Quandaries
and Queries 

If one has 3, 6 sided dice what is the probability of the numbers that are rolled to total 4 through 10 inclusively? Subsequent to this, what is the probability to do this consecutively...say
3 times? 

Hi, To work this out you need to consider (quite) a number of cases. Think of the dice as being Red, White and Blue. To get a total of 4 the pattern of the dice (R,W,B) must be (1,1,2) or (1,2,1) or (2,1,1) so this sum has probability of ^{3}/_{63} = ^{1}/_{72}. To get a sum of 5 you need to see how many ways it can happen as we did with 4: (1,1,3), (1,3,1), (3,1,1), (1,2,2),(2,1,2), (2,2,1) for a ^{6}/_{63} chance. And so on for larger sums. It is a daunting task and requires some patience to do all of the cases this way. One can get some help using the Principle of Inclusion and Exclusion but his may not be what you're after. For example, if you want to find the number of ways to get a sum of 10, it is the number of solutions to r + w + b = 10 where r,w, and b are integers between 1 and 6; this translates to solving r + w + b = 7 where r, w and b are non negative integers each at most 5. To solve r + w + b = 7 where r, w and b are non negative integers is a standard combinatorics problem and the answer is 9C2 = 36 (9 choose 2) but you now have to exclude form these the solutions that have r, w or b > 5. If one die was 5 (3 ways) the other two have to total 2 (3 ways) and if one die was 6 (3 ways) the other two have to total 1 (2 ways); 15 ways in all, so (please check my arithmetic) it looks like 21 ways to get a total of 10. As for doing it say 3 times in a row, these are independent trials so you need only to cube your anwser for one toss of the three dice. Penny 

