Tuesday, 30 March 2010

Introduction to Mathammer

The problem with statistics is that the real world is different to your perfect statistical model. However, mathammer should still give some steer as to how to actually play the game. While you can never guarantee to kill a unit or blow something up, you can get a very good idea of which actions should (in that perfect world) give you the more favourable outcomes. Never forget the best way to get the result you want is always to roll more dice. The more shots you take, the more hits you get and the more wounds you cause. The more wounds you cause, the more saving throws your opponant has to make and the more chance they will fail. I'll talk in depth with more specific examples later on but for now I'm going to mention one of those occasions where the mathammer failed. Everyone who has ever played the game will have had a similar experience I guarantee but I'm going to discuss the time that a Scout Sergeant with a Power Fist took out Bjorn The Fell Handed.

So, he has two attacks and needs 4+ to hit. Then he needs a 6 to pen. Then Bjorn needs to fail his 5+ inv. save. Then the sergeant needs a 5+ to wreck or destroy and then another 5+ after the venerable re-roll.

P(Kill Bjorn) = Number of attacks x P(Hit) x P(Pen) x P(Fail) x P(Damage) x P(damage re-roll)
= 2 x 1/2 x 1/6 x 2/3 x 1/3 x 1/3
= 1/27

Very unlikely but it still happened. Always with any mathammer you might do, remember two things, statistics is nothing without context and, in the words of Arthur Conan Doyle, "Once you remove the impossible, whatever remains, no matter how improbable, must be the truth" The Scout killing Bjorn was very improbable but it was definitely the truth.


  1. Looks even better with the odds there in black and white! Wonder how it stacks up to Bjorn's Assault Cannon wrecking my Land Raider a turn earlier eh? ;)


    Actually I just worked it out and it looks like that's about a 1/16 chance, still pretty unlikely though.

  2. I made it slightly better than that:

    P(Kill) = Shots x P(Hit) x P(Rend) x P(Destroy)
    = 4 x 5/6 x 1/6 x 1/3
    = 5/27

    I didn't include his misses hit on a re-roll of 6 because of BS6 thing becuase I don't think that would have made much difference.

    Although, as my old stats techer used to say, if it's already happened it has a probablity of 1 lol

  3. Thredomancy! Ok I was having these thoughts the other day. Say you want to calculate the chances of killing a battlewagon with a penetrating hit. Since it's open topped it will die on a 4,5 and six. Safe to say that it's a 1/2 chance. Now do 2 penetrating hits guarantee the kill?
    Should we calculate like this 2 x 1/2 = 1 ? So you are absolutely 100% sure that it works like that? Unfortunately not. The real chance we have to calculate here depends on the dice. So the real number we are looking for is the chance that either one of the dice will roll above sure. That brings us down to 3/4 = 75% chance of killing it with 2 penetrating hits.
    Now if we had 3 dice then the chance would be 7/8 which is 88%. It is better than 75% but not as much significantly so as to commit to it. Further more penetrating hits make even less than a difference. 4 dice would be 15/16 = 93% which is as good as 88% in game terms.
    So in reality we have to calculate the chances of dice rolling and not the probabilities of an event happening. It get extremely complicated when a lot of dice or rolls are involved and I hope someone can find a patter for it (certainly not me :P)



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