Don't Order The Same

The denominator of the probability will be the total number of ways for guests to choose their meals. Since there are 3 choices for each guest, this is $3^6.$ The numerator will be the number of ways to choose entrées such that adjacent entrées are different.

Fix the sequence of guests at a point. Now, we're looking for the number of ways to compose a string of $n$ letters using the letters S, B, and L such that adjacent letters are different and the first and last letters are different. For example:

$\text{SBLBSB}$

Construct a recurrence relation. Let $d(n)$ be the number of ways to compose a string of $n$ letters such that adjacent letters are different and the first and last letters are different. Let $s(n)$ be the number of ways to compose a string of $n$ letters such that adjacent letters are different and the first and last letters are the same. Then,

$\begin{aligned} d(n) &= d(n-1)+2s(n-1) \\ s(n) &= d(n-1) \end{aligned}$

This gives $d(n)=d(n-1)+2d(n-2).$ Now observe that $d(1)=0$ and $d(2)=6.$ Using the recurrence relation, we find $d(6)=66.$ Thus, the probability is $\frac{66}{3^6}=\frac{22}{243},$ and so $a+b=\boxed{265}.$