Dice and the Laws of Probability
by
Edward D. Collins
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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?
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. 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: 
When two dice are rolled, there are now 36 different and unique ways the dice can come up. This figure is arrived at by multiplying the number of ways the first die can come up (six) by the number of ways the second die can come up (six). 6 x 6 = 36. This graphic shows this very nicely. I've used two different colored dies to help show a roll of 21 is different from a roll of 12. 
If you use the above graphic and count the number of times is 6 appears when two dice are rolled, you will see the answer is eleven. Eleven times out of 36 or 30.5 %, slightly less than the 33.3% (2/6) Kent thought. When you roll two dice, you have a 30.5 % chance at least one 6 will appear. This figure can also be figured out mathematically, without the use of the graphic. One way to do so is to take the number of ways a single die will NOT show a 6 when rolled (five) and multiply this by the number of ways the second die will NOT show a 6 when rolled. (Also five.) 5 x 5 = 25. Subtract this from the total number of ways two dice can appear (36) and we have our answer...eleven. So, let's use this same method to answer our question and determine the chances of at least one 6 appearing when three dice are rolled. Take the chances of a 6 NOT appearing on the first die...
and multiply this by the chances of a 6 NOT appearing on the second die...
and multiply this by the chances of a 6 NOT appearing on the third die...
So, there are 125 out of 216 chances of a 6 NOT appearing when three dice are rolled. Simply subtract 125 from 216 which will give us the chances a 6 WILL appear when three dice are rolled, which is 91. 91 out of 216 or 42.1 %, not quite the 50% Kent originally thought. Here is a table showing the fractions and percentages of a 6 appearing (or any other single digit for that matter) and not appearing with several different numbers of dice: 



...with just one die 


























Notice you have just over a 50 % chance (51.77) of rolling at least one 6 when rolling four dice. With six dice it's just over 66 %. Also notice even when you roll nine dice, rolling a 6 is far from a certainty  you have just a little better than 80 % chance of rolling this 6. This means, on the average, even when you roll nine dice, two times out of ten you will not roll a single 6! Hope
this helped clear it up, Kent! 