Odds of winning lottery

Question

I tried to work out the odds of winning the UK National Lottery last night and got myself into a muddle. Please put me out of my mysery someone!
public class WinTheLottery{
double dd;
java.text.DecimalFormat df;
public static void main(String[] args){
new WinTheLottery();
WinTheLottery(){
df = new java.text.DecimalFormat("#,###");
int factor = 6;
odds(factor);
dd = 1/dd;
     System.out.print(dd);
public void odds(int factor){
if(factor==1){
double tmp = factor/(43+factor);
if(dd==0) dd+=tmp;
else dd *=tmp;
odds(factor-1);
}

Answer 1

>
The Irish lotto has 42 numbers, and you need 6 to
win.
So the odds of winning are
( 42 * 41 * 40 * 39 * 38 * 36 ) / ( 6 * 5 * 4 * 3 * 2
* 1 )
cheers,
OwenI'm not convinced by the math there: eg;-public class WinTheLottery{
   double dd;
   public static void main(String[] args){
      new WinTheLottery();
   WinTheLottery(){
      java.text.DecimalFormat df  = new java.text.DecimalFormat("#,###.#");
      double factor = 6;
      odds(factor);     
      dd  = 1/dd;
      System.out.println("AlphaNumerics's method: "+df.format(dd)+" to 1 against");
      dd = (49 * 48 * 47 * 46 * 45 * 44)/ ( 6 * 5 * 4 * 3 * 2 * 1 );
      System.out.println("Owen's Method: "+df.format(dd)+" to 1 against");
   public void odds(double factor){
      if(factor!=0){
         double tmp = factor/(43+factor);
         if(dd==0) dd+=tmp;
         else dd *=tmp;
         odds(factor-1);
}Result
AlphaNumerics's method: 13,983,816 to 1 against
Owen's Method: 2,053,351 to 1 against
I know for sure its nearer 14 million to 1 rather than 2. In fact I think I've heard the figure of 12.5 million to 1 before, which suggests both figures could be out.