Math Help

[Jaif]

but what is meant by C(D N)?

The -C-ombination of D events taken N at a time.

Let's say you want to figure out the probability that you get 2 successes (exactly).  Step 1 is to figure out the odds of the first & second die both being successes, and the remainder failures.  Step 2 is realizing that the chance of the first and second die being successes is the same as the 3rd and 4th, the 2nd and 5th, or any other combination of 2 dice out of the 5.  Then you find the number of combinations, and multiply the probability by that number.

In a math text book, the "(D N)" part would be vertical, but I lack the skills to make that happen in the messages. :-)

Now, a mistake I found.  I switched the "t-1" and "11-t" parts, so the forumula should be:

C(D N) * ((11-T)/10)^N * ((T-1)/10)^(D-N)

Again, I think.  I made an elementary spreadsheet to check this out, and the numbers are looking good, but this is a stab at something I did ages ago, so please check.

-Jeff

[Mike Holmes]

Hey, Shane, as you can see from the math in Jeff's post (and my own stupidity), it gets a bit complicated. Why are you interested, exactly? Is it just curiosity, or do you have some practical application that you are trying to apply? If the latter, if you let us know what it was about, we might be better able to help you with whaterver it is.

Mike

[Mike Holmes]

but what is meant by C(D N)?

Ah, I get it now. What I was missing, and probably Durgil as well is that you were doing the entire operation for C(D N) and substituting.

C(5 0) = 5!(0!(5-0)!) = 1

C(5 1) = 5!(1!(5-1)!) = 5

The full equation with numbers sunbstituted for variables would be:

1
- 5!(0!(5-0)!) * ((5)/10)^0 * ((5)/10)^5 ----- 0 successes
- 5!(1!(5-1)!) * ((5)/10)^1 * ((5)/10)^4 ----- 1 success

which evaluates exactly as he has it. Once again, Jeff's right on it.

Mike

[ShaneNINE]

Hey, Shane, as you can see from the math in Jeff's post (and my own stupidity), it gets a bit complicated. Why are you interested, exactly? Is it just curiosity, or do you have some practical application that you are trying to apply? If the latter, if you let us know what it was about, we might be better able to help you with whaterver it is.

1. Simple curiosity. I've always gotten into the probability and statistics and such of every RPG I've ever owned. Sort of a side hobby. TROS (with Exalted) is my first die-pool game and the mechanics are alien to me. I just want to get a handle on the probability and the math behind it.

2. I want to plug a formula into Excel where I can change the TN, the number of dice, and the number of required successes and see what the base percent chance of getting the required successes is.

[Durgil]

Check out what I found on Excel a little earlier today:
=BINOMDIST(N,D,(11-T/10),FALSE)
Now, if I wanted to find the chance of getting at least 3 successes with a TN of 6 and throwing 5 dice, I would add up the chances of 0, 1, and 2 successes, as found by the above formula, and subtract that from 1?

1-BINOMDIST(0,5,(11-6)/10,FALSE)-BINOMDIST(1,5,(11-6)/10,FALSE)-BINOMDIST(2,5,(11-6)/10,FALSE) = 0.5

[Durgil]

Hey, Shane, as you can see from the math in Jeff's post (and my own stupidity), it gets a bit complicated. Why are you interested, exactly? Is it just curiosity, or do you have some practical application that you are trying to apply? If the latter, if you let us know what it was about, we might be better able to help you with whaterver it is.

Mike

Knowing exactly how the math works behind the game makes it much easier to create House Rules for things that the author didn't include, possibly change an already present rule to better fit your own tastes or campaign, or just improve your ability to Game Master by properly awarding bonuses or penalties during play without throwing off the balance of the game.

For example, is it better to change the number of dice or change the target number?  By knowing how the chance for successes and blunders changes by changing those numbers, you will become a better GM IMHO.

[Mike Holmes]

Good point Tony.

There is another, pretty obvious technique that you can use when you don't know the math off the top of your head, and you have a spreadsheet. I call it the brute force method. Simply use the RAND function (with ROUNDUP or RANDBETWEEN if you have it) an have the machine count the dice that come up this way or that (using other stat functions like Countif, etc). Can be easier in the short run for coming up with approximations on such things as house rules. Just do a few thousand iterations, and you'll get the idea of where a rule is headed quickly.

Mike

[Durgil]

...use the RAND function (with ROUNDUP or RANDBETWEEN if you have it) an have the machine count the dice that come up this way or that (using other stat functions like Countif, etc). Can be easier in the short run for coming up with approximations on such things as house rules. Just do a few thousand iterations, and you'll get the idea of where a rule is headed quickly.

That's an excellant idea!  I've never thought about doing that before, Thanks.

[Durgil]

Hi all, if anyone is interested, I've created an Excel Workbook that computes the odds, on one page, of getting a specific number of successes, and, on the second page, finds the probability of getting at least a certain number of successes or more.  This is done for pools of ten-sided dice from 1 to 20 and for target numbers up to 20, and also shows the odds of rolling a botch or fumble.  The sheets are protected so that only the target number may be changed, but I didn't use a password to protect them with so if you want to change anything, go right ahead.  If anyone's interested, then just send me a short email to the address below, and I'll attach the workbook to the reply.

[Thirsty Viking]

As i understand it,  you want the average number of succes given a #d and a TN?

ok first use the earlier formulas  for

x = the % chance of sucess on 1 die

x =  (11-tn) /10     when    tn < 11
or    (21-tn)/100    when    10 < tn < 21
or    (31-tn/1000   when    20 < tn < 31  For the ALWAYS A CHANCE people

y = # of dice rolled

x * y =  Average number of success at TN for y dice

if you put these into a spread sheet,   you will then have the comparative effects of changing either the TN or the # of  dice.  in terms of the AVERAGE # success per role

hope this answers what you were looking for.
send me your email,  and i'll send you my works spread sheet for this.

Code:

                                   TN
d10 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2
3.0 2.7 2.4 2.1 1.8 1.5 1.2 0.9 0.6 0.3
4.0 3.6 3.2 2.8 2.4 2.0 1.6 1.2 0.8 0.4
5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5
6.0 5.4 4.8 4.2 3.6 3.0 2.4 1.8 1.2 0.6
7.0 6.3 5.6 4.9 4.2 3.5 2.8 2.1 1.4 0.7
8.0 7.2 6.4 5.6 4.8 4.0 3.2 2.4 1.6 0.8
9.0 8.1 7.2 6.3 5.4 4.5 3.6 2.7 1.8 0.9
10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0

A warning,  players will expect a little consistency,   make decisions on trends,  not current situation....   lie about opponents die roles  not how a situation is handled  when the story demands it.