对于一个包含10个整数的列表,有10!种可能的顺序或排列。为什么在只尝试5000次后,random.shuffle会产生重复的结果?
>>> L = range(10)
>>> rL = list()
>>> for i in range(5000):
... random.shuffle(L)
... rL.append(L[:])
...
>>> rL = [tuple(e) for e in rL]
>>> len(set(rL))
4997
>>> for i,t in enumerate(rL):
... if rL.count(t) > 1:
... print i,t
...
102 (7, 5, 2, 4, 0, 6, 9, 3, 1, 8)
258 (1, 4, 0, 2, 7, 3, 5, 9, 6, 8)
892 (1, 4, 0, 2, 7, 3, 5, 9, 6, 8)
2878 (7, 5, 2, 4, 0, 6, 9, 3, 1, 8)
4123 (5, 8, 0, 1, 7, 3, 2, 4, 6, 9)
4633 (5, 8, 0, 1, 7, 3, 2, 4, 6, 9)
>>> 10*9*8*7*6*5*4*3*2
3628800
>>> 2**19937 - 1
431542479738816264805523551633791983905393 [snip]
>>> L = list()
>>> for i in range(5000):
... L.append(random.choice(xrange(3628800)))
...
>>> len(set(L))
4997
编辑:FWIW,如果对于单个对而言不具有相同概率的概率为: p = (10!- 1) / 10! 并且组合数为: C = 5000!/ 4998!* 2!= 5000 * 4999/2 那么具有重复的概率为:
>>> import math
>>> f = math.factorial(10)
>>> p = 1.0*(f-1)/f
>>> C = 5000.0*4999/2
>>> 1 - p**C
0.96806256495611798