Afterworld: the Whiff Factor

[Marhault]

Afterworld is beginning to come together, but I have some fairly serious concerns which I hope the community here can help me address.  This thread is intended to address the first of these.  It is also the first thread I’ve started on the game and I’m a little nervous, so please, bear with me.

I suppose I should start with a quick overview of the mechanics.  Everything that is quantified in the game has a Rank.  The Ranks are 2, 4, 6, 8, 10 and 12, each corresponding to the type of die that is rolled when the ability/skill/whatever is used.  The defending player (including the GM) rolls a die corresponding to the Rank of an applicable quality to serve as the ‘difficulty.’  The results are compared, and the higher roll wins.  If there is no opposing quality, or it has not been defined, the GM assigns a rank to serve as difficulty on the spot.  The greater the difference between the rolls, the more spectacular the success or failure, with the acting player winning ties.

My problem comes in that I am having a little bit of difficulty determining what the actual probability of winning a given contest is.  i.e., What are the odds that a player rolling a d6 will beat the GM rolling a d4?  a d8? a d12?  Obviously, I have a rough idea, but that’s not really good enough when trying to set up the system.

Characters are intended to be somewhat exceptional individuals, so I want to make sure to keep the whiff factor under control.  My current thought is that a d4 is a fairly difficult task, or a fairly competent level of skill or whatever, and that anything higher than that is pretty impressive.  PCs, then, would have lots of things Ranked at d4, a few at d6, and only one or two at d8 or higher.  Most ordinary people would have, at best, one or two d6 Ranked abilities.

What I am really hoping for is a formula (or a case by case breakdown) of the probability of winning a contest with a single die of any type against a single die of any type.  Can anybody help me out with this?

[ffilz]

Ok, for equal sized dice (dN), a tie happens 1/N*100 percent of the time, and either player wins (1-1/N)/2*100 percent of the time, or (N-1)/2N*100 percent of the time.

For two different sized dice (dN and dM where M is larger):

A tie happens N/M*1/M*100 percent of the time.
N wins (N-1)/2M*100 percent of the time.
M wins (2M-N-1)/2M*100 percent of the time.

Frank

[Marhault]

Frank:

Thanks for the reply and the formulae!  I'm still having some trouble with the results, though.  Examples:

If M=12 and N=2.
N Wins = (2-1)/(2*12) = 1/24 = .041667
M Wins = (24-2-1)/(2*12) = 21/24 = 0.875
Tie = 2/12 * 1/12 = 2/144 = 1/72 = 0.013889

Which totals 67/72, or 0.93056, instead of adding up to 1.

If M=8 and N=6
N Wins = (6-1)/(2*8) = 5/16 = 0.3125
M Wins = (16-6-1)/(2*8) = 9/16 = 0.5625
Tie = 6/8 * 1/8 = 6/64 = 3/32 = 0.09375

Which totals 62/64, or 0.96875, instead of adding up to 1.

I did a couple of the other combinations and they come out similarly, just a little bit below 1.  What am I doing wrong?

[NN]

The chance of a tie is 1/M

[ffilz]

Oops... I knew I would goof up on something...

It's easy to see, tie should have been N/(M*N) which of course is 1/M...

Frank

[Marhault]

Yeah!  That did it, everything balances now!  Thanks, to both of you.

Time to start building that thank you page. . .

[ffilz]

It's pretty stupid that I made that mistake since I got it right in this thread when I noticed that 1 skill die didn't do all that much to improve the chance of doubles (a tie).

I should have also cross checked my work against the brute force I did for d6/d8 (well, I think I did crosscheck, but somehow I didn't pay close enough attention).

By the way, I solved this by doing the brute force for d6/d8 and then figuring out the pattern.

Ah well...

Frank