average swift stride roll

[ju1ce]

Mathematically what is the average swift stride roll?
thx
j

[SpellArcher]

Roughly 8 1/2 isn't it?

Don't quote me on this!

[b4z]

2D6 Average = 7
3D6 Pick The Two Highest Average = 9

High Elves on Foot (+5) = 12" Average
High Elves on Fast Elven Cavalry (+9) = 18" Average
High Elves on Barded Elven Cavalry (+8) = 17" Average

[Wicksi]

On the other hand I think most of us can agree to say that statisticly statistics usually doesnt apply ingame

[Andrew_uk]

2D6 Average = 7
3D6 Pick The Two Highest Average = 9

High Elves on Foot (+5) = 12" Average
High Elves on Fast Elven Cavalry (+9) = 18" Average
High Elves on Barded Elven Cavalry (+8) = 17" Average

I'd like to know how you worked out the odds on that, I mean admittedly I don't really play much 8th edition so I have no idea if theres some mini clause that I haven't considered. But I am a maths teacher and if theres one thing I know its theoryhammer.

The odds when you roll three dice and keep the highest two are as follows;

p(2)=1/216
p(3)=3/216
p(4)=7/216
p(5)=12/216
p(6)=19/216
p(7)=27/216
p(8)=34/216
p(9)=36/216
p(10)=34/216
p(11)=27/216
p(12)=16/216

Mean (average) result comes in as being 8.458333333 so closer to 8 than 9

[NightHawk45]

+1 to Andrew_uk. As a fellow mathematician, I salute you

[AlexCrute]

But don't elves use magic to alter the immutable laws of mathematics?
1+1=25 if Teclis rolls it

[Andrew_uk]

But don't elves use magic to alter the immutable laws of mathematics?
1+1=25 if Teclis rolls it

Teclis is indeed a true mathmagician... for those geeks around here though see if you can spot the error in this working

y=x
multiply both sides by y
y^2=xy
so y^2-xy=x^2-y^2
factorising gives y(y-x)=(y+x)(y-x)
so y=y+x

Doesn't appear to make sense unless x=0, which makes y=0, which means all numbers MUST equal 0... but theres a mistake in the logic, spot it

[slug]

y=x
multiply both sides by y
y^2=xy
so y^2-xy=x^2-y^2
factorising gives y(y-x)=(y+x)(y-x)
so y=y+x

y^2-xy=0 regardless of the value of x and y
factorising gives y(y-x)=(x+y)(x-y), which is opposite to the way you have it. Your maths is wrong as well, or is that the error?

Slug

[DarkWingDuck]

2D6 Average = 7
3D6 Pick The Two Highest Average = 9

High Elves on Foot (+5) = 12" Average
High Elves on Fast Elven Cavalry (+9) = 18" Average
High Elves on Barded Elven Cavalry (+8) = 17" Average

I'd like to know how you worked out the odds on that, I mean admittedly I don't really play much 8th edition so I have no idea if theres some mini clause that I haven't considered. But I am a maths teacher and if theres one thing I know its theoryhammer.

The odds when you roll three dice and keep the highest two are as follows;

p(2)=1/216
p(3)=3/216
p(4)=7/216
p(5)=12/216
p(6)=19/216
p(7)=27/216
p(8)=34/216
p(9)=36/216
p(10)=34/216
p(11)=27/216
p(12)=16/216

Mean (average) result comes in as being 8.458333333 so closer to 8 than 9

How exactly do you roll 8.4583333?

I think the original poster was taking the Median, not Mean (which in this case is 9).

[hewhorocks]

But don't elves use magic to alter the immutable laws of mathematics?
1+1=25 if Teclis rolls it

Teclis is indeed a true mathmagician... for those geeks around here though see if you can spot the error in this working

y=x
multiply both sides by y
y^2=xy
so y^2-xy=x^2-y^2
factorising gives y(y-x)=(y+x)(y-x)
so y=y+x

Doesn't appear to make sense unless x=0, which makes y=0, which means all numbers MUST equal 0... but theres a mistake in the logic, spot it

Shame on you trying to divide by zero! Shame! Shame! Economists FTW!

[Andrew_uk]

y=x
multiply both sides by y
y^2=xy
so y^2-xy=x^2-y^2
factorising gives y(y-x)=(y+x)(y-x)
so y=y+x

y^2-xy=0 regardless of the value of x and y
factorising gives y(y-x)=(x+y)(x-y), which is opposite to the way you have it. Your maths is wrong as well, or is that the error?

Slug

Yes that was a mistake by me, its supposed to be factorised to y(y-x)=(x+y)(x-y) then cancel to
y=-(x+y)

which still makes it seem like an oddity

Shame on you trying to divide by zero! Shame! Shame! Economists FTW!

This man spotted it, divide by zero and the whole world implodes on you

I think the original poster was taking the Median, not Mean (which in this case is 9).

Yeh the median is 9. Generally talking the mean is considered the fairest average to use in most circumstances (theres some boring theory behind why which I don't fully remember from university and which is completely irrelevant to anything I ever have to teach at even A level statistics). The general idea of the theory is that other ways of calculating averages don't take into account the weighting as fairly so distributions such as this which are skewed give a fairer average value.

[grantmepower]

But don't elves use magic to alter the immutable laws of mathematics?
1+1=25 if Teclis rolls it

Nope. He is still required to meet the casting value, just like the Book of Hoeth.

[possumman]

p(2)=1/216
p(3)=3/216
p(4)=7/216
p(5)=12/216
p(6)=19/216
p(7)=27/216
p(8)=34/216
p(9)=36/216
p(10)=34/216
p(11)=27/216
p(12)=16/216

This is clearly correct, and as a fellow maths teacher I also salute you.
However, the biggest change is not the average moving from 7 to 8.5 - the biggest change is how skewed the distribution is. On 2D6, the odds of a 2 or 12 are the same - here, a 12 is SIXTEEN times more likely! Same with 3 and 11; rather than being even, an 11 is NINE times more likely than a 3! This distribution has a massive skew towards the higher and, and this isn't reflected just in the average moving up by 1.5.
Swift is HUGELY better, and HUGELY more reliable, than a normal charge move.

[DarkWingDuck]

I think the original poster was taking the Median, not Mean (which in this case is 9).

Yeh the median is 9. Generally talking the mean is considered the fairest average to use in most circumstances (theres some boring theory behind why which I don't fully remember from university and which is completely irrelevant to anything I ever have to teach at even A level statistics). The general idea of the theory is that other ways of calculating averages don't take into account the weighting as fairly so distributions such as this which are skewed give a fairer average value.

As the poster above points out, the distribution is not normal, it's heavily skewed to the high end. As a gamer, the true average doesn't really matter much. The median is a much more interesting number (that's the distance where over 50% of my charges will at least reach).

A better way to think about it, is what are the odds of me making a specific charge. As gamers, that's what we care about.

So what are these odds ... here they are:
P(3" charge): 215/216 = 99.5%
P(4" charge): 212/216 = 98.1%
P(5" charge): 205/216 = 94.9%
P(6" charge): 193/216 = 89.4%
P(7" charge): 174/216 = 80.6%
P(8" charge): 147/216 = 68.1%
P(9" charge): 113/216 = 52.3%
P(10" charge): 77/216 = 35.6%
P(11" charge): 43/216 = 19.9%
P(12" charge): 16/216 = 7.4%

[elvennoob]

Nope. He is still required to meet the casting value, just like the Book of Hoeth.

I'm sorry, where does this say that (in reference to the Book of Hoeth)? My apologies for basically derailing this into a different rules question, but that just threw me for a loop.

[grantmepower]

If you check the errata for high elves on the games workshop website it replaces the wording for the book of hoeth in the army book with

Any spell cast by the bearer will be cast with irresistible force on any successful casting roll which is a double, however it will only be considered a miscast if the roll includes a double 6.

(bold mine)
The errata can be found here: http://www.games-workshop.com/gws/conte ... ue&start=2