有缺陷的随机数发生器？

Question

我使用了这个加权随机数发生器。

import random

def weighted_choice(weights):
    totals = []
    running_total = 0

    for w in weights:
        running_total += w
        totals.append(running_total)

    rnd = random.random() * running_total
    for i, total in enumerate(totals):
        if rnd < total:
            return i

如下：

# The meaning of this dict is a little confusing, so here's the explanation:
# The keys are numbers and values are weights of its occurence and values - 1
# are weights of its disoccurence. You can imagine it like biased coins
# (except for 2 which is fair coin).
probabilities = { 0 : 1.0, 1 : 1.0, 2 : 0.5, 3 : 0.45, 4 : 0.4, 5 : 0.35,
                    6 : 0.3, 7 : 0.25, 8 : 0.2, 9 : 0.15, 10 : 0.1
                  }
  numberOfDeactivations = []
  for number in probabilities.keys():
    x = weighted_choice([probabilities[number], 1 - probabilities[number]])
    if x == 0:
      numberOfDeactivations.append(number)
  print "chance for ", repr(numberOfDeactivations)

我经常看到7 ， 8 ， 9 ， 10中的结果。

是否有一些证据或保证这对概率论是正确的？

Answer 1

编辑：作为旁注：我认为你的代码相当于

import random
probabilities = { 0 : 1.0, 1 : 1.0, 2 : 0.5, 3 : 0.45, 4 : 0.4, 5 : 0.35,
                    6 : 0.3, 7 : 0.25, 8 : 0.2, 9 : 0.15, 10 : 0.1}
numberOfDeactivations=filter(
         lambda kv:random.random()<=probabilities[kv] , probabilities)

原始答案：

方法是正确的。 下面是一个完整的示例，创建频率表并将其与请求的权重进行比较。

通过100000次迭代，没有任何迹象表明您没有得到您的要求。 'expected'是您请求的概率，'got'是您实际获得该值的一小部分。 比率应接近1，它是：

  0, expected: 0.2128 got: 0.2107 ratio: 1.0100
  1, expected: 0.2128 got: 0.2145 ratio: 0.9921
  2, expected: 0.1064 got: 0.1083 ratio: 0.9825
  3, expected: 0.0957 got: 0.0949 ratio: 1.0091
  4, expected: 0.0851 got: 0.0860 ratio: 0.9900
  5, expected: 0.0745 got: 0.0753 ratio: 0.9884
  6, expected: 0.0638 got: 0.0635 ratio: 1.0050
  7, expected: 0.0532 got: 0.0518 ratio: 1.0262
  8, expected: 0.0426 got: 0.0418 ratio: 1.0179
  9, expected: 0.0319 got: 0.0323 ratio: 0.9881
 10, expected: 0.0213 got: 0.0209 ratio: 1.0162

 A total of 469633 numbers where generated for this table. 

这是代码：

import random

def weighted_choice(weights):
    totals = []
    running_total = 0
    for w in weights:
        running_total += w
        totals.append(running_total)
    rnd = random.random() * running_total
    for i, total in enumerate(totals):
        if rnd < total:
            return i


counts={ k:0 for k in range(11)}
probabilities = { 0 : 1.0, 1 : 1.0, 2 : 0.5, 3 : 0.45, 4 : 0.4, 5 : 0.35,
                    6 : 0.3, 7 : 0.25, 8 : 0.2, 9 : 0.15, 10 : 0.1
                  }

for x in range(100000):
  numberOfDeactivations = []
  for number in probabilities.keys():
    x = weighted_choice([probabilities[number], 1 - probabilities[number]])
    if x == 0:
      numberOfDeactivations.append(number)
  for k in numberOfDeactivations:
    counts[k]+=1.0

sums=sum(counts.values())
counts2=[x*1.0/sums for x in counts.values()]

print "ratio expected frequency to requested:":

# make the probabilities real probabilities instead of weights:
psum=sum(probabilities.values())
for k in probabilities:
    probabilities[k]=probabilities[k]/psum

for k in probabilities:
    print "%3d, expected: %6.4f got: %6.4f ratio: %6.4f" %(k,probabilities[k],counts2[k], probabilities[k]/counts2[k])

Answer 2

这在数学上是正确的。 这是逆变换采样的应用（尽管它在这种情况下工作的原因应该是相对直观的）。

我不知道Python，所以我不能说是否有任何细微之处使得这个特定的实现无效。