Beginners Query - Simple counter problem

[mensanator at aol.com]

On Sep 6, 1:44 pm, Carsten Haese <cars... at uniqsys.com> wrote:
> On Thu, 2007-09-06 at 11:24 -0700, Ian Clark wrote:
> > Carsten Haese wrote:
> > > def d6(count):
> > >     return sum(random.randint(1, 6) for die in range(count))
>
> > My stab at it:
>
> >      >>> def roll(times=1, sides=6):
> >      ...     return random.randint(times, times*sides)
>
> That produces an entirely different probability distribution if times>1.
> Consider times=2, sides=6. Your example will produce every number
> between 2 and 12 uniformly with the same probability, 1 in 11. When
> rolling two six-sided dice, the results are not evenly distributed. E.g.
> the probability of getting a 2 is only 1 in 36, but the probability of
> getting a 7 is 1 in 6.
>
> --
> Carsten Haesehttp://informixdb.sourceforge.net

Why settle for a normal distribution?

import random

def devildice(dice):
  return sum([random.choice(die) for die in dice])

hist = {}

for n in xrange(10000):
  the_key = devildice([[1,2,3,10,11,12],[4,5,6,7,8,9]])
  if the_key in hist:
    hist[the_key] += 1
  else:
    hist[the_key] = 1

hkey = hist.keys()
m = max(hkey)
n = min(hkey)
histogram = [(i,hist.get(i,0)) for i in xrange(n,m+1)]
for h in histogram:
  print '%3d  %s' % (h[0],'*'*(h[1]/100))

##    5  **
##    6  *****
##    7  ********
##    8  ********
##    9  ********
##   10  *******
##   11  *****
##   12  **
##   13
##   14  **
##   15  ******
##   16  ********
##   17  ********
##   18  ********
##   19  ********
##   20  *****
##   21  **

They're called Devil Dice because the mean is 13 even
though you cannot roll a 13.