Stealth and Scattering Armor bonus

[Wardad]

SHIELD DEPLEATERS are smaller and more effective than normal weapons. They are the big equalizers for Shields vs Armor debate.

2 shots at 50% accuracy do not equal 1 shot at 100% accuracy. Two shots at 50% should break down to 75% for one hit, 25% for both hits, and 25% both miss. It should be the same as formula for repeated dice throws. That is why ECM and Combat Sensors are so important.

Is anyone here a Mathamagician at probability and odds? I can't find the link I was looking for.

[Ward]

Wardad:
This is the math you look for. Yesterday I have spent two hours inventing this. Just check what all those abbreviations mean.
((TH sqared with AAW)!)/(100 sqared with AAW!))*AWW +CAM -CDM + E

For your 50% chance:

s=sqared
50%s(2weapons)!
---------------
100%s(2weapons)!

50s(2) 50s(1)
------ + --------- = 0.75
100s(2) 100(1)

[Ward]

Wardad:

SHIELD DEPLEATERS are smaller and more effective than normal weapons. They are the big equalizers for Shields vs Armor debate.

Yes I counted with it. For comparison, here is a rough and simplificated formula for components effective attack: Damage/Space used. That's all.

[Suicide Junkie]

Two shots at 50% should break down to 75% for one hit, 25% for both hits, and 25% both miss.

I believe you meant to type 50/25/25, so it adds up to 100%.

(25% x 2) + (50% x 1) + (25% x 0) = 100%

So you can expect an average of one hit per volley of double-shots @ 50%

If you fire N shot volleys, at 1/N accuracy, you'll get 1 hit per volley on average.

[tbontob]

Originally posted by Suicide Junkie:
Aw, man. Can you believe I'm in a math program at university?

A math major!

No wonder I am having some difficulty understanding the underlying premises/conclusions of some of the Posts being made by the participants.

Even with my having a good math background.

I read what you guys are posting with a great deal of interest.

This is great stuff.

[ March 21, 2003, 16:19: Message edited by: tbontob ]

[Slick]

Originally posted by Ward:
Abbreviations:
!: "faktorial"; I don't know the right word here. 5! means 5+4+3+2+1, 1! is 1, 10! is 10+9+8...+2+1. (3!)! is (3)+(2)+(1).

Just to be clear, the ! operator in math, called factorial, is a multiplicative, not additive. So 5! = 5x4x3x2x1 = 120, not 5+4+3+2+1 = 15.

I didn't check your usage of it so if in fact you intended to add the numbers the answer might still might be right.

There are math symbols for doing addition as you suggest, but they are different from the ! operation.

Slick.

[DirectorTsaarx]

BTW, if you want a fast formula for calculating the sum of "n+(n-1)+(n-2)+...+2+1", it's n*(n+1)/2.

The idea is that you're finding the sum of "n" terms of an arithmetic progression; the generic formula is:

Sum=(n/2)*(A+L)

Where:
A is the first term in the series
L is the Last term in the series
n is the number of terms



[ March 21, 2003, 16:25: Message edited by: DirectorTsaarx ]

[Ward]

I admit I didn´t remember the symbol is multiplicative, not additive. What is important is the usage. I also didn´t know how to say the wole formula is to be repeated AAW times with % changing. But then - it´s about five years I did some math.