considering a aDb+c dice throw
now, we want to calculate 2 arrays, the first is the array with the list of all possible throw in order: lets call this a1 it start at a+c and finishes at ab+c so
a+c, a+c+1, ..., ab+c-1, ab+c
how long is this array ? ab-a+1
now let's consider a second array (a2), as long as the first, and fill it with the corresponding values of the probability of the throw. We discovered that we can calculate the values of the probability from the diagonals of the Pascal's triangle. the values are exactly the values of the diagonal number a, up to the value b The additional values, up to the half of the array lenght or half of the array leght +1 (if the array is odd) are calculated as : number of the pascal's triangle-(number of the 1st pascal triangle at the same diagonal*a), number of the pascal's triagle-(number of 2nd pascal triangle at the same diagonal*a)
moreover the sum of all the numbers of a2 = b^a
Example, step by step: lets take 3d5 (aDb) lenght of the array is: ab-a+1=13 a1=3,4,5,6,7,8,9,10,11,12,13,14,15
the a (so the third) pascal's triangle diagonal is: 1,3,6,10,15,21,28,36,…. so let's put the numbers of the diagonal up to b in the a2 a2=1,3,6,10,15,x,y,x,15,10,6,3,1 x will be 21-(a*1)=18 where 1 is the first number of the same diagonal y will be 28-(a*3)=19 where 3 is the second number of the same diagonal as a check of our calculations: 1+3+6+10+15+18+19+18+15+10+6+3+1=125 and 125 = 5^3 (b^a)
Resolved dices thows, as full examples:
2d6+1 3,4,5,6,7,8,9,10,11,12,13 1,2,3,4,5,6,5,4,3,2,1 3d6+1 4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19 1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1 4d3 4,5,6,7,8,9,10,11,12 1,4,10,16,19,16,10,4,1 3d3 3,4,5,6,7,8,9 1,3,6,7,6,3,1 3d4 3,4,5,6,7,8,9,10,11,12 1,3,6,10,12,12,10,6,3,1 5d4 5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 1,5,15,35,65,101,135,155,155,135,101,65,35,15,5,1
Resources: an interactive pascal's triangle http://mathforum.org/dr.cgi/pascal.cgi?rows=20
to get the diagonal
all negative terms are 0
the 0 term is always 1
the nth term is the ∏ from i=1 to n of (a + i - 1)/i
number of the pascal's triangle-(number of the 1st pascal triangle at the same diagonal*a), number of the pascal's triangle-(number of 2nd pascal triangle at the same diagonal*a)
Just to clarify this you can have more than two extra numbers they just follow the same pattern.
lets do some pseudo code to fill out a2 Very Happy
lets define N(x,y) as the y term on the x diagonal lets define the current terms index as t lets define the array length as l
for every t up to and including l/2 - 1 + l%2 { the t term is N(a,t)-(a*N(a,t-b)) the l - t - 1 term is the same as the t term }
how does that look (might be an off by one error but that is the basic look)
There's an excellent online dice roll probability calculator at http://topps.diku.dk/torbenm/troll.msp It has tons of features and will graph the results for you.
Simple statistics unfortunately don't tell the whole story. Because Transcendence uses integer math lower damage weapons can lose a great deal to roundoff error while higher damage weapons lose proportionally less.
Take for example the basic 1d4 laser cannon and 3d4 turbolaser cannon.
Let's assume basic armor. Armor that has no defined resistances still has resistances and vulnerabilities derived from its level. (see shield and level damage adjustment curves1.01)
The laser resistance by level is in the hpBonus terminology used in most places in the game data
Level | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|
Armor Laser Resistance | +0 | +25 | +67 | +150 | +300 | +614 | +1150 | +2400 | +4900 | *** |
The formula in question is 100 / (100 + hpBonus) = percentile_resistance = 1 - damage_adjustment. Remembering that percent is literally per hundred and to get a real ratio you simply move the decimal two places more significant.
First let's look at the laser colimnator. It adds +75% damage.
The laser is easy. Just 1d4.
possible rolls | 1 | 2 | 3 | 4 | average |
---|---|---|---|---|---|
laser | 1 | 2 | 3 | 4 | 2.5 |
colimnated laser | 1 | 3 | 5 | 7 | 4 |
As you can see it only actually gains +60%.
The turbolaser is more troublesome with multiple dice.
possible rolls | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | average |
---|---|---|---|---|---|---|---|---|---|---|---|
probability | 1/64 | 3/64 | 3/32 | 5/32 | 3/16 | 3/16 | 5/32 | 3/32 | 3/64 | 1/64 | - |
turbolaser | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 7.5 |
colimnated turbolaser | 5 | 7 | 8 | 10 | 12 | 14 | 15 | 17 | 19 | 21 | 12.75 |
Now we're getting +70%.
Of course doing that in a table is cumbersome. Unfortunately you cannot algebraicly simplify a series that contains the floor function. Machine assistance is suggested.
Because I don't have anything like that set up right now I'm only going to do one example. Well, two but since level 1 armor takes full rated damage from lasers that one's trivial.
possible rolls | 1 | 2 | 3 | 4 | average |
---|---|---|---|---|---|
laser raw damage | 1 | 2 | 3 | 4 | 2.5 |
against light titanium | 1 | 2 | 3 | 4 | 2.5 |
against light plasteel | 0 | 1 | 1 | 2 | 1 |
Rather than doing 60% as much damage against level 3 armor as against level 1 the laser is doing 40%, a 50% drop in effectiveness due to roundoff.
possible rolls | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | average |
---|---|---|---|---|---|---|---|---|---|---|---|
probability | 1/64 | 3/64 | 3/32 | 5/32 | 3/16 | 3/16 | 5/32 | 3/32 | 3/64 | 1/64 | - |
turbolaser raw damage | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 7.5 |
against light titanium | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 7.5 |
against light plasteel | 1 | 2 | 3 | 3 | 4 | 4 | 5 | 6 | 6 | 7 | 4.09375 |
Roundoff still hurts dropping the damage from 60% to just under 55%, but that's about a 9% reduction in effectiveness, much better than the 50% drop the laser suffers from theoretical against the same armor.
Further examples will wait until I've made a script to do the math.