In Which I Consider The Number Of Successes
In most versions of the Year Zero System, if players roll more than one 6, they can spend them for added benefits. (Mutant Year Zero and the Alien RPG calls them “stunts”; Forbidden Lands has no special name and only certain skills benefit.)
So how likely are you to roll more than one 6? Not very. The average number of successes is N/6, where N is the number of dice.
Below is a table created by a Lua program at the end of this post.
| Dice | Success | No Stunts | 1 | 2 | 3 | 4 | 5 | 6 | 7 | > 7 |
|---|---|---|---|---|---|---|---|---|---|---|
| 1d | 16.67% | 16.67% | ||||||||
| 2d | 30.56% | 27.78% | 2.78% | |||||||
| 3d | 42.13% | 34.72% | 6.94% | 0.46% | ||||||
| 4d | 51.77% | 38.58% | 11.57% | 1.54% | 0.08% | |||||
| 5d | 59.81% | 40.19% | 16.08% | 3.22% | 0.32% | 0.01% | ||||
| 6d | 66.51% | 40.19% | 20.09% | 5.36% | 0.80% | 0.06% | <0.01% | |||
| 7d | 72.09% | 39.07% | 23.44% | 7.81% | 1.56% | 0.19% | 0.01% | <0.01% | ||
| 8d | 76.74% | 37.21% | 26.05% | 10.42% | 2.60% | 0.42% | 0.04% | <0.01% | <0.01% | <0.01% |
| 9d | 80.62% | 34.89% | 27.91% | 13.02% | 3.91% | 0.78% | 0.10% | <0.01% | <0.01% | <0.01% |
| 10d | 83.85% | 32.30% | 29.07% | 15.50% | 5.43% | 1.30% | 0.22% | 0.02% | <0.01% | <0.01% |
| 11d | 86.54% | 29.61% | 29.61% | 17.77% | 7.11% | 1.99% | 0.40% | 0.06% | <0.01% | <0.01% |
| 12d | 88.78% | 26.92% | 29.61% | 19.74% | 8.88% | 2.84% | 0.66% | 0.11% | 0.01% | 0.02% |
| 13d | 90.65% | 24.30% | 29.16% | 21.38% | 10.69% | 3.85% | 1.03% | 0.21% | 0.03% | 0.03% |
| 14d | 92.21% | 21.81% | 28.35% | 22.68% | 12.47% | 4.99% | 1.50% | 0.34% | 0.06% | 0.07% |
| 15d | 93.51% | 19.47% | 27.26% | 23.63% | 14.18% | 6.24% | 2.08% | 0.53% | 0.11% | 0.13% |
| 16d | 94.59% | 17.31% | 25.96% | 24.23% | 15.75% | 7.56% | 2.77% | 0.79% | 0.18% | 0.21% |
| 17d | 95.49% | 15.32% | 24.52% | 24.52% | 17.16% | 8.93% | 3.57% | 1.12% | 0.28% | 0.35% |
| 18d | 96.24% | 13.52% | 22.99% | 24.52% | 18.39% | 10.30% | 4.46% | 1.53% | 0.42% | 0.53% |
| 19d | 96.87% | 11.89% | 21.41% | 24.26% | 19.41% | 11.65% | 5.44% | 2.02% | 0.61% | 0.79% |
| 20d | 97.39% | 10.43% | 19.82% | 23.79% | 20.22% | 12.94% | 6.47% | 2.59% | 0.84% | 1.13% |
In Which I Consider Pushed Rolls
The table above doesn’t include the chances of success for a Pushed Roll in Mutant Year Zero or Forbidden Lands, because that would make the table (and the math) much more complicated. (The mix of Base Dice, Skill Dice, and Gear Dice is also a factor.)
To calculate pushed rolls in Coriolis, Tales from the Loop, Things from the Flood, and Vaesen, where a roll of 1 on a die has no effect, use the row that’s double the size of the pool. e.g. a pushed pool of 6d has a total1 probability of 12d.
In Alien, if Stress is less than 3, take the number of Base Dice and Stress Dice, double it, and add 1 for the new point of Stress. Once Stress reaches 4 or more, the chance of a Panic Action superseding the player’s intended action starts affecting the probabilities.
BTW, I’ve worked out a success table for the Alien RPG as a function of the number of Base Dice and Stress Dice, because of course I would. However, I’m not entirely sure it’s correct. (I actually tried calculating it two different ways, and the numbers of each are close, but not exact, which makes me suspicious.) When I’m more confident about the calculations I’ll print them for the amusement of whatever masochist reads this blog, if any. And my own, of course.
In Which I Show Yet More Code
#!/usr/bin/env lua
-- Chances of success on a single die
local SUCCESS = 1/6
local function factorial(n)
if n <= 1 then
return 1
end
return n * factorial(n-1)
end
local function choose(n, k)
if n < 1 or k < 0 or k > n then
return 0
end
return factorial(n) / (factorial(k) * factorial(n-k))
end
local function binomial(n, k, p)
return choose (n, k) * (p)^k * (1-p)^(n-k)
end
-- chance of rolling at least exactly `k` successes on `n` dice
local function success_equal(n, k)
return binomial(n, k, SUCCESS)
end
local function success_greater(n, k)
local result = 0
for i = k+1, n do
result = result + success_equal(n, i)
end
return result
end
local function success(n)
return (1 - (1 - SUCCESS)^n)
end
local function format_percent(p)
if p <= 0 then
return string.rep(' ', 8)
elseif p < 0.0001 then
return ' <0.01% '
end
return string.format(" %5.2f%% ", p * 100)
end
local function print_table(maxdice, maxstunts)
local md = (maxdice or 10)
local ms = (maxstunts or ((md-1) // 2))
local rowbuf
-- print header
local gtcol = ms < md
local cols
if gtcol then
cols = ms + 3
else
cols = ms + 2
end
rowbuf = { "Dice ", " Success ", " No Stunts ", " 1 " }
for n = 2, ms do
table.insert(rowbuf, string.format(" %-6d ", n))
end
if gtcol then
table.insert(rowbuf, string.format(" > %-4d ", ms))
end
table.insert(rowbuf, "")
print(table.concat(rowbuf, '|'))
rowbuf = { ":---:" }
for n = 1, cols do
table.insert(rowbuf, "-------:")
end
table.insert(rowbuf, "")
print(table.concat(rowbuf, '|'))
-- print each row
for n = 1, md do
rowbuf = { string.format(" %2dd ", n) }
-- success totals
table.insert(rowbuf, format_percent(success(n)))
-- 0 - `ms` stunts
for k = 0, ms do
table.insert(rowbuf, format_percent(success_equal(n, k+1)))
end
-- > `ms` stunts
if gtcol then
table.insert(rowbuf, format_percent(success_greater(n, ms)))
end
table.insert(rowbuf, "")
print(table.concat(rowbuf, '|'))
end
end
print_table(20, 7)
-
I.e. including the probabilities of the original roll and the reroll. If the original roll failed, the reroll has the same probability as the original; if the original roll succeeded, the column for N indicates the probability of scoring N + 1 additional stunts. ↩︎