Will using a module favor high numbers?

Question

Will 6 random unique numbers in the range 0-32 be added and the execution of the module by the result in favor of a large number?

Example: 9 +10 +11 +18 +25 +28 +32 = 133% 20 = 13

Answer 1

Not. Its even or at least skew does not exceed 0.05%.

Despite the fact that the range of possible numbers is not uniformly displayed on mod (192% 20 = 12), the distribution range is much larger than the mode, so it works. Here is my mileage of 1,000,000.

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 

Answer 2

As interesting aside, there is a powerful method that can be used to calculate this manually or very quickly (instead of using brute force) on a computer using the concept of Creating Functions .

(Warning: long post)

You work in a range from 0 to 19, but get this by randomly generating numbers from 0 to 32.

If the probability of getting the number i is equal to p (i) [Note, p (0) = p (1) = p (2) = ... = p (12) and p (13) =. = p (19) and p (0) = 2p (13)).

Now we are interested in the possibility of obtaining a specific amount by generating random numbers 6 times and adding them.

This can be modeled by calculating the coefficients of the sixth power of the polynomial

P (x) = p (0) + p (1) * x + p (2) * x ^ 2 + ... + p (r) * x ^ r + ... + p (19) * x ^ 19

Thus, we look at the coefficients for (P (x)) ^ 6.

For this task, we can ignore the factor 1/33 (to compare which amount is more likely) and p (0) = 2, p (1) = 2, ..., p (19) = 1.

Thus, we look at P (x) = 2 (1 + x + x ^ 2 + ... + x ^ 12) + x ^ 13 + x ^ 14 + .. + x ^ 19.

Now we just need to calculate the coefficients of its sixth degree, take the indicators modulo 20 and add them up. Here you can use fast multidimensional multiplication algorithms such as FFT.

In fact, we could probably do this manually, using some algebra with complex numbers and / or prove the probability distribution assertions with conviction.

Answer 3

Answer: It depends. The following sample program will print the average values ​​for various module values. Obviously, this is not a mathematical proof, but it should already give you a sense of how the average values ​​behave:

 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()); } } 

Generated Output:

 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 

Answer 4

Here is a small python program to calculate probability distribution

 # 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))) 

This gives the following result:

 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 

those. high numbers are indeed a little more likely, but the differences are very small.

Answer 5

Counterexamples:

 9 +10 +11 +18 +25 +28 +32 = 133 % 2 = 1 9 +10 +11 +18 +25 +28 +32 = 133 % 200 = 133 

Which perhaps suggests that you could usefully explain or clarify your question.