Savage Math

I'm about to partake in a Savage Worlds game, and I was reminded that I came up with the calculation to be able to work out the chance of succeeding a roll. And since I didn't post that before (that I can find), I'm going to try to explain it now. [So, yes, there will be formulae.]

First some terminology. When you roll, you are going for Target Number. You roll a Skill Die (d4, d6, d8, etc...) which might have a Modifier. Because you are a PC, you also have a Wild Die (usually a d6), which might also have a Modifier. In Savage Worlds, if you roll the highest number on the dice, you roll again and add to that number (it explodes!). You roll both the skill and wild dice and as long as you make the target number on either dice (you take the highest (exploded) roll), you succeed. [Getting higher is better, but here I am only considering making the target number.]

Letter time. TN = Target Number. SDT = Skill Die Type. SDM = Skill Die Modifier. WDT = Wild Die Type. WDM = Wild Die Modifier. As an example, let TN=8, SDT=4, SDM=-2, WDT=6, WDM=-2. [If you are unskilled, your skill die is d4-2 and your wild die is d6-2.]

First, let's consider just the Skill Die and work out the probability of success for that. What you are in fact trying to roll is TN-SDM (8--2 = 10). For this you will need to roll the die ceiling((TN-SDM)/SDT) times. Ceiling((8--2)/4) = ceiling(10/4) = 3 [Ceiling means 'round to the next highest integer'.] Call this SDR = skill die rolls. You need the die to explode SDR-1 times and then need to roll whatever the difference is between the target number and the lowest multiple of the skill die. In this example, we need to explode twice, then roll a 2 or better. Exploding is a 1 / 4 chance, and then we need to roll ((TN-SDM)-4*(SDR-1))/4. [I prefer, and will do so, to write that as (4*SDR-(TN-SDM-1))/4] What we get from putting this together, is that the chance (SDChance) is (SDT*SDR-(TN-SDM-1)) / (SDT^SDR). [Yep, that's SDT to the power of SDR.]

Now the Wild Dice Chance is the same, so first WDR = ceiling((TN-WDM)/WDT) and then WDChance = (WDT*WDR-(TN-WDM-1)) / (WDT^WDR).

However, we need to combine these chances. And need to consider that it is possible for both dice to succeed. This can either be written as SDChance + WDChance - SDChance*WDChance, or as SDChance + (1 - SDChance) * WDChance.

And there you are. Do the math, and an unskilled person trying to get a target of 8 has... just over 12% chance.

Or you use this TN Calculator I created...

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