; \sum_{n \ge 1} \mathbb{P}[L \ge n] \; = \; \sum_{n \ge 1} \big(\tfrac23\big)^{n-1} \; = \; \boxed{3}$