We're going to play a game. Here are the rules:
- A game position is a pair $(x,\bar y)$ of a natural number $x$ and a bag of positive numbers $\bar y$.
- To make a move, pick $y \in \bar y$. If $x>0$, the next game position is $(y-1, (\bar y\backslash\{y\})\cup\{x\})$; otherwise don't insert $x$ and arrive at the next game position of $(y-1,\bar y\backslash\{y\})$.
- We win the game by getting to game position $(0,\{\})$.
To reduce brackets, let's write e.g. 3,2579 as an abbreviation for $(3,\{2,5,7,9\})$. Here's some example moves we could make in this game:
- 3,57 $\to$ 4,37 by picking 5
- 3,57 $\to$ 6,35 by picking 7
- 2,1223 $\to$ 0,2223 by picking 1
- 2,1223 $\to$ 1,1223 in two different ways by picking either of the two 2's in the bag
- 2,1223 $\to$ 2,1222 by picking 3
- 0,2222222 $\to$ 1,222222 in seven different ways
The game may seem a bit odd at first, but it corresponds exactly to drawing balls from an urn without ever seeing two of the same color ball twice in a row. The numbers in $\bar y$ represent how many balls of each color are in the urn; the number $x$ represents how many balls there are of the color we drew last. So drawing a ball decrements one of the numbers in the sequence (that isn't the color we drew last), and puts all the balls of the color we drew last back in the urn.
With this setup, we can phrase your question this way: how many ways can we win from position 0,222? Using $\#p$ for the number of ways to win from position $p$, we can calculate:
$$
\begin{align*}
\#0,222
&= 3\cdot\#1,22 \\
&= 3\cdot2\cdot\#1,12 \\
&= 3\cdot2\cdot(\#0,12 + \#1,11) \\
&= 3\cdot2\cdot((\#0,2 + \#1,1) + 2\cdot\#0,11) \\
&= 3\cdot2\cdot((\#1, + \#0,1) + 2\cdot2\cdot\#0,1) \\
&= 3\cdot2\cdot((0 + 1) + 2\cdot2\cdot1) \\
&= 30
\end{align*}
$$
Meanwhile, to compute the total number of ways we can draw balls from the urn (both with and without back-to-back duplicates), there are
$$ \frac{6!}{2!\cdot2!\cdot2!} = 90 $$
distinguishable sequences of draws. This gives a $30/90=1/3$ chance of not seeing doubles (or $2/3$ chance of seeing doubles).