Sunday, December 11, 2011

Calculating the chance to kill a vehicle: The Correct Way

Ok, so I've seen a couple of instances of bad math out there lately, and I thought I would take a moment to put out some information which apparently hasn't made its way out into the general knowledge pool yet.

Most people have a pretty easy time figuring out the "number of wounds" of a given unit firing against another unit. What they are actually calculating in most cases is the mean number of wounds, or the average. It's important to remember that this calculated mean will not necessarily be the number of dead Spesh Mareenz after you have unleashed your volley of DOOM on them.

Anyways, more complicated than calculating the "number of wounds" or "average number of destroyed vehicles results" is calculating the "chance to destroy" against a vehicle when firing a multiple weapon barrage. This is actually a really important thing to understand how to do, because the "average destroyed results" and the "chance to destroy the vehicle" are NOT the same, even though sometimes the "average destroyed results" might appear to be a "percentage-like number" and make you think it's actually a percentage chance.

Here's an example everyone should know and understand:

A Psyfleman walks into a bar, and shoots at a Rhino (standing at the other end of the bar, out of cover).

How many hits, pens, and destroyed results should that Psyfleman average?

Well, we know that a twin-linked BS 4 weapon hits 8/9 of the time (2/3)+((2/3)*(1/3)) = .889

Teaching moment aside:

Eight-ninths. That's one of those fractions that's handy to have memorized. Anything you ever need a "3" to accomplish, that gets a re-roll, has an eight-ninths chance of success. This comes up a lot.

Memorized? Good, moving along.

End teaching moment.

Ok, so with an 8/9 chance to hit, and 4 shots, we will average 3.56 hits, and since I only care about first salvo destroyed results (Nurglitch, don't get mad, I know that there is a chance to have a glance-to-death hallelujah moment occur, but I'm trying to keep this simple) I am just going to look at pens. So with 3.56 hits, and half of those penetrating on average, we should average 1.77 penetrating results. Now, we know that, the vehicle damage chart being what it is, only one THIRD of those pens will result in a destroyed vehicle (if ONLY Psyflemen were AP 1... lol *eyeroll*). So take that 1.77 pens, multiply by a third, and we get .59 destroyed results.

So... that means that if I shoot at a Rhino with my Psyfleman, I have a 59% chance to wreck it, right?


If that were true, and we fired TWO Psyflemen at a Rhino, we would have (with eight shots) 1.18 average destroyed results. So... 118% chance to kill it? That doesn't make sense, now does it?

I know that the education system isn't what it used to be, but hopefully we all recognize that there can never be greater than 100% chance of something happening.

Hopefully we also have all had the pleasure of watching a Rhino shrug off the effects of 15 missiles, rolling Shaken, Stunned, Stunned, Stunned, Shaken, Weapon Destroyed, Shaken... etc. until you are staring at this Rhino, sitting insolently on an objective, with every anti-tank weapon in your army glowing red-hot and all your ammunition exhausted... and you just want to fucking cry.

No? Never had that happen?

Umm... me neither. Moving along.

Ok, so we know that the .59 we got from the first equation is not actually the chance to destroy the Rhino. What's the real chance?

Well, it gets a bit trickier. I have to use Exponents to get the answer. It's not that bad, but if you don't remember what "raising to the third power" means, it basically means multiplying your original number by itself twice. So if your number was 2, and you raised it to the third power, you get:

2 x 2 x 2 = 8

(Notice there are three twos, thus the "third power").

Ok, so this time I'm going to start by focusing on just ONE of the four shots that are going downrange. I'm going to do everything I did last time, but with just ONE shot.

So eight-ninths to hit, one half to pen, one third to kill.

(8/9)*(1/2)*(1/3) = .148 average kills for one shot (and please note, if you are only firing one weapon, this is BOTH the "average destroyed results" AND the "chance to destroy." It's only when you fire more than one shot that these numbers diverge).

Ok, so we now need to take one minus one minus the number of average kills raised to a power equal to the number of shots fired, which will give us our percent chance destroyed. Everyone got that? ;D

So 1 - ((1 - .148)^4) = chance to destroy.

What we're really doing here is determining the chance for the Rhino to NOT die, and then from that we know what the chances that it WILL die are.

Anyways, on to the answer. One minus .148 is .852. This is the chance that one shot will NOT kill a rhino. Now we raise this to the fourth power: .852^4 is the same as .852 x .852 x .852 x .852 which equals .53, which is the chance that the Rhino will NOT die if you fire four shots at it.

Now, we know that if the rhino will NOT die 53% of the time, then that means... it WILL die 47% of the time. So 1 - .53 = .47 .

If you run this whole process as one calculation, keeping all of the long strings of digits, you actually get .473429 .

Yeah, .47 was probably close enough. ;D

Ok, so now we have to realize something. Before we said that a Psyfleman could expect to average .59 destroyed results every time it fired. But now we know something better: every time a Psyfleman fires it has a 47% chance of killing a Rhino-like target.

Ok, so part two of this educational bonanza will be talking about Bell Curves and meltaguns.