In the game of craps, a \"Pass Line\" bet proceeds as follows. The first roll of

Question

In the game of craps, a "Pass Line" bet proceeds as follows. The first roll of the two, six-sided dice in a craps round is called the "come out roll." The bet immediately wins when the come out roll is 7 or 11, and loses when the come out roll is 2, 3, or 12. If 4, 5, 5, 8, 9, or 10 is rolled on the come out roll, that number becomes "the point." The player keeps rolling the dice until either 7 or the point is rolled. If the point is rolled first, then the player wins the bet. If the player rolls a 7 first, then the player loses.

Write a program that plays craps using those rules so that it simulates a game without human input. Instead of asking for a wager, the program should calculate whether the player would win or lose. Create a function that simulates rolling the two dice and returns the sum. Add a loop so that the program plays 10,000 games. Add counters that count how many times the player wins, and how many times the player loses. At the end of the 10,000 games, compute the probablity of winning, as (Wins + Loses), and output this value. Over the long run, who is going to win more games of craps, you or the house? Comments must be included in your code.

Explanation / Answer

#include #include #include using namespace std; int main() { int lose = 0; int win = 0; srand(time(NULL)); int i; for (i = 1; i < 101; i++) { int roll = rand()%11 + 1; if(roll == 7 || roll == 11) { cout

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.