Throw 100 throws calculate the number of occurrences of each number C # source case attached to the second simple solution

 
Console.WriteLine ("the experiment was thrown 100 times:");//Prompt Information
Random Randomnum =NewRandom ();//Create a random number
            intNUM1 =0;//define the number of occurrences of 1
            intnum2 =0;//define the number of occurrences of 2
            intNUM3 =0;//define the number of occurrences of 3
            intNUM4 =0;//define the number of occurrences of 4
            intNUM5 =0;//define the number of occurrences of 5
            intNUM6 =0;//define the number of occurrences of 6


             for(inti =1; I <= -; i++)//For statement Loop
{
                intnum = Randomnum.next (1,7);//define a variable to accept random numbers

                Switch(num)
{
                     Case 1:
{
num1++;
                             Break;
}
                        
                     Case 2:
{
num2++;
                             Break;
}
                     Case 3:
{
num3++;
                             Break;
}
                     Case 4:
{
num4++;
                             Break;
}

                     Case 5:
{
num5++;
                             Break;
}

                     Case 6:
{
num6++;
                             Break;

}
                    default:
                         Break;

}

Console.WriteLine ("the number of {0} times is: {1}", I, NUM);
}


Console.WriteLine ("1 has occurred {0} times", NUM1);
Console.WriteLine ("2 has occurred {0} times", num2);
Console.WriteLine ("3 has occurred {0} times", num3);
Console.WriteLine ("4 has occurred {0} times", NUM4);
Console.WriteLine ("5 has occurred {0} times", NUM5);
Console.WriteLine ("6 has occurred {0} times", NUM6);

            //verify whether it is 100 times
            intsum = num1 + num2 + num3 + num4 + num5 + num6;
Console.WriteLine (sum);




The second method of solution:

namespace_2011._12._3
{
    classProgram
{
        Static voidMain (string[] args)
{
            int[] Count =New int[7] ;
Random Randomnum =NewRandom ();
            intnum =0;
             for(inti =0; I < -; i++)
{
num = Randomnum.next (1,7);
Count[num] + +;
}

Console.WriteLine ("1 occurrences are: {0}", count[1]);

Console.WriteLine ("2 occurrences are: {0}", count[2]);

Console.WriteLine ("3 occurrences are: {0}", count[3]);

Console.WriteLine ("4 occurrences are: {0}", count[4]);

Console.WriteLine ("5 occurrences are: {0}", count[5]);

Console.WriteLine ("6 occurrences are: {0}", count[6]);


Console.WriteLine ("altogether threw {0} times! ", count[1]+count[2]+count[3]+count[4]+count[5]+count[6]);
}
}
}

Throw 100 throws calculate the number of occurrences of each number C # source case attached to the second simple solution