Eric Morton DiceRolling-Lab

src/Answers-.md

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+,## Dice Rolling Lab 
+### Eric Morton 
+### Feb 3 , 2015 
+
+
+**Question 1 :** There is a 1/6 chance of rolling a 7, and a 1/36 chance of rolling a 2. 
+
+**Question 2:** Once 100 rolls were rolled, I believe that 100 rolls is not sufficient enough to determine the relative frequency of all the possible rolls becuase though 100 is a large number, it does not give an accuracte frequency for all of the posssible rolls. 
+
+**Question 3:** By increasing the number to 500, there is more of a chance for the coder to see if they can accuratley predict the chances of rolling their die. 
+
+**Question 4:** It would take approximately 400,000 rolls before we can predict the porbablity of rolling a 7 or a 3. For finding the proabality of a 2, that would take 200,000 rolls. 

src/Histogram.java

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+public class Histogram {
+ private int[] values = new int[13];
+ private int count = 0;
+  int q = 0;
+
+  public void tally(int roll) {
+   values[roll] +=1; 
+     count +=1;
+ }
+  public int getCount(int roll) {
+    return values[roll];
+  }
+  public double getRatio(int roll) {
+    return values[roll] / (double) count;
+  }
+  public void print() {
+    int val = 2; 
+    while(val <=12) {
+      System.out.println(val + "/" + getCount(val) + "*"+ getRatio(val));
+       val+= 1; 
+    }
+  }
+}

src/Main.java

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+public class Main 
+{
+  public static void main(String [] args)
+  {
+    Histogram h = new Histogram();
+    int q = 0;
+    int total = 0; 
+
+    Die d1 = new Die();
+    Die d2 = new Die();
+
+    while (q != 100) {  
+
+    d1.roll();
+    d2.roll(); 
+
+    total = d1.getUpValue() + d2.getUpValue();
+
+    h.tally(total);
+    h.print();
+    q += 1;
+
+    }
+  }
+}
+
+