Mike Holmes - question @ success %

[unodiablo]

Hi Mike,
I'm working on changing the game mechanics to Dead Meat... And I know this is your forte'.

Can you tell me what the success chance is if you have a three die pool -d6's, and your target # is 4 or higher on each die? How often would you get at least one, two or all three successes?

Also, how hard is it to figure out percentages if I give you multiple target number for the same 3d6 pool?

Thank you kindly,
Sean

[Epoch]

I'm not Mike Holmes.  As I'm sure you figured out.

The odds of complete failure (zero successes) in 3d6, TN 4, are 1/8.

The odds of exactly one success are 3/8

The odds of exactly two successes are 3/8

The odds of exactly three successes are 1/8

Cheers!

[Added in an edit:

The above can be taken to show a valuable lesson about die pools: they're an inherently ineffecient use of dice if you don't alter the size of the pool and/or the target number, or read the spots as something other than success failure.  The proposed system could be profitably reduced to rolling a single d8.

Second corollary:  You can use three coins of different types (nickle, dime, penny, for example) to simulate a d8 roll.
]

[ This Message was edited by: Epoch on 2001-11-07 17:49 ]

[unodiablo]

Hi Epoch,
Just to clarify, do you mean two successes = 2/8?

What I'm looking for, and wanting to utilize, is two things that I wouldn't get with a d8. Style-wise, I want to use the 'd666' mechanic, odes to Fulci's Gates of Hell, and the DOTD quote "When there's no more room in Hell...", etc...

And with the three dice, the player will get to see degrees of success upon which to base their narration. Later in the game, the first chance / degree of success will come easier, as will the last.

Thanks,
Sean

[Epoch]

Nope.  I mean 2 successes, 3/8.  That way, y'know, it totals out to 8/8.  :smile:

A single d8 could have a degree of success mechanic.  It works like this:

If you roll a 1, you fail.

If you roll a 2-4, you get a marginal success.

If you roll a 5-7, you get a success.

If you roll an 8, you get a complete success.

A 666/111 mechanic, of course, would require the 3d6 method.

[Le Joueur]

Epoch wrote:

I'm not Mike Holmes.  As I'm sure you figured out.

Likewise, I'm sure.

unodiablo wrote:

I'm working on changing the game mechanics to Dead Meat... And I know this is your forte'.

(Just to quibble, but wouldn't forté mean "in a loud, forceful manner" like in classical music?  I think you mean his forte, without the ', but then I don't know Mike personally.)

Can you tell me what the success chance is if you have a three die pool -d6's, and your target # is 4 or higher on each die? How often would you get at least one, two or all three successes?

Also, how hard is it to figure out percentages if I give you multiple target number for the same 3d6 pool?

Epoch is right that with three six-sided dice:

TAR  0SUC    1SUC    2SUC    3SUC
 4  12.50%  37.50%  37.50%  12.50%

But to extend it to all target numbers:

TAR  0SUC    1SUC    2SUC    3SUC
 6  57.87%  34.72%   6.94%   0.46%
 5  29.63%  44.44%  22.22%   3.70%
 4  12.50%  37.50%  37.50%  12.50%
 3   3.70%  22.22%  44.44%  29.63%
 2   0.46%   6.94%  34.72%  57.87%

Each column lists the percentage chance of getting exactly that number of successes.  (You'll have to add across to get a total of X successes or more.)  Zero successes are, of course, failure.  And a target number of 1 results in having 100% chance of succeeding on all three dice; there will be no other alternatives.

If you want a further breakdown, such as "if the shooter has one success (with a target number of 4), what are the odds of how many he will also have with the same dice at target number 2?"  Just ask.

Fang Langford

(Personally we chose the 'two ten-sided dice taken from a target rating' mechanic for Scattershot.  It gives a nice sigmoid probability graph, easy enough math, a small range of success ratings for our FitM system and centers out at ten roughly.)

[ This Message was edited by: Le Joueur on 2001-11-08 13:46 ]

[Mike Holmes]

Good to know that when I'm out with bronchitis that there are plenty more good Samaritan statisticians available out there.  :smile:

Mr. Sullivan makes a good point about die pools and rolling methods in general. Quite often designers seem to not know what the odds are on their mechanics. This sometimes has the ironic effect of creating systems that are more complicated than they have to be, systems that could reduce handling time by redesigning just a bit.

Alternately, there are lots of cool things that you can do with die mechanics that some people miss entirely, things that can really add to what you are trying to accomplsh. Given the multitude of ways to handle things, I advocate trying to determine what sort of curves and expected outcomes you are looking for first. Then when you know what you want, find a mechanic that fits around that. Too often I see designers enamored of a rolling mechanic trying to fit their system around that mechanic, when they should be designing a die mechanic that works with the system.

Mike

[unodiablo]

Hey Guys,

Been thinking about this this morning... And wouldn't your first success % be at least 50%? With a TN of 4, 4-6 is a success, and that's 1-3 fail / 4-6 succeed. Looks like 50/50 to me...

And, uh, sorry about the improper ' in forte. :smile:

Sean

Or is Fang's chart failure %'s? Makes more sense that way...

[ This Message was edited by: unodiablo on 2001-11-08 11:20 ]

[Epoch]

The odds of getting a success on any die is 50%, yes, unodiablo.  The odds of getting at least one success on three dice are 7/8 (that is, it's the odds of getting 1 success plus the odds of getting 2 successes, plus the odds of getting three successes).

No, Le Jouer's chart is not listing the chance of failure, unless you mean in the first column.