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Message no. 1
From: MEYFROIDT Andy <ameyfroi@****.MOBISTAR.BE>
Subject: [OT] I think
Date: Fri, 15 Jan 1999 09:57:02 +0100
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.

thanks

Krypton
Message no. 2
From: Micheal Feeney <Starrngr@***.COM>
Subject: Re: [OT] I think
Date: Fri, 15 Jan 1999 04:40:05 EST
In a message dated 99-01-15 04:07:36 EST, you write:

> 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.
>

::SIGH:: and here I thought that statistics class was a waste of time.

OK, first of all, this system only works well with TN's of six or less.

Subtract the TN from 7, multiply it by .16. That gives you the probobility of
success for one die.

Divide the number of dice to be rolled by the number of successes your
computing for. Multiply this by the first number and you should have the
probability you want.

I think it also works if you multiply the probability for one success by the
number of dice your rolling then divide it by the number of successes.


As a rule of thumb, for a TN of two, five out of every six dice will give you
a success. for a 3, two thirds of them. for a four, Half, for a 5, one
third, and for a six, only one out of every six dice rolled.

Interestingly enough, the odds of rolling a 7 are just the same as rolling a 6
as long as the rule of 6 applies. 1 out of every six dice would give you a
success. To carry this analogy on into the higher TN's (I won't bother trying
to think about TN's larger than 13!) take the number of dice you have to roll.
First remove five of every six dice you have to roll, then follow the rule of
thumb I gave you for TN's less than six for the remaining dice

At any rate.... If anyone is still awake at this point in my discourse...
Please, tell me how you did it... this even put ME to sleep trying to type it
up!!

--
Starrngr -- Now with an UPDATED webpage:
Ranger HQ
<A HREF="http://hometown.aol.com/starrngr/index.htm">;
HTTP://hometown.aol.com/starrngr/index.htm</A>;

"You wear a Hawaiian shirt and bring your music on a RUN? No wonder they call
you Howling Mad..." -- Rabid the Pysad.
Message no. 3
From: Paul Gettle <RunnerPaul@*****.COM>
Subject: Re: [OT] I think
Date: Fri, 15 Jan 1999 10:51:18 -0500
-----BEGIN PGP SIGNED MESSAGE-----

At 09:57 AM 1/15/99 +0100, you 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.

I'm not so keen on the formulas, but there's a page out there that is
a chart of the odds of getting at least one success, given number of
dice, and target number.

The explanation of how to use the chart is at:
http://www.logicrucible.com/diceguid.htm

and the actual chart is at:
http://www.logicrucible.com/d6reroll.htm


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--
-- Paul Gettle, #970 of 1000 (RunnerPaul@*****.com)
PGP Fingerprint, Key ID:0x48F3AACD (RSA 1024, created 98/06/26)
C260 94B3 6722 6A25 63F8 0690 9EA2 3344
Message no. 4
From: "D. Ghost" <dghost@****.COM>
Subject: Re: [OT] I think
Date: Fri, 15 Jan 1999 14:47:50 -0600
On Fri, 15 Jan 1999 09:57:02 +0100 MEYFROIDT Andy
<ameyfroi@****.MOBISTAR.BE> writes:
>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.

I've never had a statistics class (it was covered in algebra or something
...) but here's how I think it goes:

1) If the T# is 6 or less:
a)Count the number rolls that will succeed (ie, if the target number is
4, the successful rolls are {4, 5, 6} so the number is 3.)
b)Divide that number by 6.

2) If the T# is 7 or more:
-First, start with a base chance of 1.
a)Divide the base chance by six (to represent the chance of rolling a
six) and decrease the T# by 6.
b)Repeat step 2a until the T# is 6 or less.
c)Multiply the curent base chance the result of using the current target
number in step 1.

3) Multiply the number you got in above (the chance to roll the given
target number on one die.) by the number of dice available. This gives
you your average number of successes and should give you an idea of your
chance to succeed.

Hope that helps.

--
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)

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Message no. 5
From: Mongoose <m0ng005e@*********.COM>
Subject: Re: [OT] I think
Date: Sat, 16 Jan 1999 14:14:38 -0600
: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.
:
:thanks
:
: Krypton


Theres been a couple replies here, all wrong
(afiak), so I'll try myself, from scratch. I may be overlooking
something here, but I think this method works:

I assume you are good enough to figure out how to calculate the chance
of getting 1 success of a given TN with X dice*? If you can do that, you
can then calculate the chance of getting another with the remaining X-1
dice , and another with the remaining X-2 dice... until you have the
desired successes. The Chance of getting that many success is the chance
of getting the first, AND the second, AND the third...
Making up numbers, say you have 6 dice and need 3 sixes.
Getting 1 success is 1-[(5/6)^6] likely. Thats .665
Getting another on the remaining 5 is .665 x (1- [(5/6)^5]) likely. .598
Getting the third; .398 x (1- [(5/6)^4]) , or .398
So your chances of getting 3 sixes (together) with 6 dice is .665 x .
598 x .398 = .158

Hmm, that seems suprisingly good. Did I do something wrong?

Mongoose

*Since it seems a lot of people DON'T seem to know how to calculate the
chance of getting 1 success on x dice:
Find the chance of getting 1 failure on a single die, at that TN. Then
raise that to the a power equal to the dice rolled (giving the chance that
all dice are failures), and subtract that from 1 (giving the chance that
at least 1 is a success).
Message no. 6
From: Mongoose <m0ng005e@*********.COM>
Subject: Re: [OT] I think
Date: Fri, 15 Jan 1999 16:29:37 -0600
:OK, first of all, this system only works well with TN's of six or less.
:
:Subtract the TN from 7, multiply it by .16. That gives you the
probobility of
:success for one die.
:
:Divide the number of dice to be rolled by the number of successes your
:computing for. Multiply this by the first number and you should have the
:probability you want.
:
:I think it also works if you multiply the probability for one success by
the
:number of dice your rolling then divide it by the number of successes.

Well, yes, those would be mathmaticallly equivalent, but its still
wrong. A simple demo : two dice (A+B) and a tn of 4, 50-50 chance of each
die making it. You want a single 4:

You could get:
25%- no 4=s
25%- die A 4+, B less.
25% B 4+. A less
25% B 4+ and A 4+

That makes a 75% cance of getting one 4 on two dice. (I think you all
knew that one...)

By yout system, its (2/1)x.48. Thats 96%, or even 100% without the
rounding in your step 1!

Heck, youv'e got the chance of getting 2 4's on 8 dice is .48x8/2, or
1.92!

:At any rate.... If anyone is still awake at this point in my
discourse...
:Please, tell me how you did it... this even put ME to sleep trying to
type it
:up!!


Yes, apparently it did. :)

Mongoose
Message no. 7
From: K in the Shadows <Ereskanti@***.COM>
Subject: Re: [OT] I think
Date: Sun, 17 Jan 1999 10:38:45 EST
In a message dated 1/16/1999 3:03:31 PM US Eastern Standard Time,
m0ng005e@*********.COM writes:

> So your chances of getting 3 sixes (together) with 6 dice is .665 x .
> 598 x .398 = .158
>
> Hmm, that seems suprisingly good. Did I do something wrong?
>
Maybe .. that end number there, the .158 in terms of "1 in..." what is it?

-K (who just wants this clarified, as he finds it interesting that Mongoose is
right (alone) and three others he's been given all had the basically same
words and expressives and are wrong (by him))
Message no. 8
From: Mon goose <landsquid@*******.COM>
Subject: Re: [OT] I think
Date: Sun, 17 Jan 1999 12:25:32 PST
>In a message dated 1/16/1999 3:03:31 PM US Eastern Standard Time,
>m0ng005e@*********.COM writes:
>
>> So your chances of getting 3 sixes (together) with 6 dice is .665
x .
>> 598 x .398 = .158
>>
>> Hmm, that seems suprisingly good. Did I do something wrong?
>>
>Maybe .. that end number there, the .158 in terms of "1 in..." what
is it?



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Message no. 9
From: Mon goose <landsquid@*******.COM>
Subject: Re: [OT] I think
Date: Sun, 17 Jan 1999 12:28:19 PST
>From: K in the Shadows <Ereskanti@***.COM>
>> So your chances of getting 3 sixes (together) with 6 dice is .665
x .
>> 598 x .398 = .158
>>
>> Hmm, that seems suprisingly good. Did I do something wrong?
>>
>Maybe .. that end number there, the .158 in terms of "1 in..." what
is it?

There is 1 event- the roll. .158 of that event will come up the way
desired. Its like batting averages, which are emperically extrapolted
proablities.


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Message no. 10
From: Mon goose <landsquid@*******.COM>
Subject: Re: [OT] I think
Date: Sun, 17 Jan 1999 12:37:02 PST
>From: K in the Shadows <Ereskanti@***.COM>

>> So your chances of getting 3 sixes (together) with 6 dice is .665
x .
>> 598 x .398 = .158
>>
>> Hmm, that seems suprisingly good. Did I do something wrong?
>>
>Maybe .. that end number there, the .158 in terms of "1 in..." what
is it?

.158 out of 1; its like a batting average. In other terms, that's
15.8%, or about 1 in 6. Better than I noticed when chucking 6d6,
looking for 3 sixes (Took me 12 rolls to get that just once). But I
maybe just did not continue rolling long enough to hit statistical
average.
The other formula posted today looks like it could be right, though I
haven't puzzled out the method behind it; it may be a compressed version
of my procedure- If both are right, they should be mathmatically
equivalent.

Mongoose

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Message no. 11
From: Scott Harrison <Scott_Harrison@*****.COM>
Subject: Re: [OT] I think
Date: Sun, 17 Jan 1999 21:24:24 -0500
Mon goose <landsquid@*******.COM> made my mailer see:
-> >From: K in the Shadows <Ereskanti@***.COM>
->
-> >> So your chances of getting 3 sixes (together) with 6 dice is .665
-> x .
-> >> 598 x .398 = .158
-> >>
-> >> Hmm, that seems suprisingly good. Did I do something wrong?
-> >>
-> >Maybe .. that end number there, the .158 in terms of "1 in..." what
-> is it?
->

The correct answer is 671/46656 or about 0.144 (1.44%). This translates
to about 1 in 70.

-> .158 out of 1; its like a batting average. In other terms, that's
-> 15.8%, or about 1 in 6. Better than I noticed when chucking 6d6,
-> looking for 3 sixes (Took me 12 rolls to get that just once). But I
-> maybe just did not continue rolling long enough to hit statistical
-> average.
-> The other formula posted today looks like it could be right, though I
-> haven't puzzled out the method behind it; it may be a compressed version
-> of my procedure- If both are right, they should be mathmatically
-> equivalent.
->
-> Mongoose
->
-> ______________________________________________________
-> Get Your Private, Free Email at http://www.hotmail.com

It should be correct. Basically the theory behind it relies on two
principles. The first is that the odds of getting something to happen on
all dice is equal to the change on the first die times the change on the
second die, etc. The second is that sometimes it is easier to figure out
the negative and arrive at the answer one seeks. For example, the chance
to roll two sixes on two dice is equal to 1/6 times 1/6 = 1/36. But the
chance to roll one six on two dice is more easily found by looking at the
negative: i.e., what is the chance that no sixes will be rolled and take
that away from 1. Therefore, 5/6 times 5/6 = 25/36 and taking that from
1 results in 11/36. One can easily show this is correct by creating a
table of all 36 possible combinations of rolling two dice and seeing that
there are 11 that have at least 1 six in them.

Therefore, one uses this technique to proceed with the series. One can
actually show it is true by saying something like: the chances of rolling
two sixes on three dice is equal to the chance of rolling one six on one
die times the chance of rolling one six on the remaining two dice. Which
turns out to be 1/6 times 11/36 = 11/216. This is fact in the answer
using the formula I provided.

By the way, I actually had to work out that formula by hand since I saw
the postings and knew there was something amiss. I spent a good hour
making sure it is correct. I hate having to dredge up the old statistics
stuff since I usually rely on a computer providing me with all the good
information I need (including all my dice rolling while gaming). Note
that when using a computer for gaming most of the rolls are where one
would expect in the bell curve. It seems to make life more like it
should be with only a minimum of bizarre incidents (which makes things
really fun when they happen)! :-)

--Scott
Message no. 12
From: Logan Graves <logan1@*****.INTERCOM.NET>
Subject: Re: [OT] I think
Date: Sun, 17 Jan 1999 21:32:50 -0500
In our last episode, MEYFROIDT Andy wrote:
>
> Hi,
> Does anyone know the formulas needed to calculate dice roll
> probabilities in shadowrun.
>
<etc etc etc>

Just in case anyone's still paying attention to all the MATH stuff,
here's the URL to the SHADOWRUN PROBABILITY charts:

http://www.ccs.neu.edu/home/ratinox/srprob.html

It's really amazing what one finds lying around in Bookmark folders...

--Fenris
_______________________________________________logan1@*****.intercom.net
(>) Point-and-click interface??!! Ya stoopid chiphead,
I got yer point-n-click interface RIGHT HERE!!
(>) Fenris, drawing his "Harbringer of the Void™" plasma cannon
Message no. 13
From: Mon goose <landsquid@*******.COM>
Subject: Re: [OT] I think
Date: Sun, 17 Jan 1999 20:00:59 PST
>Just in case anyone's still paying attention to all the MATH stuff,
>here's the URL to the SHADOWRUN PROBABILITY charts:
>
> http://www.ccs.neu.edu/home/ratinox/srprob.html




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Message no. 14
From: Mon goose <landsquid@*******.COM>
Subject: Re: [OT] I think
Date: Sun, 17 Jan 1999 20:04:02 PST
>Just in case anyone's still paying attention to all the MATH stuff,
>here's the URL to the SHADOWRUN PROBABILITY charts:
>
> http://www.ccs.neu.edu/home/ratinox/srprob.html


That chart gives the average number of succeses (which IS found by
multiplying the chance for one success by the # of dice rolled), but
that will not help you to determine the chance of getting a certain # of
succeses, which is what the original poster wanted to know.

Mongoose

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Message no. 15
From: Robert Watkins <robert.watkins@******.COM>
Subject: Re: [OT] I think
Date: Mon, 18 Jan 1999 14:29:36 +1000
Mongoose writes:
> That chart gives the average number of succeses (which IS found by
> multiplying the chance for one success by the # of dice rolled), but
> that will not help you to determine the chance of getting a certain # of
> succeses, which is what the original poster wanted to know.

Simply put, there is no simple function. It involves combinations, which I'm
a bit rusty on.

It's a binomial situation, Success or Failure.

There's a pattern for working out binomial possibilities. It goes like this:

Actually, snip that... after writing it up, I realised I can't remember it.
If you really want to know the maths, look it up in a book.

(Not very helpful, but it's been over 8 years since I did this stuff. Look
up the Binomial Theorem, as part of probabilities, and combinations and
permutations. It's a fairly simple formula to look at, but very
computationally expensive to work out, as it involves iterating through it)

--
Duct tape is like the Force: There's a Light side, a Dark side, and it
binds the Universe together.
Robert Watkins -- robert.watkins@******.com
Message no. 16
From: "Ojaste,James [NCR]" <James.Ojaste@**.GC.CA>
Subject: Re: [OT] I think
Date: Wed, 20 Jan 1999 09:49:33 -0500
Scott Harrison wrote:
> -> >> So your chances of getting 3 sixes (together) with 6 dice is
> .665
> -> x .
> -> >> 598 x .398 = .158
> -> >> Hmm, that seems suprisingly good. Did I do something wrong?
> -> >Maybe .. that end number there, the .158 in terms of "1 in..."
what
> -> is it?
> The correct answer is 671/46656 or about 0.144 (1.44%). This translates
> to about 1 in 70.
>
Let me try.

This is a perfect situation for combinatorics.
Writing "X choose Y" as "XcY", where "XcY" means
X!/((X-Y)!Y!)
S = rolling a 6
F = rolling a 1,2,3,4 or 5
P(S) = 1/6
P(F) = 5/6
D = number of dice = 6
X = number of successes = 3

So the probability of rolling exactly 3 sixes on six dice is:
DcX * P(S)^X * P(F)^(D-X)
= 20 * (1/6)^3 * (5/6)^3
= 0.053584

and the probability of rolling 3 or more sixes is the same as the
sum of the probabilities of rolling 3, 4, 5 or 6 sixes, OR 1 -
the sum of the probabilities of rolling 0, 1 or 2 non-sixes. I'll
go with the first since it isn't much more work.

P(3 sixes) = 0.053584
P(4 sixes) = 15 * (1/6)^4 * (5/6)^2
= 0.008038
P(5 sixes) = 6 * (1/6)^5 * (5/6)^1
= 0.000643
P(6 sixes) = 1 * (1/6)^6 * (5/6)^0
= 2.1433E-05
Total = 0.06228567

or about 1 in 16.

James Ojaste
Message no. 17
From: "Ojaste,James [NCR]" <James.Ojaste@**.GC.CA>
Subject: Re: [OT] I think
Date: Wed, 20 Jan 1999 15:55:15 -0500
I wrote:
> > -> >> So your chances of getting 3 sixes (together) with 6 dice is
> Let me try.
[snip]
> So the probability of rolling exactly 3 sixes on six dice is:
> DcX * P(S)^X * P(F)^(D-X)
> = 20 * (1/6)^3 * (5/6)^3
> = 0.053584
>
> and the probability of rolling 3 or more sixes is the same as the
[snip]
> Total = 0.06228567
>
> or about 1 in 16.
>
And then I read the Probability thread... *sigh* That'll teach me
to let 1200 unread messages build up in my ShadowRN folder. Although
I *do* think that my formula was the neatest. ;-)

We *really* need some way of drawing equations neatly on this list...

James Ojaste

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