c++ Problem A: Dice game (20 points) Suppose we are playing a 2-dice rolling gam

Question

c++

Problem A: Dice game

(20 points)

Suppose we are playing a 2-dice rolling game with the following rules:

Start with 0 points.

Roll 7 = add one point and roll again

Roll anything else (not 7) = add that many points to the current amount and finish the game

The “Roll 7” rules can happen multiple times. For example, if they rolled 7, then 7, then 9... they

would get 1+1+9 = 11 points. Figure out the distribution of points for this game by simulating it

happening 1,000,000 times then display the results. Do not show the probability of any values you

never saw.

Example 1:

Chance of rolling 2 is 2.7915%

Chance of rolling 3 is 6.0461%

Chance of rolling 4 is 9.3637%

Chance of rolling 5 is 12.7405%

Chance of rolling 6 is 15.9513%

Chance of rolling 7 is 2.6614%

Chance of rolling 8 is 14.2741%

Chance of rolling 9 is 13.4987%

Chance of rolling 10 is 10.5849%

Chance of rolling 11 is 7.2984%

Chance of rolling 12 is 3.9861%

Chance of rolling 13 is 0.6729%

Chance of rolling 14 is 0.1081%

Chance of rolling 15 is 0.0182%

Chance of rolling 16 is 0.0036%

Chance of rolling 17 is 0.0004%

Chance of rolling 19 is 0.0001%

pls give a correct answer

Explanation / Answer

/** c++ code for playing playing a 2-dice rolling game ****/

#include <iostream>
#include <vector>
#include <stdio.h>
#include <stdlib.h>
#include <fstream>
#include <iomanip>
#include <vector>

using namespace std;

int main()
{
srand (time(NULL));
int dice1;
int dice2;
int sum = 0;
int i =0;
int d;
double a[1000000]= {0};
int points =0;
while(i < 1000000)
{

dice1 = rand() % 6 + 1;
dice2 = rand() % 6 + 1;
sum = dice1 + dice2;
if( sum == 7) points++;
else
{
d = points+ sum;
a[d]++;
points = 0;
}
  
i++;
}

for (int i = 0; i < 1000000; ++i)
{
if(a[i] != 0) a[i] = a[i] / 1000000;
}

for (int i = 2; i < 1000000; ++i)
{
if(a[i] != 0) cout << "Chance of rolling " << i << " is "<< a[i]<< endl;
}
return 0;
}

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