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 12, you move down to the next die.
When you roll a 12 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 12 on
the D8 after the second charge is used, a 12 on the D6 after the third charge
is used, and a 12 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 12 on each attempt; on a D6, I have a 33.3%
chance of rolling 12 on each attempt; on D4, I have a 50% chance of rolling
12 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 38), 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 12 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
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.
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 
Very interesting. I need to think on this. Glad I got some of your juices flowing again.
ReplyDeleteI have a whole draft for a game going now  because I cannot stop myself. I'll send it to you soon.
ReplyDeleteAwesome, I look forward to it.
ReplyDelete