Way back in high school, I used to do this competition where they would occasionally ask simple probability questions, and one of the most common questions to ask was about dice rolls. I would usually quickly scratch out a table that looked like this:
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
The numbers in bold represent the number showing on each particular die, and of course the table lists out sums. It’s helpful to know that there are 36 possible sums (this is just \(6^2\)), and to have a decent grasp of how to simplify fractions with 36 in the denominator.
Once you have these basics down though, it’s really easy to calculate the chances of any particular roll of two 6-sided dice. How likely is rolling a 7? I count six sevens in the table, and \(6 / 36 = 1 / 6\). What are the chances of rolling a 10 or higher? Again: six numbers 10 or higher in the table, so the chances are \(1 / 6\).
The questions never got this strange, luckily, but I guess you could also figure out the chances of, say, rolling a prime number. The primes here are 2, 3, 5, 7, and 11. Just find the length of the list \([2, 3, 3, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 11, 11]\) and divide by 36. (It’s \(15 / 36 = 5 / 12\).)
This little table might also be useful in some board games or making bets with friends. Sometimes it’s just convenient to have a prepared model of something ready to use whenever it’s needed.