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Two dice and the sum of seven

Two dice can land in 36 equally likely ways, and 6 of them add up to 7, more than any other sum.

Bet on a sum of 2, 7 or 12: which would you pick?

Before reading on, try this. You roll two dice and add them, and you may bet on a sum of 2, 7 or 12. Which do you pick? And how many times as often do you expect your sum to come up as each of the others? Write your guess down. One honest guess is that all three are equally likely, since each is just one number between 2 and 12. Another is that 7 wins, but only by a little. Counting settles it, and the answer is not a little.

Board games that move by the sum of two dice feel as if 7 comes up all the time. It does, and counting shows why. Paint one die red and the other blue. The red die has 6 possible faces and so does the blue one, so the pair has 6 × 6 = 36 outcomes, from (1, 1) to (6, 6). Each of the 36 is equally likely. The sums are not: a sum of 2 needs (1, 1) and nothing else, while a sum of 7 can be (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) or (6, 1).

All 36 outcomes of two dice with their sums. Red numbers down the side are the red die, blue numbers across the top the blue die. The six orange cells on the diagonal are the ways to make 7.

Reading the grid by diagonals gives the number of ways for each sum: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for the sums 2 to 12. They add up to 36, as they must. So P(sum = 7) = 6/36 = 1/6, about 0.167, while P(sum = 2) and P(sum = 12) are only 1/36 each, about 0.028. That answers the opening question: 7 comes up six times as often as 2, and six times as often as 12. The distribution of the sum has a tent shape: it rises in equal steps to 7 and falls in equal steps after it.

Sum	Ways	Probability
2	1	1/36
3	2	2/36
4	3	3/36
5	4	4/36
6	5	5/36
7	6	6/36
8	5	5/36
9	4	4/36
10	3	3/36
11	2	2/36
12	1	1/36

Twenty-one unordered pairs are not equally likely

A common mistake is to say that 7 can be made from 1 and 6, 2 and 5, or 3 and 4, so it has three ways, and 2 has one way, out of 21 unordered pairs. Those 21 pairs are not equally likely. The pair "3 and 4" can happen as red 3, blue 4 or as red 4, blue 3, while "2 and 2" can happen only one way. Only the 36 ordered outcomes are equally likely, so they are the ones to count. Even if the two dice look identical, nature still treats them as two separate objects.

BackgroundEarly in the 1600s, Galileo was asked to explain something that dice players had noticed over many games with three dice. The sums 9 and 10 can each be made from six different sets of numbers, so why did players find that 10 came up more often than 9? Galileo's short essay, "On the investigations of dice", answers it exactly as we did with two dice. A set of three different numbers, such as 6, 3 and 1, can fall in six orders; a set with two equal numbers, such as 6, 2 and 2, in three orders; and three equal numbers in only one. Adding up the orders, he found that 10 can be thrown in 27 ways and 9 in only 25, out of 216.

Galileo's numbers give P(10) = 27/216 = 0.125 and P(9) = 25/216, about 0.116. The difference is about one throw in a hundred, so the gamblers must have played a lot to notice it. With three dice the shape of the sum is no longer a tent. It is rounder, with flat tops at 10 and 11, already looking like the bell curve you will meet on the Galton board.

Drag to turn. Scroll or pinch to zoom.

Bars show how often each sum has come up; the thin line across each bar is the exact probability from the 36 outcomes. Roll 10,000 and the bars line up with the tent. Switch to 3 dice to check Galileo's 27 and 25.

Try it: choose 3 dice, roll 10,000, and compare the bars at 9 and 10. The exact values are 25/216 ≈ 0.116 and 27/216 = 0.125, so with 10,000 rolls the gap is usually visible but small.

Outcomes for two dice	6 × 6 = 36 ordered pairs
P(sum = 7)	6/36 = 1/6
P(sum = 2)	1/36
Three dice: ways to make 10 and 9	27 and 25 out of 216 (Galileo)

Two dice. What is P(sum = 7)?
Two dice. What is P(sum = 2)?
Two dice. In how many ways can the sum be 8?
Two dice. What is P(sum is 10 or more)?
Two dice. What is P(doubles)?
How many equally likely outcomes do three dice have?
Three dice. In how many ways can the sum be 10?
Three dice. What is P(sum = 10) as a decimal?
Why can you not count the 21 unordered pairs as equally likely?
- Dice are never fair
- A pair of different numbers can happen in two orders, a double in one
- There are really 30 pairs
- Sums are all equally likely
What is the average sum of two fair dice over many rolls?

Common mistakes

"There are 11 possible sums, 2 to 12, so each has probability 1/11, and 2 and 12 are as likely as 7." The 11 sums are not equally likely; the 36 ordered outcomes are. A sum of 7 has 6 of them (6/36), while 2 and 12 have one each (1/36). So 7 is six times as likely as either, and 1/11 (about 0.091) is wrong for every sum.
"11 needs a 5 and a 6, and 12 needs two 6s, so 11 and 12 are equally likely." 11 comes as (5, 6) or as (6, 5), two outcomes, so P(11) = 2/36. 12 comes only as (6, 6), so P(12) = 1/36. 11 is twice as likely.
"Roll two dice 36 times and 7 will come up 6 times." 6 out of 36 is the long-run share, not a schedule. In 36 rolls you might see 7 three times or nine times. Only over many rolls does the share of sevens settle near 1/6.

Teacher's noteThis is the first lesson where ordered and unordered counting give different answers. Use two dice of different colours, or colour them, and build the 6 × 6 table together before anyone touches a formula. Filling it in by hand makes the six sevens on the diagonal visible. The Korean middle-school curriculum keeps counting to adding or multiplying two counts and leaves permutations and combinations to high school, so the table is the right tool at this stage. For three dice, 216 cells are too many for one table; showing how Galileo counted the orders of each set of numbers is enough.

Key points

Two dice: 36 equally likely ordered outcomes.
Ways for sums 2 to 12: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1. P(7) = 1/6.
Count ordered outcomes, even for identical-looking dice.
Three dice: 216 outcomes; 10 in 27 ways, 9 in 25 (Galileo).

Sources for this course (9)

Curriculum: 교육부 고시 제2022-33호 [별책 8] 수학과 교육과정 (ncic.re.kr/inv/org/download.do?year=2022&seq=10003559&orgType=ogi4, PDF read 2026-10-19: 초등학교 5~6학년군 [6수04-04]~[6수04-06] 가능성을 말과 0, 1/2, 1 같은 수로 나타내기; 중학교 1~3학년군 [9수04-03] 상대도수, [9수04-05] 경우의 수(두 경우의 수를 더하거나 곱하는 정도), [9수04-06] 확률(상대도수로서의 의미와 경우의 수의 비율로서의 의미, 각 경우의 가능성이 같다는 가정), 용어 상대도수·사건·확률; 공통수학1 [10공수1-03-01]~[10공수1-03-03] 합의 법칙·곱의 법칙·순열·조합, 기호 nCr; 확률과 통계 [12확통01-03] 이항정리, 용어 이항계수·파스칼의 삼각형; [12확통02-01]~[12확통02-06] 확률의 기본 성질·덧셈정리·여사건의 확률·조건부확률·독립과 종속·곱셈정리, 용어 시행·통계적 확률·수학적 확률·여사건·배반사건·조건부확률·종속·독립·독립시행과 기호 P(A), P(B|A); [12확통03-03] 이항분포, [12확통03-04] 정규분포와 이항분포의 관계, [12확통03-05]·[12확통03-06] 임의추출·표본평균과 모평균, 용어 큰 수의 법칙·이항분포·정규분포·모평균·표본평균과 기호 B(n, p), N(m, σ²), X̄)
Textbook: OpenStax, Introductory Statistics 2e by Barbara Illowsky and Susan Dean, 3.1 terminology, equally likely outcomes, dime and nickel, law of large numbers, Belgian one euro coin 56% heads in 250 trials, conditional probability, Table 3.1 handedness of 100 people (openstax.org/books/introductory-statistics-2e/pages/3-1-terminology)
4.3 binomial distribution
7.1 central limit theorem for sample means, example mean 90, SD 15, n = 25, standard error 3 (.../pages/7-1-the-central-limit-theorem-for-sample-means-averages)
Yut rules for the Korean example only: 국립민속박물관 한국민속대백과사전 「윷놀이」, 도·개·걸·윷·모 by how many of the four sticks land 뒤집어진 (folkency.nfm.go.kr/kr/topic/detail/1595, read 2026-10-19)
Kerrich story: John Kerrich, 10,000 tosses, 5,067 heads (JMP case study JMP014 by DeWayne Derryberry, jmp.com/en/academic/case-study-library/is-a-coin-fair, citing Baldi and Moore; Wikipedia "John Edmund Kerrich")
Galileo story: Sopra le scoperte dei dadi, sum 10 in 27 ways and 9 in 25 ways with three dice (English translation, University of York history of statistics, york.ac.uk/depts/maths/histstat/galileo.pdf)
Pascal story: Pascal to Fermat, 29 July 1654, one six in four throws "as 671 to 625" (American Physical Society, aps.org/apsnews/2009/07/pascal-letters-fermat-points; Gaston Sanchez, De Mere's games; Statistical Science 2024 "Antoine Gombaud, Chevalier de Méré")
Galton story: Francis Galton, Natural Inheritance of 1889, the quincunx (Wikipedia "Galton board")

Checked: 2026-10-17