Dice from Changing Random Generator

Python, 31 chars

Similarly to scleaver, define the generator like this:

from random import randint
n=0
def r():
    global n;n+=1
    return randint(1,n)

Then a function to return dice rolls:

D=lambda:eval('r(),'*6)[-1]%6+1

Call D() any time you need a uniformly random dice roll.

Scala 23

def s={r;r;r;r;r;r%6+1}

The method r can be (approx.) implemented like this:

var cnt = 0 
val rnd = new util.Random 

def r = {
  cnt %= 1000
  cnt += 1
  rnd.nextInt (cnt)
}

a rough test:

scala> (1 to 6).map (i => ((1 to 600) map (_=>s)).filter (_ == i).size)
res26: scala.collection.immutable.IndexedSeq[Int] = Vector(110, 105, 91, 96, 106, 102)

Every 6th call should produce an equal distribution over the 6 values, so I throw away 5.

GolfScript (15 chars)

This assumes that the number of rolls required is supplied on stdin and lists that many results to stdout.

# The free increasing random function
0:N;{N):N rand)}:r;

~{r{;r}5*6%)n}*

Online demo

While I could get the 10 point bonus for using fewer than 1000 rolls to generate 1000 numbers, it would cost me far more than 10 characters. The trivial approach of extracting suitable entropy when N is a multiple of a power of 2 or 3 falls well short because the number of results available mod 3 is only 333 + 111 + 37 + 12 + 4 + 1 = 498. Therefore it's necessary to take a sample-and-reject approach. Using this approach you can get an expected 2242 rolls from 1000 calls to r, but there's extra overhead from the book-keeping and base is a very long function name.