Question

please do it in C++

Write a program that simulates the rolling of two dice. The program should use rand to roll the first die and should use rand again to roll the second die. The sum of the two values should then be calculated. [Note: Each die can show an integer value from 1 to 6, so the sum of the two values will vary from 2 to 12, with 7 being the most frequent sum and 2and 12 being the least frequent sums. The image below shows the 36 possible combinations of the two dice. Your program should roll the two dice 36,000 times. Use a one-dimensional array to tally the numbers of times each possible sum appears. Print the results in tabular format. Also, determine if the totals are reasonable (i.e., there are six ways to roll a 7, so approximately one-sixth of all the rolls should be 7).

Answer 1

Code::

C++

#include <cstdlib>
#include <ctime>
#include <iostream>

using namespace std;

int main() {

/* initialize random seed: */
srand (time(NULL));
  
//to store sum between 2 to 12
int count[13]={0};
  
for (int start = 0; start < 35000; start++) {
/* generate secret number between 1 and 6: */
  
int firstdice,seconddice;
  
firstdice = rand() % 6 + 1;
  
seconddice= rand() % 6 + 1;
count[firstdice+seconddice]++;
}
  
for (int start=2; start<=12; start++)
{
  
cout<<"Sum "<<start<<" Occurs " <<count[start]<<"\n";
}
cout<<"\n";
/*
The frequencies are stored in count. and sum of them is definitely 36000.
when you want to show that they are reasonable you evaluate count[i]/36000. they must be close to the numbers I wrote in the previous reply.
count[2]/36000 ~ 1/36
count[3]/36000 ~ 2/36
.
.
.
count[11]/36000 ~ 2/36
count[12]/36000 ~ 1/36*/

for (int start=2; start<=12; start++)
{
  
cout<<"Freq "<<start<<" Occurs " <<float(count[start])/36000<<"\n";
  
  
}

 #include <cstdlib> #include <ctime> #include <iostream> using namespace std; int main() { /* initialize random seed: */ srand (time(NULL)); //to store sum between 2 to 12 int count[13]={0}; for (int start = 0; start < 35000; start++) { /* generate secret number between 1 and 6: */ int firstdice,seconddice; firstdice = rand() % 6 + 1; seconddice= rand() % 6 + 1; count[firstdice+seconddice]++; } for (int start=2; start<=12; start++) { cout<<"Sum "<<start<<" Occurs " <<count[start]<<"\n"; } cout<<"\n"; /* The frequencies are stored in count. and sum of them is definitely 36000. when you want to show that they are reasonable you evaluate count[i]/36000. they must be close to the numbers I wrote in the previous reply. count[2]/36000 ~ 1/36 count[3]/36000 ~ 2/36 . . . count[11]/36000 ~ 2/36 count[12]/36000 ~ 1/36*/ for (int start=2; start<=12; start++) { cout<<"Freq "<<start<<" Occurs " <<float(count[start])/36000<<"\n"; } } 

OUTPUT:

  
}

Result CPU Time: 0.00 sec(s), Memory: 3348 kilobyte(s) compiled and executed in 1.469 sec(s) Sum Sum 2 3 4 5 Sum 6 Sum Sum Sum Sum Sum Sum Sum 7 8 9 10 11 12 Occurs Occurs Occurs Occurs Occurs Occurs Occurs Occurs Occurs Occurs Occurs 945 1941 2867 3893 4917 5829 4853 3911 2937 1924 983 Sum Freq 2 Freq 3 Freq 4 Freq 5 Freq 6 Freq 7 Freq 8 Freq 9 Freq 10 Freq 11 Freq 12 Occurs Occurs Occurs Occurs Occurs Occurs Occurs Occurs Occurs Occurs Occurs 0.02625 0.0539167 0.0796389 0.108139 0.136583 0.161917 0.134806 0.108639 0.0815833 0.0534444 0.0273056