Math hammer

[Grish]

Hey everyone,

I wanted to start a thread on general math questions as applies to Warhammer. I personally think some good probability/theory hammer is important as a fundamental in this game, so I wanted a 'general help' for Math-specific questions. In this case, I wanted to analyze the Screaming banner, and if it is any better this edition, with the constant fear checks. I then realized I don't remember enough of Finite math to solve my own questions.

I wanted to see how effective it was in the following scenarios:
LD 7, 8, 9, 10
Same leaderships, but with BSB rerolls. I want to figure out if it's better to take out the general or BSB if possible to increase our fear checks. With combats so drawn out now, and a check each round for us and them, it's not uncommon for this item to force 6+ checks.

So can someone show me how to do it with LD10 to start with? A failure happens if they roll 56, 65, 66 with 3 dice. Once I have an example to go off of, I can take it from there.

 

[Bishop]

Well, for Ld 10 they would fail on 11 and 12.

There are 216 different die combinations that can occur when rolling 3d6 (instead of the usual 36).

To figure that out I took each set of 36 and determined how many failures occurred within each set.

example:

if the first dice is a 1, then there are 3 failures (just like a normal Ld test)

1,5,6
1,6,5
1,6,6

Then did the same for the rest of them

this gives us 3 failures for the occurrences of 1,2,3,4 on the first die. When you get to the 5 and 6 they are different. They have 11 and 20 failures respectively.

example:
5,1,6
5,2,6
5,3,6
5,4,6
5,5,6
5,6,1
5,6,2
5,6,3
5,6,4
5,6,5
5,6,6

So the answer for Ld10 failures when rolling 3d6 is

3 / 36 + 3 / 36 + 3 / 36 + 3 / 36 + 11 / 36 + 20 / 36

or

43 / 216

which is really close to 20% (it's 19.9%)

 

[Swissdictator]

Just a friendly recommendation, as I can see this threat becoming popular and long.

Could we start indexing it in the first post? Possibly even by subject? That way it can be indexed by type of rolls, with specific posts/situations linked to directly.

Hope it comes across as a friendly suggestion as intended.

 

[autti]

So, it seems I was too slow. Anyways, here's what I was about to post:


Interesting question!

As as disclaimer, I'm sure there's a prettier way for calculating this, and I'm rather sure that I've even been taught that method, probably some time during Konrad's reign, but in any case I can't recall this stuff any more So the following calculations are based on some simple logics, and are therefore subject to errors on the logical level as well as failing to take some things into account.

As you stated, failing a LD10 leadership test requires a roll containing at least two sixes or one six and one five. Let's divide this into pieces

A: Ways to roll triple sixes
(1/6)*(1/6)*(1/6) = 1/216

B: Ways to roll exactly two sixes, and one lesser number
(1/6)*(1/6)*(5/6) = 5/216
However, this can be done in three ways (66?, 6?6, ?66), so we get 3 * (5/216) = 15/216

C: Ways to roll exactly one six, one five, and one number that isn't six (another five is ok though)
(1/6)*(1/6)*(5/6) = 5/216
This combination can be thrown in six way (65?, 56?, 6?5, 5?6, ?65, ?56), so we get 30/216. However, we have two ways to roll 655, 565 and 556, so we have to substract three combinations, resulting in 27/216.

Change to fail LD10 test: A+B+C = 43/216 = 19,9%


If the leadership drops to 9, the following new combinations fails the test (the previously calculated combinations naturally fail too):

D: Ways to roll triple fives: 1/216

E: Ways to roll exactly two fives, and one number that isn't five (triple fives calculated at D) or six (calculated at C)
(1/6)*(1/6)*(4/6) * 3 = 12/216

F: Ways to roll exactly one six, one four, and one number that isn't six or five (previously calculated)
(1/6)*(1/6)*(4/6) * 6 = 24/216
Remember to substract the three duplicate combinations, we get 21/216

Change to fail LD9 test: A+B+C+D+E+F = 77/216 = 35,6%


To calculate LD8, the failure also occurs with 63? and 54?, resulting in 52,3% change of failure.
For LD7, also 62?, 53? and 44? fails, so that sums up to 68,1%, and LD6 has failure rate of 80,6%.


So let's say our choices are to feast on the enemy general, or to butcher the battle standard bearer, and see how this affects the enemy unit based on their LD.

The following table contains changes to fail the test. BSB column is for killing the bsb, the LD containing generals LD value. GENERAL column is for killing the general, the LD containing the enemy unit's LD value (with reroll):

LD    | BSB   | GENERAL
LD10: | 19,9% | 0,199^2 =  4,0%
LD9:  | 35,6% | 0,356^2 = 12,7%
LD8:  | 52,3% | 0,523^2 = 27,4%
LD7:  | 68,1% | 0,681^2 = 46,5%
LD6:  | 80,6% | 0,806^2 = 65,0%

So if general has LD10 and the target unit LD7, it's much better to kill the general. For comparison, without the screaming banner, LD10 without reroll would have 8,3% failure rate, and LD7 with reroll 17,4%. So even then killing the general would be a better choice, given that the target unit has significantly lower LD value.