Let's imagine you are playing a game which uses dice. You are about to roll three of them. You NEED to roll at least one 6. A 6 appearing on any one (or more) of the three dice will win the game for you! What are your chances? 33.3 %

42.1 %

50 %

66.6 % Quite some time ago, I was over at a friend's house watching him and another friend play a board game called Axis & Allies. At one point this exact scenario came up - Kent was planning on rolling three dice and really wanted at least one 6 to appear. He made a comment that with three dice, his chances were 3 / 6 or 50%. Kent's reasoning was, with one die, the chances of rolling a 6 were 1 / 6 which is correct. He also believed if he were to roll two dice, his chances were double this or 2 / 6. This is INCORRECT and this is where his faulty reasoning begins. Knowing a little bit about the laws of probability, I quickly knew the fraction "2 / 6" for two dice and "3/6" for three dice was incorrect and spent a brief moment computing and then explaining the true percentages.



Unfortunately, I do not believe I was successful in explaining to Kent why my figures were correct. Maybe I can do so here. Obviously, with Kent's logic above, if the chances of rolling a 6 with two dice is 2 / 6 and the chances of rolling a 6 with three dice is 3/6, then the chances of rolling a 6 with six dice would be 6 / 6 !! 100%?? Of course, this is obviously incorrect. I don't care how many dice you roll, the chances of rolling a 6 will never be 100%. When you roll just one die, there are six different ways the die can land, as shown by the following graphic: