Melta-weapons are interesting because as soon as you start throwing more than one dice during the same "gate" you get an entirely different pattern of possibilities. It's not a straight 1-6 anymore, but instead looks like this:

Dice Roll Outcome | Chance out of 36 |

2 | 1 |

3 | 2 |

4 | 3 |

5 | 4 |

6 | 5 |

7 | 6 |

8 | 5 |

9 | 4 |

10 | 3 |

11 | 2 |

12 | 1 |

There are 36 total combinations you can roll on 2d6. So you can see that the left side shows the "results" of the roll, and the right side shows the odds of getting that result out of 36. There is only one combination that will result in a result of a "2": a one and a one. Likewise there is only one combination that will result in a 12: a six and a six. But you can roll a seven the following ways:

1 plus 6

2 plus 5

3 plus 4

4 plus 3 (yes these are different!)

5 plus 2

6 plus 1

So six different ways. So if you had to guess blindly what you would roll on 2d6, your best bet is to guess you will roll 'seven.' It's the most common outcome.

Ok, so let's say that we want to find out how many meltagun shots it takes to "reliably" take down a Land Raider. For our purposes we are once again going to concern ourselves with destroying it only. Many times against a Raider Immobilized is as good as destroyed, but let's Keep It Simple, Stupid.

Anyways, the process starts out the same. We need to know what the odds are that ONE meltagun (in melta range and without cover) will destroy a Raider. First we need to pass through the 'hit gate.'

Step 1: "Hit Gate," same as before, 2/3 (unless we have Vulkan, then it's 8/9, lol).

Step 2: "Pen/Glance Gate." Ahh now things are more interesting. Well, a melta starts out base Strength 8. We need a six to glance and anything seven or higher will penetrate. So looking at our handy chart above we see that we get a penetration roll of "six" 5 times out of 36. So we glance 5/36 of the time. And we see that we will roll a "seven" or higher 21 out of 36 times. So the odds of getting a penetrating roll are 21/36.

Now, in our previous example we ignored glances because we were just looking at destroyed results and you can't destroy a Rhino with a glancing hit (except through cumulative damage, of course). But a meltagun is AP 1. That means it DOES have a chance to destroy you on a glancing hit, so we will have to calculate the glances and add them to the pens. Let's start out with the pens and come back to the glances in a bit.

Here's the damage table for AP 1 Penetrating hits:

Damage Table Roll | Damage Result AP 1 |

1 | Stunned |

2 | Weapon Destroyed |

3 | Immobilized |

4 | Wrecked |

5 | Explodes |

6 | Explodes |

So a full half of the damage results will result in a destroyed vehicle. Nifty. So 1/2 chance to destroy on a pen.

Let's review what we've got so far and then go back to glances:

Chance to hit: (2/3)

Chance to penetrate: (21/36)

Chance to destroy target on a pen: (1/2)

So (2/3) x (21/36) x (1/2) = 19.44%

So a little less than a one in five chance of destroying a Land Raider with a meltagun. Cut that in half if the target has cover.

But we had also mentioned that we might get a destroyed result off of a glance, right?

Let's see what we can see there.

Damage Table Roll | Damage Result AP 1 |

1 | Shaken |

2 | Shaken |

3 | Stunned |

4 | Weapon Destroyed |

5 | Immobilized |

6 | Wrecked |

So we can immediately see that we have a one in six chance (1/6) of getting a destroyed result on a glance. Recall from above that a meltagun will roll a "six" to penetrate (and thus get a glancing result against a Land Raider) 5/36 of the time. Let's see what that does for us.

Chance to hit: 2/3

Chance to glance: 5/36

Chance to destroy on a glance: 1/6

(2/3) x (5/36) x (1/6) = 1.54%

Meh, nothing to write home about but it shows the technique well.

So what do we do now? Well, we need to add the percent chance we got from the pens to the percent chance from the glances to get our total percent chance per meltagun fired to slag the Raider.

Chance to destroy from a pen: 19.44%

Chance to destroy from a glance: 1.54%

Total chance to destroy: 20.98%

Ok, but most of the time we aren't just firing one melta, right? Let's see what the odds of getting a destroyed result are if we fire two.

Remember from part two I explained how to calculate the chance of getting a specific result if you try more than once - such as trying to destroy a Land Raider when firing multiple shots.

We know what the chances of destroying the Land Raider are - 20.98%. That means we know what the chances of it NOT destroying the Land Raider are: 100% - 20.98% = 79.08%.

Take the chances of it not happening once and multiply it by itself to get the chances of it not happening twice (note that you need to multiply the percentages as decimals).

So .7908 times .7908 = .6254 or 62.54%. Since we know the chances that it WON'T happen, we also know the chances that it will.

100% - 62.54% = 37.46%

Note that this is lower than the 41.96% we would have gotten had we (incorrectly) multiplied the odds of a destroyed result from a meltagun by two.

Well, how about three? Easy enough. Just take the .7908 number we got above and multiply it by itself twice (you could also say you're raising it to the third power), like so:

.7908 x .7908 x .7908 = .4945

So, 49.45% chance that it WON'T happen, which means a 50.55% chance that we WILL wreck the Raider!

Looking pretty good. Note AGAIN that this is now QUITE A BIT less than the 62.94 that we would have gotten if we had just taken the base number and multiplied by three.

Ummm... ok, I think that's about everything for the basics. If you have any questions or if you catch any errors... let me know!