Why 7 owns the two-dice table
Two dice produce 36 equally likely pairs, and the sums are not created equal. A sum of 2 needs exactly one pair (1,1); a sum of 7 can be built six ways (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Probability is counting with the total as denominator, so P(7) = 6/36 = 0.1667, the peak of the distribution. The counts rise 1, 2, 3, 4, 5, 6 and fall symmetrically: a triangle. Add a third die and the triangle softens into the beginnings of a bell, the central limit theorem assembling itself one die at a time.
Two dice: P(sum ≤ 4) counts 1 + 2 + 3 = 6 pairs, so 0.1667, the same as P(sum ≥ 10) by symmetry. Three dice: 216 outcomes, mean 10.5, and P(exactly 10) = 27/216 = 0.1250, while a low total like P(sum ≤ 5) is only 10/216 = 0.0463. Craps, Monopoly and backgammon strategy all reduce to reading tables like the one the tool prints.
Means and variances of sums
One die averages (1+2+3+4+5+6)/6 = 3.5 with variance 35/12 ≈ 2.9167. Independent dice add both quantities: n dice average 3.5n with variance 35n/12, so the standard deviation grows like √n while the mean grows like n. That mismatch, averages steadying as counts grow, is the law of large numbers in miniature, and it is why casino games priced on single rolls behave so predictably over a night.
Assuming the 11 sums of two dice are equally likely (probability 1/11 each). They are not: the pairs are equally likely, the sums inherit their counts. This is the single most common dice error, and it silently breaks every game calculation built on it.
Frequently asked questions
What is the most likely sum of two dice?
7, with probability 6/36 = 0.1667. Six of the 36 pairs build it, more than any other sum.
What is the probability of rolling doubles?
6 pairs out of 36, so 0.1667 with two dice: the same headline number as a sum of 7, by coincidence of counts.
How does the distribution change with three dice?
216 outcomes, sums from 3 to 18, mean 10.5, and a shape already close to a bell curve: the centre sums (10 and 11) take 27/216 each.