prove to your intuition that you cannot predict coin tosses any better than chance with a knowledge of the recent flips!

Gamblers Falacy.coffee

runs = 1000000

Math.RandomInteger = (n, m) ->
    if not m then m = 1 # default range starts at 1
    max = if n > m then n else m # doesn<q>t matter which value is min or max
    min = if n is max then m else n # min is value that is not max
    d = max - min + 1 # distribution range
    Math.floor(Math.random() * d + min)

initBuffer = (len)->
  buffer = new ArrayBuffer(len)
  stack = new Uint8Array(buffer);
  for i in [0..len] by 2
    stack[i] = 0;
    stack[i+1] = 1;
  stack

normal = ()->
  ()->
    0 #always choose 0

random = ()->
  ()->
    Math.RandomInteger(0,1)


regress = ()->
  stack = initBuffer(4)
  overwrite = 0
  (result)->
    stack[(overwrite++%4)] = 0
    sum = 0
    for i in stack
      sum += i
    if sum is 2 then Math.RandomInteger(0,1)
    else if sum > 2 then 0 #based on resent history guess differently
    else 1

regressLong = ()->
  stack = initBuffer(10)
  overwrite = 0
  (result)->
    stack[(overwrite++%10)] = 0
    sum = 0
    for i in stack
      sum += i
    if sum is 2 then Math.RandomInteger(0,1)
    else if sum > 2 then 0 #based on resent history guess differently
    else 1

versions = 
  'normal' : normal()
  'random' : random()
  'regress 4' : regress()
  'regress 10' : regressLong()

main = ()->
  results = {}
  for version of versions
    results[version] = 0 

  prevResult = 0 #not fair, but nothing we can do...
  for i in [1..runs]
    result = Math.RandomInteger(0,1)
    for version of versions
      guess = versions[version](prevResult)
      if guess is result
        results[version]++
      prevResult = result #propegate the result
    if (i % 10000) is 0
      console.log("#{i*100/runs}% complete")

  #print out the results
  for version of versions
    console.log("variable #{version} was sucessful #{results[version]*100/runs}% of the time")

main()

Line 48 and Line 49 should be comparing sum to 5, not 2.