What is the probability that on $3$ successive button presses, you get $3$ consecutive numbers in any order 
A machine has a button. When the button is pressed, it tosses a virtual die repeatedly. This happens automatically until a multiple of $3$ is obtained. Then the machine displays the number of die rolls performed. Only after the number is displayed, you can press the button again. What is the probability that on $3$ successive button presses, you get $3$ consecutive numbers in any order (like $2,4,3$ or $5,6,7$) ?

I have solved it by summing infinite exclusive events' probability of obtaining $\{1,2,3\}$ or $\{2,3,4\}$. To find $p(r,r+1,r+2)$, I have noted they can occur in $6!$ ways, then used multiplication rule. Basically I have solved it but just want to confirm my answer. Please help.
 A: The probability of getting your first multiple of three on the $k$'th roll on the first time the machine is asked to start followed by getting your first multiple of three on the $k+1$'st roll the second time the machine is asked to start, and then the $k+2$'nd roll for the third time will be $$\left(\frac{2^{k-1}}{3^k}\right)\left(\frac{2^k}{3^{k+1}}\right)\left(\frac{2^{k+1}}{3^{k+2}}\right)$$
For instance with $k=1$, this corresponds to the machine displaying the numbers $1,2,3$ in that order as the number of die rolls it performed to get its first multiples of three in each of the first, second, and third times asking the machine to run.
Multiplying by $6=3!$ gives the probability of the numbers being consecutive but in any order.
Adding over all possible values of $k$ gives our final answer:
$$\sum\limits_{k=1}^\infty 6\cdot \left(\frac{2^{k-1}}{3^k}\right)\left(\frac{2^k}{3^{k+1}}\right)\left(\frac{2^{k+1}}{3^{k+2}}\right)$$
$$=6\sum\limits_{k=1}^\infty\left(\frac{2^{3k}}{3^{3k+3}}\right)=\frac{6}{3^3}\sum\limits_{k=1}^\infty\left(\frac{8}{27}\right)^k$$
$$=\frac{6}{27}\cdot \frac{\frac{8}{27}}{1-\frac{8}{27}}=\frac{16}{171}$$
