Re: [HOE] Math question

[Jason Young <jason_d_young@yahoo.com>]

Okay, if you only account for i=0 to 2, the formula for 1dy looks like
this:

((y+1)/2)+((y+1)/2y)+((y+1)/2y^2)

That could probably be simplified, but not by me at 2:30 am. It checks
exactly against my brute-force spreadsheet method, though. Multiply by x
to get your answer to the average of xdy.

Someday I'll bother to calculate the odds and averages for xdy where
you're allowing for aces but only taking the best die. It's basically the
same process as the Wild Card calculations on that page below, but I
haven't been motivated enough to do it. I imagine it would be pretty
valuable for Deadlands and HOE/LC scenario planning, though.

--- Jason Young <jason_d_young@yahoo.com> wrote:
> Robert,
> 
> Your series looks correct to me. I haven't done high-end math in about
> 12 
> years, so I'm not going to be any help with the simple formula. I wanted
> to do something like that, too, but realized I would spend hours simply
> trying to resurrect my math skills and still might end up without an
> answer. I chose not to do that.
> 
> What I just happen to have, though, is a page I wrote for Savage Worlds.
> If you look at the chart for Averages for Extras (the top, left chart on
> the page), you'll get the approximate results for 1dy. You can then
> multiply for x's larger than that.
> 
> http://www.aracnet.com/~jnryoung/j/sw/swdice.html
> 
> I'm not good at math, but I'm pretty quick with a spreadsheet. I
> calculated each probability for each die out to a result of 40 and took
> a
> weighted average to determine the numbers reported in that chart.
> 
> Sometimes, it's faster to be dumb. :-)
> 
> Incidentally, a quick check of your series for x=1, y=4, and i=0 to 2
> gives an average of 3.3 (actually 3.28). My chart has 3.3 as the average
> of 1d4. Hmmm, very suspicious.
> 
> --- Robert Holland <robert@macauley.com> wrote:
> > Is there a simple formula for calculating the average of a damage (not
> > skill)
> > roll? I *think* the average of xdy damage (x rolls of a die with y
> > sides) could
> > be expressed as 
> > 
> > x * (the sum for i = 0 to infinity of ( (1 / (y^i) ) * ( (y + 1) / 2 )
> )
> > )
> > 
> > Where ( (y + 1) / 2 ) is the average of a single, non-open-ending die
> of
> > y
> > sides, and (1 / (y^i) ) is the probability of the die open-ending i
> > times.
> > 
> > I don't remember enough calculus to find a simple formula for that
> > infinite
> > series, though.
> > 
> > Of course, for values of i greater than 2 the term is less than 1/20th
> > of a
> > point in all cases (assuming standard die types), so just finding the
> > formula
> > for i = 0 to 2 would yield sufficiently accurate results for any
> purpose
> > I can
> > think of.
> > 
> > --Robert Holland


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