Posted on April 06, 2018, 09:25:23 PM by Jubal
Axes and Arithmetic 1: Percentages and Probability
Axes and Arithmetic 1: Percentages and Probability
These Chaos Dwarves may not care where their rocket lands, but you might do.
IntroductionI originally wrote this article under the title "Math-hammer" for the now defunct e-zine A Call To Arms, which I ran in 2010 and 2011, and was based at my high school's gaming club, covering wargaming news, the club's internal news, and a wide range of articles mostly covering Warhammer and related topics. As the e-zine is long since nonfunctional, I figured it was time to give some of the articles I wrote a new lease of life, and as such I'm re-publishing this series under the new title of Axes and Arithmetic. It starts pretty basic, but ultimately covers a good deal of A Level mathematics topics including statistics, mechanics, and decision maths in a way that should hopefully be useful to gamers and game designers alike.
Many wargames, ultimately, are games of chance. It’s all the luck of the dice. Or is it? The factors that determine whether you’ll win or lose are all based, in some way, on probability – if it was pure luck, the game would of course be boring. Instead, it’s a case of army composition, picking your fights, and so on. These things (combat calculations in particular) can all be based on simple mathematical tools that can let you work out where you want to be fighting, and can help with factors such as weapon selection going into combats no end. Of course, chances are you guesstimate such things anyway; but have you got the mental distinction between “big choppy axe” and “two little choppy axes” mathematically nailed down? If not, here’s how you do it.
Probability and FractionsA fraction, as we probably all know, is just a representation of a number divided by another number, usually used when the lower number (the denominator) is higher than the upper one (the numerator), and thus the result is less than one. We’ll be dealing only with fractions in that normal form, since we’re using them to express probability. A probability, put simply, is the likelihood that something will happen. One, or a one hundred percent chance, is a certainty – something that will happen no matter what you do. Zero is also a certainty, but a negative one – whatever you do, that thing will not happen. The numbers in between are probabilities, the higher the more likely.
I’m going to use fractions for this for a simple reason; we’re working in sixths almost entirely. Whilst there are lots of possibilities for polyhedrals, and computer game designers can have much freer choice on their random number generators, it's still basically the case that a large percentage of wargaming rolls are done on a humble D6 - a cubic, six-sided die. As such, the probability of many useful numbers or rolls is something over 6. Remember, the probability is the number of possible “good” results. So a 4+ roll succeeds on a 4, 5, or 6; the probability is 3 over 6. If you have the choice between that or a 5+ roll, (5 or 6, probability 2 over 6), you’ll obviously go for the 4+, since the higher probability has a higher chance of success. But what about when you have multiple dice rolls to do?
Multiple RoundsOk, let’s get some gaming into this - we'll be using old-school Warhammer rules as a vaguely representative gaming system. I’m facing an Orc, and I’ve got the option of attacking with a human swordsman or another human with a flail (we’ll keep them singular for now, doing calculations for whole units of different sizes will be a future article). I want to decide which of my two men will take on the beast. The Orc is Weapon Skill (WS) 3, and Toughness (T) 4, with a 4+ Armour Save. My swordsman is WS4, and Strength (S) 3 with a sword; my flailer has WS3, but S5 with his flail.
And here comes the maths. Rolling to hit, the rulebook chart shows us that the swordsman only needs a 3+. That means a 3, 4, 5, or 6 will succeed and the chance is thus 4/6. The flailer needs a 4+, and so only has a 3/6 chance. From here it looks like the swordsman is the better option. However, we still need to factor in the second "To Wound" roll and the armour save. We do this by multiplying across the tops AND bottoms of the fractions. The flailer needs a 3+ to wound, so that’s 4/6. We multiply the 4/6 to wound by his previous 3/6 to hit making a 12/36 chance to wound across both rolls. The swordsman now needs a 5+, so 2/6 by 4/6 means he only has an 8/36 chance to wound across both rolls. Because we multiplied both in the same way, as you can see, the numbers are still comparable. Finally, the armour save. Since we want the Orc to fail his save we count as "ours" the scores he fails on. The S5 flail is a high strength weapon that leaves him just a 6+ chance to save, so that’s 5/6 chance of him failing, final total 60/216. The sword gives no armour bonuses so the Orc has only a 3/6 chance of failure, final total 24/216. Whilst it’s clear already who the better candidate is, remember that you can divide the top and bottom of fractions by the same number to “cancel” them and make them easier to look at, and that to compare 2 fractions at a glance the denominator (bottom number) should be the same. In this case both can be divided by 12; the flail-man has a 5/18 chance (27.8%) of wounding the orc, his sword-armed counterpart just 2/18 (11.1%).
ConclusionSee, we're just one article in and maths can already help kill orcs with a big spiky flail! You can also use exactly this technique for working out the likelihood of causing a wound from a shooting attack, or in reverse for the chance of surviving (not being wounded by) an attack. If you're a game designer, of course, you can work in reverse - work out how likely you want something to be, and then see if your current requirements and die rolls actually facilitate that.
But what happens with multiple attacks, and when you want to know about rolling lots of dice? You can’t just add the probabilities, since that could end up with a number like 20/18 if the above flailman had 4 attacks (and probabilities can’t be higher than 1). The answer has to be to distribute the probabilities somehow... and that’s what I’ll be going on to next article. Stay tuned!