Dice rolls: Counting number of rolls that equals 6 or 7

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carl123
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The following calculates the number of times the sum of two dice (randomly rolled) equals six or seven.

Code:
#include <iostream>
#include <cstdlib>

using namespace std;

int main(){ 
int i = 0; // Loop counter iterates numRolls times
int numRolls = 0; // User defined number of rolls
int numSixes = 0; // Tracks number of 6s found
int numSevens = 0; // Tracks number of 7s found
int die1 = 0; // Dice values
int die2 = 0; // Dice values
int rollTotal = 0; // Sum of dice values

cout << "Enter number of rolls: " << endl;
cin >> numRolls;

srand(time(0));

if (numRolls >= 1) {
// Roll dice numRoll times
for (i = 0; i < numRolls; ++i) {
die1 = rand() % 6 + 1;
die2 = rand() % 6 + 1;
rollTotal = die1 + die2;

// Count number of sixs and sevens
if (rollTotal == 6) {
numSixes = numSixes + 1;
}
else if (rollTotal == 7) {
numSevens = numSevens + 1;
}
cout << endl << "Roll " << (i + 1) << " is "
<< rollTotal << " (" << die1
<< "+" << die2 << ")";
}

// Print statistics on dice rolls
cout << endl << endl;
cout << "Dice roll statistics:" << endl;
cout << "6s: " << numSixes << endl;
cout << "7s: " << numSevens << endl;
}
else {
cout << "Invalid rolls. Try again." << endl;
}

return 0;
}

QUESTION

Create different versions of the program that:

1) Calculates the number of times the sum of the randomly rolled dice equals each possible value from 2 to 12.

2) Repeatedly asks the user for the number of times to roll the dice, quitting only when the user-entered number is less than 1. Hint: Use a while loop that will execute as long as numRolls is greater than 1. Be sure to initialize numRolls correctly.

3) Prints a histogram in which the total number of times the dice rolls equals each possible value is displayed by printing a character like * that number of times, as shown below.

Dice roll histogram:

2: ******
3: ****
4: ***
5: ********
6: *******************
7: *************
8: *************
9: **************
10: ***********
11: *****
12: ****
 
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Can you show what you'e tried? It's hard to know how to help you when you merely post problems with no effort shown. Our goal is to guide, not do your work. Please add your work to each thread you've posted where you have not already done so, and you will find people will be much more likely to help. :)