(*Roll a die and record the result, 1000000 times. Plot results in a histogram.*)
Histogram[RandomInteger[{1,6},1000000]]

(*Roll a pair of dice and record the sum of the results, 10000 times. Plot results in a histogram.*)
L={}
For[i=0,i<10000,i++,L=Append[L,RandomInteger[{1,6}]+RandomInteger[{1,6}]]]
Histogram[L]

(*Roll a whole bunch of dice (in this case, 100) and record the sum of the results. Do this experiment 100000 times. Plot the results in a histogram.*)
L={}
For[i=0,i<100000,i++,L=Append[L,RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]+RandomInteger[{1,6}]]]
Histogram[L]

(*BEAN MACHINE: http://vis.supstat.com/2013/04/bean-machine/
The Bean Machine simulates a "Galton board," in which at each moment, a ball is dropped from the top center. At each triangular peg, the ball can go left or right with equal probability. This continues until the ball has fallen into one of the bins at the bottom. We repeat this with another ball, and so on. 
WHY IS THIS RELEVANT TO SUMS OF INDEPENDENT RANDOM VARIABLES AND THE CENTRAL LIMIT THEOREM: 
Suppose there are N bins, and the machine has a height of N.
How many pathways are there to get from the top to the bin which is in position k from the end? 
The ball has to go right k times and left the other N-k times.
The ball has N "decisions" to make. 
So there are N-choose-k = N!/(k!(N-k)!) ways 
		to choose which of the N times you go left.
Therefore the probability of a ball ending up in that bin is 
	N-choose-k divided by the total number of choices that can be made, 
		which is 2^N.
  So on average, if we have, say, 100000 balls total then we'd expect 
		100000 (N-choose-k)/2^N
   balls in the bin which is k from the left. 
  The numbers N-choose-k are largest when k is as close as possible to N/2. 
   This is why the middle bins have more balls in them than the side bins.
	 
  This is the same idea behind the sums of dice rolls.
  If we have, say, just 2 dice, then the most likely outcome is 7. 
	The reason this is so much more likely than, say, 2, is 
	because there are more ways for two dice to sum to 7
	((1,6),(2,5),(3,4), (4,3), (5,2), (6,1)) than to 2 (just (1,1)). *)