Skullmantle and Braingobbla combo. Help with probability

someone done the maths on the likelyness of fleeing units with the braingobbler spell 1) without skullmantle 2) with skullmantle ?

propability of fleeing units with Ld 10-9-8, and the hightened propability when its skullmanteled to 9-8-7?

Well... math or no math, casting braingobler with a skullmantle butcher is sure more convenient and your opponent has higher probabilities of failing the test.

But, if numbers is what you're looking for, here are the ratios:
Ld7 (with or without Skullmantle): 58% probs of passing
Ld8 (with or without Skullmantle): 72% probs of passing
Ld9 (with or without Skullmantle): 83% probs of passing
Ld10: 91% probs of passing


cheers

More probabilities: Leadership Calculations

I could be wrong (I'm good at maths, but hate change calculations )

First a list of all different outcomes of 2d6 rolls (they got to be different, cause a 1-2, is the same as 2-1 but the dices turned in another order):

2:  1-1
3:  1-2
4:  1-3, 2-2
5:  1-4, 2-3
6:  1-5, 2-4, 3-3
7:  1-6, 2-5, 3-4
8:  2-6, 3-5, 4-4
9:  3-6, 4-5
10: 4-6, 5-5
11: 5-6
12: 6-6

This means that there are 21 different outcomes. To calculate the chance of something one uses the following formula: chance = effective outcomes / total outcomes.

Ld6, 9 effective outcomes. 9 / 21 = 0,43 (43%)
Ld7, 12 effective outcomes. 12 / 21 = 0,57 (57%)
Ld8, 15 effective outcomes. 15 / 21 = 0,71 (71%)
Ld9, 17 effective outcomes. 17 / 21 = 0,81 (81%)
Ld10, 19 effective outcomes. 19 / 21 = 0,90 (90%)

You were close Ikim (or I could be wrong)
btw, a friend of mine is making a whole essay about warhammer and chances, I helped him lately and we found out that a normal orc boy has 0,92 (92%) chance to survive an attack from an empire swordsmen. And the swordsmen has far less chance (about 0,50 (50%)) to survive the orc boy's attack. Rather freaky

you are slightly wrong on your probability equations Vargh - there are 36 different ways of the results turning up on 2 dice, not just 21   Here are all the possible results for dice 1 and dice 2 to come up with on a roll.

2  (1-1)
3 (1-2) (2-1)
4 (1-3) (3-1) (2-2)
5 (1-4) (4-1) (2-3) (3-2)
6 (1-5) (5-1) (2-4) (4-2) (3-3)
7 (1-6) (6-1) (2-5) (5-2) (3-4) (4-3)
8 (2-6) (6-2) (3-5) (5-3) (4-4)
9 (3-6) (6-3) (4-5) (5-4)
10 (4-6) (6-4) (5-5)
11 (5-6) (6-5)
12 (6-6)

Ikim has the odds correct

but a 2-1 is the same as a 1-2. isn't it ?
I mean I roll two dices, the one is a 1 and the other is a 2. Then who's going to tell me if it's 1-2 or 2-1 ?

think of the 2 dice being different colours if that helps - one red and one blue.

the red dice rolling 1 and the blue dice rolling a 2 is one possible outcome from the roll.

the red dice rolling 2 and the blue dice rolling a 1 is another distinct possibility from the roll.

to work out every possible combination of rolling the 2 dice, and therefore calculating the true odds, you need to take each possible roll into account - it deosn't matter that the result is the same as this is what you then do:

rolling a total of 3 can happen in 2 ways, I could roll a 1 on the red dice and a 2 on the blue, or the other way around.

so rolling a 3 has a probability of happening 2 in every 36 rolls (or 1/18 - it's the same thing) - these odds can then be worked out as a percentage (approx 9% in this case)

quickly wrote a program on my calculator to roll 2 dices 5000 times, it gave an outcome of 0,589 to be higher lower then 8 (Ld7). So I guess you're right. Now I remember why I hate chance calculations

  personally I love probability and game theory maths - but then I'm an out-and-out geek

wow, nice work!

the best i can do is working out that miscast/irresistable force is 1/36 =D

moehaha, helped my friend with his essay a bit more. An ogre bull will surely kill a High Elf spearmen in a one-to-one combat.

The high elf has 0,08 (8%) chance of wounding the ogre.
Then the ogre has 0,83 (83%) chance of killing the high elf.

This means that the ogre will most likely kill the elf in one or two turns where it is impossible for the elf to kill the ogre in 2 turns.
Now isn't that brutal 

But look at the points comparisons. An elf isn't meant to be able to kill an ogre in one turn. You should check out the odds when equal points are involved (probably 2-3 elves Vs 1 ogre)

Thanks
Corai

Corai is right, and when are you even going to encounter a single elf, fighting a single ogre.
Usually elfs go around in groups (read: units) and so do ogres. To have real ratios, you should consider real situations.
A good comparison would be: 3 ogre bulls, w/ XHW (117 pts) vs 10 spear elfs with muso (116 pts).
Using Avians combat calculator ( http://folk.ntnu.no/tarjeia/avian/calculations/unit_combat.php ), and considering both situations, in which the bulls have charged and the elfs have charged, here are the stats:

Ogres charge:
Side.............Attacks...Hits...Wounds.Unsaved..Unwarded....Result
Ogres............12.00....6.00....4.00......3.33.........3.33..........3.33
Elfs.................6.67....4.44....1.48.......1.48.........1.48.........1.48

Elfs charge:
Side....Attacks..Hits....Wounds....Unsaved.......Unwarded....Result
Elfs.......5.00....3.33.....1.11............1.11...........1.11...........1.11
Ogres...10.52...5.26.....3.51............2.92...........2.92...........2.92


So, ogres would still win in CC both if charging or being charged.

cheers

I know that it is not fair.
But he chose to do his essay about one-to-one fights.