With the outcome of almost every ingame event determined by the toss of dice, a grasp of probability is an essential tool in Warhammer 40,000. How do you give yourself the best chances if you can’t work out the odds? This post will give you basics of the probabilities associated with common rolls. But for now, I’ll open with a question:
5 space marines rapid fire their bolters at a chaos marine – what is the probability that the chaos marine dies? Scribble down (or just remember) your gut instinct answer now, and check it against the results at the end, it may be quite different…
1d6 – What are the odds?
On a 1d6 roll you can get results from 1 to 6, and each result is as likely as any other i.e. ones are as likely as sixes (though sometimes it doesn’t feel that way!). Here’s a chart showing the odds:
This may look confusing at first, but let’s take a moment to examine the chart. The 1 to 6 across the top is the number on the die, and each row below shows the odds of a particular type of outcome.
The row starting with the = (equals) symbol is the odds of that particular result. So as we read across we see that all of them are the same odds (16.67% or 1 in 6, which makes sense if you think about it).
The row starting with the (greater than) symbol shows the odds of a result that is higher than the number at the top of the column. So, for example, the odds of getting greater than 3 on a d6 is 50%, which makes sense as half the possible results are higher than 3.
The row starting with the <= (less than or equal to) symbol shows the odds of a result that is equal to or lower than the number at the top of the column. So while the odds of getting a 2 is 16.67% (check the = row), the odds of getting a 2 or less is 33.33% (check the = (greater than or equal to) symbol. This shows the odds of getting a particular number or above. So the odds of getting a 3 or more on a d6 is 66.67%. This bottom row is the most interesting for Warhammer 40,000 as this is generally how we think of the game, e.g. 4 or higher to hit at BS3, or 3+ to wound on Strength 5 vs Toughness 4.
At this point you are either thinking ‘duh this is bloody obvious’ or you are utterly confused. Perhaps it’s time for an example.
A space marine shoots his bolter at a chaos marine. He is BS4 so he hits on a 3 or better (66.67% from our bottom row). The bolter is strength 4 and the chaos marine is toughness 4 so it’s 4 or higher to wound (50% from the bottom row). The odds of the chaos marine making his save is 66.67% as his save is a 3+. BUT if we want to calculate the probability of the kill, then we need to look at the odds of the chaos marine FAILING his save. To do that we simply look up the odds of him getting less than (<) 3, which our table shows as 33.33%.
How do we add it all up to get a result? Well for a start you don’t add them, you MULTIPLY them:
(odds of hitting) x (odds of wounding) x (odds of failing the save) = odds of kill
So based on our example above (66.67%) x (50%) x (33.33%) = 11.11%
But what about that opening question? There are 5 marines in rapid fire range of that lone chaos marine. So we get 10 bolter shots. Most players will multiply the odds of the kill (11.11%) for one shot, by the number of shots (10). This is wrong! the fact that it gives an answer over 100% should be your first clue (nothing can have odds greater than 100%).
So how do we work out the real answer? You have to look at the odds of the chaos marine surviving. The odds of him surviving one shot is 100% minus the odds of the kill (11.11%) so the odds of survival for one shot is 88.89%. The odds of him surviving all 10 shots is:
88.89% x 88.89% x 88.89% x 88.89% x 88.89% x 88.89% x 88.89% x 88.89% x 88.89% x 88.89%=30.79%
This means 10 bolter shots has (100%-30.79%) = 69.21% chance of killing that marine. So it’s quite likely, but definitely not certain!
Hopefully I’ve explained this all clearly – but if anything is unclear then say so in the comments section and I’ll clarify.
Next time I’ll cover the nuances of the 2d6 roll.