About the Lorenn's Duplicates, I'm not sure how you want to do it.
All the 6 attacks have a chance to hit a Duplicate. So they must do 6 rolls unless some Duplicates are destroyed in the process.
But maybe you decided to make both attacks on the same target, that why you did only 3 rolls?
In that case 1 Duplicate is destroyed (I supposed the last 6 attacks are for Lorenn):
1D20+5 = [4]+5 = 9
1D20+5 = [7]+5 = 12
[1D20 = [12] = 12, Target is a
Duplicate (AC14) and it's a
miss]
1D20+5 = [8]+5 = 13
1D20+5 = [9]+5 = 14
[1D20 = [5] = 5, Target is
Lorenn (AC19) and it's a
miss]
1D20+5 = [3]+5 = 8
1D20+5 = [12]+5 = 17
[1D20 = [9] = 9, Target is a
Duplicate (AC14) and it's a
hit]
Or we need 3 more rolls (for the moment, no Duplicates are destroyed):
1D20+5 = [4]+5 = 9 [1D20 = [12] = 12, Target is a
Duplicate (AC14) and it's a
miss]
1D20+5 = [7]+5 = 12 [1D20 = [5] = 5, Target is
Lorenn (AC19) and it's a
miss]
1D20+5 = [8]+5 = 13 [1D20 = [9] = 9, Target is a
Duplicate (AC14) and it's a
miss]
1D20+5 = [9]+5 = 14
1D20+5 = [3]+5 = 8
1D20+5 = [12]+5 = 17
Or maybe I missed something in your post and you have another solution. Anyway, the result might be the same at the end (14 and 17 have good chance to destroy 2 Duplicates), but you never know. With a sleeping Dracolich at 30ft., you take what you can get.
