# "Exploding Dice" vs. "Non-Exploding Dice"

For a roleplaying game I'm designing, I'm using a core resolution mechanic that comes down to counting "successes" on rolled dice. Characters roll a handful of dice (six-sided) and count each die that generated a 4, a 5, or a 6. Dice that generated a 1, a 2, or a 3 are discarded. Further, for dice that generated a 6, you roll one additional die and count that die as if it was part of the original roll - this is called "exploding."

Example: A character rolls 6 dice. He rolls the following: 1, 1, 3, 4, 6, 6. So far, three successes (4, 6, 6). The two 6s explode, giving him the following rolls: 2, 5. In total, the character scores four successes (4, 6, 6, 5).

Question: What are the consequences of removing a character's ability to "explode" 6s compared to the consequences of allowing a character's 6s to "explode" twice (where each 6 adds two additional dice to the roll). I'm hoping that the probabilistic outcome of these two processes are similar, but I can't figure out if they actually are.

• What do you mean with similar? $\mathrm{Score}(No\ Exploding)\leq\mathrm{Score}(Exploding)\leq\mathrm{Score}(Exploding\ twice)$, if you understand what I mean... Commented Dec 3, 2015 at 14:33
• @EricS. I suspect that the question is whether $\langle \textrm{Exploding}\rangle - \langle \textrm{No Exploding}\rangle \approx \langle \textrm{Double Exploding}\rangle - \langle \textrm{Exploding}\rangle$
– Sten
Commented Dec 3, 2015 at 14:44
• Ah okay, that would indeed make sense. Hmm, interesting question. Commented Dec 3, 2015 at 14:46

Let's call $x$ the average of successes per dice. First, let's examine the case in which they can not explode 6s. 3 possible successes, 3 possible fails, so $x=1/2$
Now the case where they explode twice. There are 3 possible successes and 3 possible fails, but also 1 case where 2 additional dices are rolled (so and additional $2x$ successes in that case). Therefore, the successes are $x={1\over2} +{2 \over 6} x$
You get then that $x={3 \over 4}$
Edit: In the standard explode system (1 for each 6) you would get $x={1 \over 2}+{1 \over 6} x$, which implies $x={3 \over 5}$