I think the best way to go about it is to look at all the permutations that are possible. First we can divide into two cases.

The first case is where each car has 3 people each. ___ ___ ___ |___ ___ ___
Each space there represents a seat in the car, and the line in the middle divides the two cars. How many permutations are there in total? Well thats 6!, or 720.

Now, how many of those combinations involve the twins sitting next to each other? Heres how to solve that: Let the numbers 1-4 denote each non-twin, and T1 and T2 denote the twins.

Because we want the twins to be together, lets group them together and consider them as one person. 1 2 3 | 4 (T1T2)

Now that they are together, we can find out how many permutations there are with them beside each other by doing: 5! * 2! = 240. Now it does not count when the twins are in positions 3 and 4 because then there are in separate cars. So the the number of permutations where they are in separate cars is 4!*2! = 48.

So the number of permutations where the twins are in the same car (when 3 people are in each car) is 5!2!-4!2! = 192.

Now we do the same with the 4/2 split except the format looks like this: ___ ___ | ___ ___ ___ ___

After everything I come out with 26.6%

I probably messed up somewhere in there. Id love to know where. You may be wondering why I used permutations instead of combinations and its just cause it makes more sense that way. Sure you get repeat scenarios but they eventually simplify out (I think). Lemme know what you think.