Probability Help

[dyjoots]

I'm again reworking the resolution system for my little RPG, and I was wondering if any of the math wizards here could help out with an equation or chart.

The mechanic i'm looking at is roll a pool of dice and try to get one under the target number, but as close to the TN as possible.  So, for example, roll 3d10, and keep the closest one to the target number of 8 without going over.

I haven't quite decided if I want to roll a pool of d6s and keep 2 (added together) or a pool of d10s and keep 1.

Either way, I'm looking for a way to determine the average if you keep the highest AND the average if you keep the lowest.


Sorry if that's confusing...

[M. J. Young]

Dice pools are notoriously difficult to crunch without calculus. The way I recommend in Multiverser's appendix on basic dice curves is to work from the chance of failure. If you need one die to come in to a particular range, find the chance that it will fail to do so--e.g., if you must roll under 8 on a d10, you've got 7 chances to succeed and three to fail, so your chance of failure on one die is 30%. Now, that chance of failure is reduced by a percentage equal to the chance of success for each additional die. Thus for two dice, that 30% is reduced by 70%, which gives (30x.7=)21%; add a third die, and again it reduces by 70%, (21x.7=)14.7%. A fourth die reduces it again to (14.7x.7=)10.29%.

The chance of success is thus determined by subtracting the chance of failure from one hundred, thus:[list=1][*]70%[*]79%[*]85.3%[*]89.71%[/list:o]
However, to find what you want, you probably want to write a program that crunches the numbers for you--someone like Epoch is better able to handle that than I.

--M. J. Young

[bluegargantua]

The mechanic i'm looking at is roll a pool of dice and try to get one under the target number, but as close to the TN as possible.  So, for example, roll 3d10, and keep the closest one to the target number of 8 without going over.

I haven't quite decided if I want to roll a pool of d6s and keep 2 (added together) or a pool of d10s and keep 1.

Either way, I'm looking for a way to determine the average if you keep the highest AND the average if you keep the lowest.


Sorry if that's confusing...

 Er...it is a little confusing.  A couple of quick answers might clear things up:

 1.)  What's the range of Target Numbers?  If you can have a TN of 11, then no matter how big your d10 pool is, you won't hit it.

 2.)  If you *must* take the two highest or two lowest, then you may not be able to get as close to the target number as possible without going over.  If the TN is 4 and you roll 6, 6, 1, 1 and take the two highest, you're going to blow it.  Is this what you want?

Tom

[dyjoots]

The mechanic i'm looking at is roll a pool of dice and try to get one under the target number, but as close to the TN as possible.  So, for example, roll 3d10, and keep the closest one to the target number of 8 without going over.

I haven't quite decided if I want to roll a pool of d6s and keep 2 (added together) or a pool of d10s and keep 1.

Either way, I'm looking for a way to determine the average if you keep the highest AND the average if you keep the lowest.


Sorry if that's confusing...

 Er...it is a little confusing.  A couple of quick answers might clear things up:

 1.)  What's the range of Target Numbers?  If you can have a TN of 11, then no matter how big your d10 pool is, you won't hit it.

 2.)  If you *must* take the two highest or two lowest, then you may not be able to get as close to the target number as possible without going over.  If the TN is 4 and you roll 6, 6, 1, 1 and take the two highest, you're going to blow it.  Is this what you want?

Tom

1) The target nuber is roll equal to or under, so a range from 1 to 10 is normal.

2) As I currently envision it, you take whatever die or dice you want from the pool.  I just wanted to be able to figure out the probabilities for the lowest (to see how easy it is to succeed) and the highest (to see how easy it is to get a higher resolution value).


Also,  thanks, MJ Young.  That ought to help.

I've decided to use pools d10s because I think they are a little less opaque than the two curves that come up with d6s.

[dyjoots]

Dice pools are notoriously difficult to crunch without calculus. The way I recommend in Multiverser's appendix on basic dice curves is to work from the chance of failure. If you need one die to come in to a particular range, find the chance that it will fail to do so--e.g., if you must roll under 8 on a d10, you've got 7 chances to succeed and three to fail, so your chance of failure on one die is 30%. Now, that chance of failure is reduced by a percentage equal to the chance of success for each additional die. Thus for two dice, that 30% is reduced by 70%, which gives (30x.7=)21%; add a third die, and again it reduces by 70%, (21x.7=)14.7%. A fourth die reduces it again to (14.7x.7=)10.29%.

Now that I have had some time to think about it, something seems a little weird about this.  Perhaps I'm doing the math wrong, but it seems like having a smaller target number to roll under is MORE beneficial than a high one if you have the more dice... which seems very, very wrong.

But, like I said, I'm probably doing the math wrong.

Let's say two people are both rolling 2d20.  One of them (Person A) is trying to roll under a 10 on at least one of the dice, and the other (Person B) is trying to roll under a 5.

Person A: 1.00 - .5(.5) = 75%
Person B: 1.00 - .75(.25) = 82%

So, person B, who intuitively should NOT have as good a chance, does.  Can anyone clue me into what I'm doing wrong here?

[Lxndr]

You're calculating the chance of failure.

So for person A, you want 0.5 x 0.5 = 0.25, or 25% failure (75% success)

for person B, it's 0.75 x 0.75, or 0.5625, which is 56.25% failure (or 43.75% success)

[dyjoots]

You're calculating the chance of failure.

So for person A, you want 0.5 x 0.5 = 0.25, or 25% failure (75% success)

for person B, it's 0.75 x 0.75, or 0.5625, which is 56.25% failure (or 43.75% success)

Ah.  Thanks.  I was panicking that my brainOS had crashed.

So, each time you add a die, you multiply the current chance of failure buy the chance of failure per die?  My impression from M. J. Young was that you multiplied the current chance of failure by the chance of success per die.

[Lxndr]

Yeah.  The short form is:

(chance of failure) to the power of (number of dice)

[Brian Leybourne]

Nope, Lxndr has it right. MJYoung made an error in his calculations. 30% chance of failure per die means that for two dice, the chance to fail is .3x.3 or .09 (9%) and thus you have a 91% chance of getting at least one success. Note that this does not calculate how many successes you'll have, just that there will be at least one. To calculate the odds of X number of successes is where it gets very tricky. For example, for the two dice in this example, die A and die B, you have to calculate the following:

A fails, and B fails (zero successes): .3 x .3 (9%)
A succeeds and B fails (one success): .7 x .3 (21%)
A fails and B succeeds (one success): .3 x .7 (21%)
A succeeds and B succeeds (two successes): .7 x .7 (49%)

So you can see that the chance of exactly one success is 42%, the chance of exactly two successes is 49% and if you add those together you get your chance of at least one success (91%).

With more than two dice, it gets really complicated, but it's easy enough to write a small program or excel spreadsheet to calculate it all for you.

Brian.