The Players Corner Archive

CF = 1/((5/100)^INT(DUR/100)*(1-(DUR-(98*INT(DUR/100)))/100))

- Lord Kranar, human Archwizard

[This message has been edited by LordKranar (edited 12-30-2001).] reg

Anyways... here's an explanation of the derivation:

Say you have a 6 sided die, and you want to know what the odds are of you rolling a 6. Well it's simple, 1 in 6 rolls will end up being 6. But let's say you want to look beyond that and want to know what the odds are of you rolling a 6 *twice* in a row, well then you have to do a multiplication. The odds of rolling a 6 the first time is 1/6, the odds of rolling a 6 the second time is also 1/6, so the odds of rolling a 6 twice in a row is (1/6)*(1/6), which comes out to 1/36. So the odds of rolling a 6 twice in a row are 1 in 36. The odds of rolling a 6 *three* times in a row are 1/6*1/6*1/6 = (1/6)^3 = 1/216. And the pattern continues, odds of rolling a 6 five times in a row is (1/6)^5 = 1 in 7776.

With breakage the same logic applies. If you have a DUR of 300, then you need to roll something above 96 three times in a row. The odds rolling 96 or higher three times in a row are (5/100)^3. If you have a DUR of 600, then you need to roll a 96 or higher 6 times in a row to break your item. In general terms it means you basically need INT(DUR/100) rolls in a row for your item to risk catastrophic failure. Having said that, the first part of the equation becomes the odds of getting a roll of 96 or above, to the power of how many rolls in a row you need to risk breaking your item, or in mathematical notation:

(5/100)^INT(DUR/100).

The second part of the equation is a lot more tricky. You need to determine the average between the 5 possible rolls that will result in an additional roll and multiply it by how many rolls are needed to break your item. Well it so happens that the average roll between 96, 97, 98, 99, and 100 is 98 (it's the number right inbetween so this should make sense). You take that average and multiply it by how many rolls in a row you need to give you another roll (remember, this is INT(DUR/100)) and that ends up giving you the average total of the rolls that can cause your item to break. In mathematical notation this is:

98*INT(DUR/100).

Third part of the equation is easier than the second part. We've so far figured out the odds of getting however many additional rolls are needed to break our item, and we've determined the average outcome of those rolls. Now we need to find out what the LAST roll has to be in order to break our item. Remember, if the item's DUR is 300, then you need atleast **4** rolls for your item to break, if your item's DUR is 600, you need atleast 7 rolls for your item to break. In general you need INT(DUR/100) + 1 rolls in order for your item to break. Well parts 1 and 2 of the equation only explained the first INT(DUR/100) rolls, we now have to fit in that last roll into our equation. For your item to break, this last and final roll has to be atleast DUR - 98*INT(DUR/100). If it is, your item goes byebye. Now in order to find out the odds of the final roll being equal or greater to the number needed to break our item, we take the number needed, and divide it by 100 and it ends up looking like this:

1 - [DUR - 98*INT(DUR/100)]

Why are we subtracting 1 from that whole bracket? Because if say we needed a roll of 40 or greater to break our item, what that means is that 60 out of 100 numbers will break our item, basically 60 percent. The only way to describe that mathematically is to write it out as 1 - 40/100 = 60 percent. So we subtract the whole bracket from 1 and we end up with the last part of the equation.

Last step is to put all the parts together:

1) Odds of getting required rolls in a row: (5/100)^INT(DUR/100)

2) The average of the rolls in part 1 will come out to: 98*INT(DUR/100).

3) Odds of *final* roll being greater than the items DUR - step 2 comes out to: 1 - [DUR - 98*INT(DUR/100)]

4) Multiply parts 1 and 3 together and you end up with the percentage of your item breaking:
(5/100)^INT(DUR/100)*[1 - (DUR - 98*INT(DUR/100))]

5) Since it's a percentage, divide 1 with that whole formula to get your odds:

CF = 1 / (5/100)^INT(DUR/100)*[1 - (DUR - 98*INT(DUR/100))]

And there you have it... the formula for calculating the odds of your item suffering catastrophic failure.

This means even at 100 percent integrety, brig can infact commonly suffer CF on the first hit.

- Lord Kranar, human Archwizard

[This message has been edited by LordKranar (edited 12-30-2001).] reg