In the game of Craps, a "Pass Line" bet proceeds as follows. Using two six-sided dice, the first roll of the 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, 6, 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 the game of Craps using the rules stated above so that it simulates a game without human input. Instead of asking for a wager, the program should just calculate if the player would win or lose. The program should simulate rolling the two dice and calculate 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 probability of winning, i.e. Wins / (Wins + Losses) and output this value. Over the long run, who is going to win the most games of Craps, you or the house?
Please help.. i am having trouble figuring this one out
int x=(int)(Math.random()*6);
int y=(int)(Math.random()*6);
int roll = x+y;
for (int i = 1; i<=10000; i++)
if (roll == 7 || roll == 11)
numWins = numWins +1;
else if (roll==2 || roll==3 || roll==12)
numLosses = numLosses + 1;
else
{
x=(int)(Math.random()*6);
y=(int)(Math.random()*6);
}
int point = x+y;
while (roll !=7 || roll !=11)
{
x=(int)(Math.random()*6);
y=(int)(Math.random()*6);
}
if (roll==point)
numWins = numWins+1;
else numLosses=numLosses+1;