How can I calculate the expected number of stunt points per attack?

Here's a basic AnyDice script:

function: stunt points for DICE:s and DRAGON_DIE:n vs TARGET:n { if DICE + DRAGON_DIE < TARGET { result: 0 } \ miss \ if 1@DICE = 2@DICE | DRAGON_DIE = DICE { result: DRAGON_DIE } \ pair \ result: 0 \ hit but no pair \ } loop TARGET over {3..18} { output [stunt points for 2d6 and d6 vs TARGET] named "hit on [TARGET]" } 

(For anyone wondering about the third line of code, 1@DICE = 2@DICE checks whether the two normal dice are the same, while DRAGON_DIE = DICE checks if the dragon die equals any of the normal dice. Obviously, this particular logic only works if there are exactly two normal dice and one dragon die.)

It gives the same averages as Ryan Thompson's R code, so I'm fairly certain it's correct:

Just out of curiosity, I also went and calculated the expected number of stunt points received if the attack hits, using the method described here:

function: stunt points for DICE:s and DRAGON_DIE:n vs TARGET:n { if DICE + DRAGON_DIE < TARGET { result: d{} } \ miss, will be rejected \ if 1@DICE = 2@DICE | DRAGON_DIE = DICE { result: DRAGON_DIE } \ pair \ result: 0 \ hit but no pair \ } loop TARGET over {3..18} { output [stunt points for 2d6 and d6 vs TARGET] named "hit on [TARGET]" } 

That graph has some interesting features. In particular, note the sudden jump between the target values 15 and 16. What happens there is that 4 + 5 + 6 = 15 is the highest number one can roll on 3d6 without getting a pair of equal dice. If you need 16 or more to hit, you'll always get some stunt points (and, in fact, at least four of them!) on each successful hit.