在范围0-32内添加6个随机唯一数字并对结果进行模运算,是否会偏向于较高的数字?
例如:9 + 10 + 11 + 18 + 25 + 28 + 32 = 133 % 20 = 13
在范围0-32内添加6个随机唯一数字并对结果进行模运算,是否会偏向于较高的数字?
例如:9 + 10 + 11 + 18 + 25 + 28 + 32 = 133 % 20 = 13
(警告:较长的帖子)
您正在使用 0 到 19 的范围,但通过随机生成 0-32 的数字来实现。
如果获得数字 i 的概率为 p(i) [注意,p(0)=p(1)=p(2)=...=p(12)和p(13)=..=p(19),并且p(0)=2p(13)]。
现在我们感兴趣的是通过生成随机数六次并将它们相加来获得特定总和的机会。
这可以通过计算多项式的第六个幂的系数来建模
P(x)=p(0)+p(1)*x+p(2)*x^2+...+p(r)*x^r+...+p(19)*x^19
因此,我们正在查看(P(x))^6的系数。
对于给定的问题,我们可以忽略1/33因子(为了比较哪个和更有可能),并且p(0)=2,p(1)=2,...,p(19)=1。using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static Random rand;
static void Main(string[] args)
{
rand = new Random();
for (int modulus = 1; modulus < 1000; modulus++)
{
calculateAverage(modulus);
}
}
public static void calculateAverage(int modulus)
{
List<int> moduloList = new List<int>(100);
for (int i = 0; i < 100; i++)
{
int sum = 0;
for (int k = 0; k < 6; k++)
{
sum += rand.Next(0, 33);
}
moduloList.Add(sum % modulus);
}
Console.WriteLine("Average for modulus {0}: {1}", modulus, moduloList.Average());
}
}
生成的输出:
Average for modulus 1: 0
Average for modulus 2: 0,49
Average for modulus 3: 1,03
Average for modulus 4: 1,47
Average for modulus 5: 1,96
Average for modulus 6: 2,55
Average for modulus 7: 3,03
Average for modulus 8: 3,42
Average for modulus 9: 4,15
Average for modulus 10: 5,06
Average for modulus 11: 4,62
Average for modulus 12: 5,9
Average for modulus 13: 5,82
Average for modulus 14: 6,8
Average for modulus 15: 7,28
Average for modulus 16: 7,8
Average for modulus 17: 8,15
Average for modulus 18: 9,34
Average for modulus 19: 9,2
Average for modulus 20: 10,36
Average for modulus 21: 9,74
Average for modulus 22: 9,41
Average for modulus 23: 11,5
Average for modulus 24: 11,51
Average for modulus 25: 11,45
Average for modulus 26: 13,05
Average for modulus 27: 12,59
Average for modulus 28: 14,92
Average for modulus 29: 13,1
Average for modulus 30: 14,1
Average for modulus 31: 15,5
Average for modulus 32: 16,46
Average for modulus 33: 16,54
Average for modulus 34: 16,38
Average for modulus 35: 19,61
Average for modulus 36: 17,26
Average for modulus 37: 15,96
Average for modulus 38: 19,44
Average for modulus 39: 17,07
Average for modulus 40: 17,73
# modulus
m = 20
# range of the random numbers 0..n-1
n = 33
# number of random numbers in sum
k = 6
# distribution of one random number
# a[i] is the probability that a random number modulo m is i.
a = [0]*m
for i in range(n): a[i % m]+= 1/n
# convolution
b = a
for i in range(1,k):
# Here b[t] is the probability that the sum of i random numbers is t.
# Compute c[t] as the probability that the sum of i+1 random numbers is t.
c = [0]*m
for i in range(m):
for j in range(m):
c[(i+j)%m] += a[i]*b[j]
b=c
# print the probability distribution of the result
for i in range(m): print(i, b[i])
# compute average
print("average", sum(i*b[i] for i in range(m)))
这将给出以下结果:
0 0.0500007971936
1 0.0499999764222
2 0.0499991633939
3 0.0499984370886
4 0.0499978679688
5 0.0499975063648
6 0.0499973824748
7 0.0499975063648
8 0.0499978679688
9 0.0499984370886
10 0.0499991633939
11 0.0499999764222
12 0.0500007971936
13 0.0500015451796
14 0.0500021452719
15 0.0500025347512
16 0.0500026702559
17 0.0500025347512
18 0.0500021452719
19 0.0500015451796
average 9.50015120662
也就是说,高数值确实可能性稍微更大一些,但差别非常小。
反例:
9 +10 +11 +18 +25 +28 +32 = 133 % 2 = 1
9 +10 +11 +18 +25 +28 +32 = 133 % 200 = 133
这或许表明您可以有用地澄清或深化您的问题。
不是的。它是偶数,或者至少偏差似乎不超过0.05%。
即使可能数字的范围不能均匀地映射到模数(192%20 = 12),分布范围比模数大得多,因此可以自行解决。 这是我运行的100万次。
MOD COUNT %
0 50098 5.00980
1 49660 4.96600
2 49832 4.98320
3 50150 5.01500
4 50276 5.02760
5 49864 4.98640
6 50282 5.02820
7 49771 4.97710
8 49886 4.98860
9 49663 4.96630
10 49499 4.94990
11 49964 4.99640
12 50155 5.01550
13 50169 5.01690
14 49829 4.98290
15 50191 5.01910
16 49887 4.98870
17 50334 5.03340
18 50139 5.01390
19 50351 5.03510