Exploding Dice - Counting a 6 as a 5

I've been fiddling around trying to figure out how to do exploding dice under the upcoming Torg: Eternity rules. As announced, the exploding mechanic for the "eternity dice" works like this - on a roll of 6, the die explodes into a reroll like any other exploding die mechanic. Subsequent 6's also result in rerolls. However, a roll of "6" (which causes the explosion) actually only counts as a 5. So a roll of "6"[5] + "6"[5] + 4 = 14, not 16. Is there any easy way to get this done in a macro?

Sounds like Penetrating Dice . /r 10d6!p

That's what I thought, too, but from what GM Matt is describing, penetrating dice give an off-by-one result each time they explode more than once. Matt's theoretical roll: 6, 6, 4 Desired result: 5 + 5 +4 = 14 Penetrating result: 6+ 4 + 3 = 13 EDIT: Nothing to see here. 

The penetrating result would actually be 6, 5 , 3, which also sums to 14 A die can penetrate multiple times but the modifier is only ever -1 to each additional die.

Wow. Basic math fail time for me. Thanks for catching that and stopping me before I "helped" again! ;-) 

Syl/Lady- Thanks to both of you for jumping in. I did have a look at penetrating dice, but couldn't tell whether the -1 that was being added to additional rolls was cumulative or not  - just not familiar with the mechanic. What I'm hearing is that, while the game mechanic/formula is not identical to what I'm looking for, the math will work out the same in the end. Correct? (That's how it appears to me, but again I'm not completely confident in my understanding of penetrating dice).

Every time a die penetrates, a -1 modifier is added to the die produced by the explosion (instead of the die that penetrated). Check out the picture I posted to see this in action. GM Matt said: What I'm hearing is that, while the game mechanic/formula is not identical to what I'm looking for, the math will work out the same in the end. Correct? Correct.