Logs for ShadowRN and Associated Mailing Lists

From:	David Buehrer <dbuehrer@******.CARL.ORG>
Subject:	Re: Success Probability
Date:	Fri, 15 Jan 1999 08:42:01 -0700

Message no. 1 For the mere cost of a Thaum, MEYFROIDT Andy wrote:
/
/ Hi,
/ Does anyone know the formulas needed to calculate dice roll
/ probabilities in shadowrun.
/
/ That is given, a target number and a number of dices available, how do I
/ calculate what the probability is that i will get a minimum of x
/ successes.

TN = target number

D = number of dice being rolled

S = number of desired successes

For target numbers of 6 or less:

(((7-TN)/6) x D)/S

For target numbers from 7 to 12:

((.16 x ((13-TN)/6)) x D)/S

For target numbers from 13 to 18:

((.027 x ((19-TN)/6)) x D)/S

Note, even though these equations will yield a result if the number of
desired successes is greater than the number of dice rolled, consider
the chances of rolling more successes than dice to be 0 :)

-David B.
--
"Earn what you have been given."
--
email: dbuehrer@******.carl.org
http://www.geocities.com/TimesSquare/1068/homepage.htm

From:	Mongoose <m0ng005e@*********.COM>
Subject:	Re: Success Probability
Date:	Sat, 16 Jan 1999 14:10:14 -0600

Message no. 2 :TN = target number
:
:D = number of dice being rolled
:
:S = number of desired successes
:
:For target numbers of 6 or less:
:
:(((7-TN)/6) x D)/S

8 dice rolled.

2 successes needed, TN 4

(((7-4)/6)x8)2= 2.

NOT. There is some chance of failure, and the chance of success sure
ain't 200%.

Mongoose

From:	Scott Harrison <Scott_Harrison@*****.COM>
Subject:	Re: Success Probability
Date:	Sun, 17 Jan 1999 00:00:23 -0500

Message no. 3 Hi all,

The formula for determining the chance of getting a
specific number of successes on a specific number of dice
with a specific target number is:

TN = target number
D = # dice rolled
C = chance of TN on 1 die
S = # desired successes

C^(S-1) * (1 - (1-C)^(D-S+1))

^ means raised to the power of

A simple table can be used for TNs, C, and (1-C):

TN C 1 - C
2 5/6 1/6
3 2/3 1/3
4 1/2 1/2
5 1/3 2/3
6 1/6 5/6
7 1/6 5/6
8 5/36 31/36
9 1/9 8/9
10 1/12 11/12
11 1/18 17/18
12 1/36 35/36
13 1/36 35/36
14 5/216 211/216
15 1/54 53/54
16 1/72 71/72
17 1/108 107/108
18 1/216 215/216
19 1/216 215/216
20 5/1296 1291/1296

--Scott

From:	K in the Shadows <Ereskanti@***.COM>
Subject:	Re: Success Probability
Date:	Sun, 17 Jan 1999 10:41:26 EST

Message no. 4 In a message dated 1/16/1999 3:06:50 PM US Eastern Standard Time,
m0ng005e@*********.COM writes:

>
> :(((7-TN)/6) x D)/S
>
> 8 dice rolled.
>
> 2 successes needed, TN 4
>
> (((7-4)/6)x8)2= 2.
>
> NOT. There is some chance of failure, and the chance of success sure
> ain't 200%.
>
Uhm, er, wait a second here. The number of successes (target number of a 4)
is correct here. However, that does NOT take into account the dice rolls
above 4 does it?

-K (statistics isn't my strong point, but this math is just fine, even if D.
Ghost's figures were off from yours)

From:	Gurth <gurth@******.NL>
Subject:	Re: Success Probability
Date:	Sun, 17 Jan 1999 19:24:24 +0100

Message no. 5 According to K in the Shadows, at 10:41 on 17 Jan 99, the word on
the street was...

> Uhm, er, wait a second here. The number of successes (target number of a 4)
> is correct here. However, that does NOT take into account the dice rolls
> above 4 does it?

Just because a formula gives the right answer for a given number you put
into it doesnot make it valid for any other number. You need to either
mathmatically prove that it works for _all_ numbers, or run all the
numbers you need through it and verify they are correct. Depending on what
you need the formula for, the latter can be easier and faster, but then,
if you know the answers why do you need a formula to calculate them?

That said, like K my strong point isn't statistics, nor is it complicated
math (or math that takes too long to work out :)

--
Gurth@******.nl - http://www.xs4all.nl/~gurth/index.html
And that's as far as the conversation went.
-> NERPS Project Leader * ShadowRN GridSec * Unofficial Shadowrun Guru <-
->The Plastic Warriors Page: http://shadowrun.html.com/plasticwarriors/<-
-> The New Character Mortuary: http://www.electricferret.com/mortuary/ <-

GC3.1: GAT/! d-(dpu) s:- !a>? C+(++)@ U P L E? W(++) N o? K- w+ O V? PS+
PE Y PGP- t(+) 5++ X++ R+++>$ tv+(++) b++@ DI? D+ G(++) e h! !r(---) y?
Incubated into the First Church of the Sqooshy Ball, 21-05-1998

From:	Scott Harrison <Scott_Harrison@*****.COM>
Subject:	Re: Success Probability
Date:	Mon, 18 Jan 1999 14:52:03 -0500

Message no. 6 I made my mailer see:
-> Hi all,
->
-> The formula for determining the chance of getting a
-> specific number of successes on a specific number of dice
-> with a specific target number is:

<SNIP>

-> C^(S-1) * (1 - (1-C)^(D-S+1))
->

<SNIP>

Folks,

Please ignore this formula as it is wrong. I should
never do complicated math late at night. I tried making the
general case and of course I found that I cheated in one of my
math steps because of the chances for each die are the same as
independent events. Therefore, this is bogus and methinks will
give odds that are too low as one gets higher in number of dice,
etc.

If I can come up with a simple formula I will post it.
Otherwise it still is the old summation game which is not fun
at all.

Sorry.

--Scott

From:	"D. Ghost" <dghost@****.COM>
Subject:	Re: Success Probability
Date:	Mon, 18 Jan 1999 15:49:07 -0600

Message no. 7 On Mon, 18 Jan 1999 14:52:03 -0500 Scott Harrison
<Scott_Harrison@*****.COM> writes:
<SNIP>
>Folks,
>
> Please ignore this formula as it is wrong. I should
>never do complicated math late at night. I tried making the
>general case and of course I found that I cheated in one of my
>math steps because of the chances for each die are the same as
>independent events. Therefore, this is bogus and methinks will
>give odds that are too low as one gets higher in number of dice,
>etc.
>
> If I can come up with a simple formula I will post it.
>Otherwise it still is the old summation game which is not fun
>at all.
>
> Sorry.

Let's see, let's try this.
First, take the chance of getting the desired target number on one die
and call it C.
Second, let's define a function, f(x) to be C raised to a power x.
Third, let's have D represent the number of dice available, and N
represent the number of successes desired, then if and only if N is less
than or equal to D, is the following true (i think):
The chance to get N successes on D dice is the sum of the series {f(N),
f(N+1), f(N+2), ... f(D)}.

if you understand that, i think it'll give you the right answer ... :)

Basicly, if N>D, chance = DNE;
if N=D, chance = C^N;
if N<D, chance = C^N + C^D + (C to all the powers between D and N).

a few examples (in all cases: D=5):
1) N=5
chance = C^5.
2) N=4
chance = C^4 + C^5.
3) N=3
chance = C^3 + C^4 + C^5.
4) N=1
chance = C^1 + C^2 + C^3 + C^4 + C^5.

Okay, for the programmers, pseudocode:
(same definitions as above)
1. chance = 0.
2. if N>D, goto 6.
3. chance = chance + C^N.
4. N = N+1.
5. goto 2.
6. END.

Sorry, for all the different versions saying the same thing, but I
tutored someone in College Algebra and had to find six different ways to
explain the binomial theorem... and he STILL didn't get it ... :)

Can someone check my work to see if it's right?

one more:
The pain-in-the-ass solution:
Use a spreadsheet (or whatever) and work out every possible combination
and count the number of entries that meet your criteria and divide that
result by the total number of possibilities.

Note: I've done this with 5 dice for the Star Wars system (it's slightly
more complicated to do the same for the Shadowrun system, but I've done
that too.) with up to 5 dice using Quatro Pro. If you are using Quatro
Pro 8, it should be noted that 1 file is a notebook, and each notebook
has up to 255 (I *THINK* that's the maximum) sheets. One sheet has
enough room to handle 5 dice. 6 dice will take 6 sheets, 7 will take
about 36 sheets, and 8 dice will take about 216 sheets. If you want to
calculate 9 or more dice, you're going to have to get creative ...

If anyone wants to go this route, I can give you the formulas (in Quatro
Pro 8 format) that I use. It's not difficult, it's just tedious. :)

--
D. Ghost
(aka Pixel, Tantrum, RuPixel)
"We called him Mother Superior because of the length of his habit" --
Trainspotting
"A magician is always 'touching' himself" --Page 123, Grimoire (2nd
Edition)

___________________________________________________________________
You don't need to buy Internet access to use free Internet e-mail.
Get completely free e-mail from Juno at http://www.juno.com/getjuno.html
or call Juno at (800) 654-JUNO [654-5866]

From:	Wordman <wordman@*******.COM>
Subject:	Re: Success Probability
Date:	Wed, 20 Jan 1999 00:17:10 -0500

Message no. 8 One of the advantages of being an old man on this list is that I've hoarded
a bunch of stuff that happened a long time ago. About eight years ago,
someone posted a chart of the average number of successes you could expect
with a given number of dice for a given target number.

Not exactly the charts we've been talking about, but fuel for the fire. I'm
positive I had probability charts at one time, but I can't find them now.

Anyway, here is the original post:

>>>>>>>
From: jwadsley@****.cc.monash.edu.au
Newsgroups: rec.games.frp.misc
Subject: Shadow Run dice probabilities
Date: 30 Jun 92 15:25:56 +1000
Organization: Computer Centre, Monash University, Australia
Lines: 37

SHADOW RUN: Average Number of Successes

No. of Dice Used)
v Target Number:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
1 0.8 0.7 0.5 0.3 0.2 0.2 0.1 0.1 0.1 0.1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
2 1.7 1.3 1.0 0.7 0.3 0.3 0.3 0.2 0.2 0.1 0.1 0.1 0.0 0.0 0.0 0.0 0.0 0.0
3 2.5 2.0 1.5 1.0 0.5 0.5 0.4 0.3 0.3 0.2 0.1 0.1 0.1 0.1 0.0 0.0 0.0 0.0
4 3.3 2.7 2.0 1.3 0.7 0.7 0.6 0.4 0.3 0.2 0.1 0.1 0.1 0.1 0.1 0.0 0.0 0.0
5 4.2 3.3 2.5 1.7 0.8 0.8 0.7 0.6 0.4 0.3 0.1 0.1 0.1 0.1 0.1 0.0 0.0 0.0
6 5.0 4.0 3.0 2.0 1.0 1.0 0.8 0.7 0.5 0.3 0.2 0.2 0.1 0.1 0.1 0.1 0.0 0.0
7 5.8 4.7 3.5 2.3 1.2 1.2 1.0 0.8 0.6 0.4 0.2 0.2 0.2 0.1 0.1 0.1 0.0 0.0
8 6.7 5.3 4.0 2.7 1.3 1.3 1.1 0.9 0.7 0.4 0.2 0.2 0.2 0.1 0.1 0.1 0.0 0.0
9 7.5 6.0 4.5 3.0 1.5 1.5 1.3 1.0 0.8 0.5 0.3 0.3 0.2 0.2 0.1 0.1 0.0 0.0
10 8.3 6.7 5.0 3.3 1.7 1.7 1.4 1.1 0.8 0.6 0.3 0.3 0.2 0.2 0.1 0.1 0.0 0.0
11 9.2 7.3 5.5 3.7 1.8 1.8 1.5 1.2 0.9 0.6 0.3 0.3 0.3 0.2 0.2 0.1 0.1 0.1
12 10 8.0 6.0 4.0 2.0 2.0 1.7 1.3 1.0 0.7 0.3 0.3 0.3 0.2 0.2 0.1 0.1 0.1
13 11 8.7 6.5 4.3 2.2 2.2 1.8 1.4 1.1 0.7 0.4 0.4 0.3 0.2 0.2 0.1 0.1 0.1
14 12 9.3 7.0 4.7 2.3 2.3 1.9 1.6 1.2 0.8 0.4 0.4 0.3 0.3 0.2 0.1 0.1 0.1
15 13 10 7.5 5.0 2.5 2.5 2.1 1.7 1.3 0.8 0.4 0.4 0.3 0.3 0.2 0.1 0.1 0.1
16 13 11 8.0 5.3 2.7 2.7 2.2 1.8 1.3 0.9 0.4 0.4 0.4 0.3 0.2 0.1 0.1 0.1
17 14 11 8.5 5.7 2.8 2.8 2.4 1.9 1.4 0.9 0.5 0.5 0.4 0.3 0.2 0.2 0.1 0.1
18 15 12 9.0 6.0 3.0 3.0 2.5 2.0 1.5 1.0 0.5 0.5 0.4 0.3 0.3 0.2 0.1 0.1
19 16 13 9.5 6.3 3.2 3.2 2.6 2.1 1.6 1.1 0.5 0.5 0.4 0.4 0.3 0.2 0.1 0.1
20 17 13 10 6.7 3.3 3.3 2.8 2.2 1.7 1.1 0.6 0.6 0.5 0.4 0.3 0.2 0.1 0.1

This is the number of successes you will get on average. That means that
half the time you will get more and half the time less. An average of 0.1
successes indicates that you will get a single success 1 roll in 10. For
example, if you roll 20 dice against a target number of 17, you will, on
average, only get 0.2 successes. This translates to 1 success every 5
attempts! However, with 6 dice and a target number of 4, you get 3 successes
on average. Each die has a 50/50 chance of succeeding.

Cheers,

James Wadsley.
<<<<<<<

Wordman

From:	Wordman <wordman@*******.COM>
Subject:	Re: Success Probability
Date:	Wed, 20 Jan 1999 00:23:31 -0500

Message no. 9 A-ha! Bingo. I knew I had the probability charts somewhere. From an old
Flashlife file:

ftp://thor.flashpt.com/pub/srun/Flashlife/16x13.txt

Wordman

Mailing List Logs for

Further Reading

Disclaimer