Thursday, July 15, 2010

Mathhammer: 8th Ed. Casting

So with the new edition of WHFB coming out, we have new rules. All wizards are allowed to use as many dice as they like (suck it druchii) so you can have a level 1 Warlock Engineer throwing six dice at the Dreaded Thirteenth spell in order to get that 25+. And it might seem like a very good idea, I mean, who wants to miscast with your (possibly) very expensive level 4 wizard? The downfall of a miscast now is much more devastating that it was in 7th edition, so what are your odds?

When we look at rolling the dice and we look at rolling double sixes it is pretty easy to understand that when you roll two dice, the chance that you'll get a six on the first die is 1/6 and the chance you'll get a six on the second is also a 1/6. The rules of probability say that in situations like this, you multiply the two results, so the chances of getting an irresistible force/miscast on two dice is 1/36.

When you move on to three dice it becomes a bit more complicated. You could get a six on the first die, a six on the second and something other than a six on the third die. Those chances are 1/6, 1/6 and 5/6 respectively and this would give you a probability of 5/216. But this just tells you the probability of the first two dice being a six and the third being something else. We haven't accounted for getting a six, not six, and then a six, or even getting not six, six and six. Now the chances of each of these three separate situations occurring is 5/216, so because these three situations are the only ways you can get two sixes with three dice, these are the only ones we have to count. So probability becomes 15/216 or roughly 6.94%.

But this still does not cover what we need because we will still get an irresistible force/miscast if we throw three sixes so we have to add in that probability (1/216) for a total of 16/216 which is roughly 7.41%.

It becomes even more complicated with four dice as you have to account for rolling two, three or four sixes. Using what is called a cumulative density function (CDF) on a calculator I came up with the following table of values for the percent chance of an irresistible force/miscast depending on how many dice you throw.

Miscast CDF
Dice Used


  1. I think my brain hurts now... I hate math.....

  2. I tried to make it simple and walk through the math in stead of "ta-dah!" here is the answer.

    Maybe I need more diagrams.

  3. @Student teacher - More diagrams and pretty graphs and pie charts, if in doubt colourful pictures help to distract people. On a side note i really like this, just getting into fantasy from 40k my temptation was to go magic crazy, now i see just how bad of an idea that may be lol

  4. huh...only 26% on 6 dice. That might make me rethink my 6D level 1 wizard bomb.