Climbing out of Burnout with Statistical Analysis

I blame Kevin Chenevert.
 
He sent me a draft of a game he’s working on (which you should definitely pick up when he releases it. Just sayin’). He borrows something in there that comes from The Black Hack – the idea of dice that exhaust. This was a new concept to me, and it took me a while to wrap my head around it… I know for some of you this will be like ‘duh, I’ve been using that for years’, but stay with me as I get caught up. Hey, I have admitted I don’t read a lot of stuff other people have written. I just stay in my own tiny corner of the gaming world and tinker over here most of the time.
 
Anyway, the system gives you a die, and when you roll a 1-2, you move down to the next die. When you roll a 1-2 on the D4, you are done for the day/week/month/whatever. You’ve exhausted that spell or ability or item. A wand might have Ud8 charges. This means it is guaranteed to have at least 4 (the first one, with a 1-2 on the D8 after the second charge is used, a 1-2 on the D6 after the third charge is used, and a 1-2 on the D4 after the fourth charge is used). I’m not a statistician or a math teacher, so I should probably ask someone (or maybe one of you will do some teaching), but basically, I assume the math works like this:
 
On a D8, I have a 25% chance of rolling 1-2 on each attempt; on a D6, I have a 33.3% chance of rolling 1-2 on each attempt; on D4, I have a 50% chance of rolling 1-2 on each attempt. Multiple rolls means that I keep multiplying the existing chance by the next chance to get the new cumulative chance. So, if I have a 75% chance of surviving the first D8 check (I stay on D8 with a roll of 3-8), I have a 56.3% chance of surviving two D8 checks (75% chance of a 75% chance), a 42.2% chance of surviving three D8 checks (75% chance of 56.3%), a 31.6% chance of surviving four D8 checks; a 23.7% chance of surviving five D8 checks; a 17.7% chance of surviving six D8 checks; a 13.3% chance of surviving seven D8 checks; a 10% chance of surviving eight checks… and I don’t get to below a 1% chance until I get somewhere around check seventeen! There is a realistic chance that I’m pushing this out three or four places before I move down to the next die.
 
On a D6, I have a 33.3% chance of rolling 1-2 on each attempt. I have a 66.6% chance of surviving the first check, a 44.3% chance of surviving the second, a 29.5% chance of surviving the third, a 19.6% chance of surviving the third, a 13.1% chance of surviving the fourth, an 8.7% chance of surviving the fifth, a 5.8% chance of surviving the sixth, a 3.8% chance of surviving the seventh, a 2.5% chance of surviving the eighth… you get the idea.
 
On a D4, I have a 50% chance of surviving each roll; this one is easiest in terms of math. Just keep dividing by 2… so the first roll is 50% chance of survival, the second is 25%, the third is 12.5%, the fourth is 6.3%, the fifth is 3.2%, the fourth is 1.6%, the fifth is less than 1%.
 
And now my statistical skills are at an end. I have no idea how to begin mixing and matching these results, but I will presume that the most likely outcome follows the 50% line through all dice… so you’d get through the D8 two times (failing on the third and moving down), the D6 two times (failing on the third and moving down), and the D4 once (failing on the second and being done); this gives you a rough average of 8 results before you’d peter out, starting with a D8. The minimum you could possibly get is 4 if you are rolling bad that day, but with a bit of luck you could end up over ten, and even into the low teens. There’s a pretty big swing, so I’d expect that we’d see a pretty wide standard deviation on this, but again… statistics. I don’t like what I perceive as a very wide deviation in results.
 
I’ve worked out an alternative. 

Because I’m me.
 
I figured what if you had a fixed die, but you tallied your rolls on that die; when you roll the total tallies or less, you are done. This means that you always get one success with any die (no tallies yet), but a 1 always fails on the second check, a 2 always fails on the third check, a 3 always fails on the fourth check… etc. The bigger the die, the more chances you have, and the less likely you are to roll low comparatively. Here’s a quick overview….
 
The tally system % chances of success

	1	2	3	4	5	6	7	8	9	10
D4	100%	75	37.5	9.4	--	--	--	--	--	--
D6	100%	83.3	55.4	27.7	4.6	<1
D8	100%	87.5	65.6	41	20.5	7.7	1.9	<1
D10	100%	90	72	50.4	30.2	15.1	6	1.8	<1	<1

 
The average number of results is 2 on D4, 3 on a D6 or D8, and 4 on a D10. The tally system is most effective with smaller dice; there is a significant difference 3 and 4 tallies on the D4, whereas the D10 has very little realistic chance of realizing tallies beyond 7. The D6 appears to give the most robust results, with a bit more variety across the results. However, we could offset this by adding a fixed number of successes before the tally system kicks in.
 
If you have access to 3+Td6 Tier 1 spells on a given day, you will be guaranteed to be able to cast 4, have an 83.3% chance of casting 5, have a 55.4% chance of casting 6, have a 27.7% chance of casting 7, have a 4.6% chance of casting 8, and have a miniscule chance of casting 9. This feels like a tighter bell curve.
 
Is the tally system worth the extra bookkeeping? I’m not sure. As a caster, it gives me one more thing to keep track of (which could be annoying), but it’s also more mysterious; it makes magic a bit of an unknown force in the game. It makes it so that not even the magician is fully aware of his or her power, which I really like from a player’s experience perspective.