If I have cards numbered 1 through 40 and I randomly draw 3 then replace them, shuffle them and draw three again what are the odds of drawing one of the numbers more than once
The first three cards are irrelevant.
For the second draw, the
p=P(repeat at least once)
= 1-P(no repeat at all)
=1 - (37*36*35)/(40*39*38)
=211/988
Odds: p : (1-p) = 211/988: 777/988
= 211 : 777