Which Total Comes Up Most with Two Dice?

One die first

Start smaller. Roll one die. A fair die has no favorite face: each comes up about one time in six. Roll it 60 times and each face comes up about 10 times, though not exactly: one face might get 7 and another 13.

So it is natural to think two dice are the same: eleven totals, 2 to 12, each about one time in eleven. The widget says otherwise. After a few hundred rolls the bars stand in a rough triangle, tall in the middle and short at the ends, and the more you roll, the neater it gets.

A hundred rolls

In 100 rolls, 7 usually comes up somewhere near 17 times and 12 about 3 times. But press Start again and roll 100 a few times: sometimes 6 or 8 beats 7, and now and then 12 doesn’t come up at all.

How sure can you be? Our script played 100 rolls 100,000 times. After 100 rolls, 7 was on top by itself only about 46 times in 100, because 6 and 8 often caught it. But 12 came up as often as 7 in only about 4 runs in 10,000. So a hundred rolls is plenty to show that 7 beats 12, but not to be sure it beats 6 and 8. After 1,000 rolls, 7 was alone on top about 9 times in 10.

Counting the ways

Why should 7 be the favorite? Riya starts a list. “12 is 6 + 6, the only way. 7 can be 1 + 6, 2 + 5 or 3 + 4: three ways. So 7 is three times as likely as 12.” A good start, but the widget shows 7 coming up about six times as often as 12. What has she missed?

Imagine one die is red and the other blue. Red 1 with blue 6 is a different landing from red 6 with blue 1: the same total, but two different things the dice can do. Ordinary dice look alike, but they are still two separate dice, each landing on its own face; painting one red just lets you see it. So list every way the pair can land, the red die’s face and then the blue die’s.

Each of the red die’s 6 faces can go with each of the blue die’s 6: 6 × 6 = 36 ways in all (the multiplying of How Many Number Plates Can a Province Make?). Now count the ways for each total.

Riya’s three ways for 7 become six once 1 + 6 and 6 + 1 count as two. But 6 + 6 is still one way: swapping the dice gives the same landing. So 7 beats 12 six to one, not three.

Outcomes, events and probability

Now the proper words. Each of the 36 ways the dice can land (red 3 and blue 5, say) is an outcome, and the list of them all is the sample space. What you are waiting for, “a total of 7”, is an event: a set of outcomes, here the six on the slanting line through the middle of the table.

The 36 outcomes are equally likely: none comes up more often than another. When outcomes are equally likely, the probability of an event, how likely it is, is

$$\text{probability} = \frac{\text{number of outcomes in the event}}{\text{number of outcomes in all}}$$

So the probability of 7 is $\frac{6}{36} = \frac{1}{6}$, and of 12, $\frac{1}{36}$.

What does “the probability of 7 is 1/6” mean? Not that every sixth roll is a 7: the rolls don’t take turns. Over many rolls, 7 comes up about one roll in six. A probability of 0 means it can never happen (Ashok’s 1); 1 means it always happens (a total from 2 to 12). The last row is the expected number, the rolls times the probability: 100 × 6/36 ≈ 17 sevens. It is what to expect on average, not a promise.

Why the 36 are equally likely

The two dice don’t affect each other: whatever the red die does, the blue die still lands on each face about one time in six. Events like these, where one tells you nothing about the other, are independent. The red die lands on 3 about one roll in 6, and in those rolls the blue die lands on 5 about one time in 6, so red 3 with blue 5 comes up about one roll in 6 × 6 = 36. The same goes for every pair: each of the 36 has probability $\frac{1}{36}$. (As a tree diagram, the red die is 6 branches, each splitting into 6 for the blue: 36 ends, each $\frac{1}{6} \times \frac{1}{6}$, as Grade 10’s tree diagrams multiply along the branches.)

Why the totals make a triangle

In the table, squares with the same total lie on a slanting line, and the line for 7 runs from corner to corner, one square in each row. The line for 8 is one shorter, since a red 1 would need a blue 7. Each step away from 7, either way, the line loses a square, until 2 and 12 are single squares in opposite corners. That is the triangle.

There is a neat reason why its two sides match. On an ordinary die, opposite faces add up to 7: 1 is opposite 6, 2 opposite 5, 3 opposite 4. Turn both dice upside down after a roll: each face f becomes 7 − f, so a total t becomes 14 − t. A roll of 6 becomes an 8, a 12 becomes a 2. Upside-down rolls are just as likely as the right way up, so 6 and 8 are equally likely, and so are 2 and 12.

Dice are very old. Terracotta dice have been dug up at Mohenjo-daro, a city of the Indus Valley civilization, from about 2600–1900 BCE. Most Indus dice are numbered in other ways, but at least one, the die in the photo, already has its opposite faces adding to seven, like modern dice. People in South Asia rolled dice for thousands of years before anyone counted the 36 outcomes.

Rolling against counting

There are two ways to find how often 7 comes up. Roll and count: 172 sevens in 1,000 rolls is a relative frequency of 17.2%, sometimes called the experimental probability. Or count outcomes: 6 of 36, 16.7%, the theoretical probability. How do they compare? Our script worked out the range the number of 7s falls in, 19 runs in 20:

The number of 7s wanders further from the expected one as you roll more, by up to about 7 in 100 rolls but about 73 in 10,000. Yet the percent closes in on 16.7%. This is the law of large numbers: the more times a chance event is repeated, the closer its relative frequency comes to its probability. It is why a casino playing millions of games can be almost sure of its takings, and a player playing ten can’t be sure of anything.

Shortcuts: “not” and “or”

Not. The outcomes that aren’t a 7 are the other 36 − 6 = 30, so the probability of not rolling 7 is $\frac{30}{36} = \frac{5}{6}$, which is 1 − $\frac{1}{6}$. The probability that something doesn’t happen is 1 minus the probability that it does.

Or. “A 7 or an 11” is the 6 outcomes for 7 and the 2 for 11: $\frac{8}{36}$. Adding works because no roll is both: the two events are mutually exclusive. When two events can happen together, don’t count the shared outcomes twice: “a double or a total of 8” is 6 doubles and 5 eights, but 4 + 4 is in both, so 6 + 5 − 1 = 10 outcomes.

The difference of two dice

Now take the smaller number from the bigger. The difference can be 0 to 5.

The lit squares lie on slanting lines the other way now. A difference of 0 is one diagonal, six squares, but every other difference has two lines, one on each side of it: 1 has two lines of 5. That is why 1 wins.

Galileo’s three dice

About 400 years ago, someone asked the scientist Galileo Galilei to explain a puzzle from Italian dice players, who played with three dice. (Galileo calls him only “him who has ordered me”, and doesn’t say who he was.) A total of 9 can be made from three numbers in six ways: 1 + 2 + 6, 1 + 3 + 5, 1 + 4 + 4, 2 + 2 + 5, 2 + 3 + 4 and 3 + 3 + 3. So can 10: 1 + 3 + 6, 1 + 4 + 5, 2 + 2 + 6, 2 + 3 + 5, 2 + 4 + 4 and 3 + 3 + 4. Yet long experience at the tables said 10 came up a little more often than 9, and 11 more often than 12. Were they imagining it? Roll, into the thousands.

A few hundred rolls won’t tell them apart. In our script, 10 was ahead after 100 rolls in only 53 runs of 100, after 1,000 rolls in 71, and after 10,000 rolls in 97. The gamblers must have watched an enormous number of games.

Galileo did what you did with the red and blue dice: he counted every way the three dice can land, 6 × 6 × 6 = 216 equally likely outcomes. The six sums are not equally likely, because they come in different numbers of orders:

• three different numbers, like 1 + 2 + 6, in 6 orders (any of the three on the red die, either of the other two on the blue, the last on the green: 3 × 2 × 1);
• two the same, like 1 + 4 + 4, in 3 (the odd one out on the red, blue or green die);
• all the same, 3 + 3 + 3, in 1.

So 10 comes up in 27 of the 216 outcomes and 9 in 25: the gamblers were right, by a whisker. The difference is the 3 + 3 + 3 that 9 needs, which can only be rolled one way.

Where the counting goes wrong

The rule “count the equally likely outcomes” is simple; being sure they really are equally likely is the hard part, and great minds have slipped. Gottfried Wilhelm Leibniz, who invented calculus at about the same time as Newton, wrote that with two dice a total of 11 is as likely as 12, since each can be made in only one way, 5 + 6 and 6 + 6. You can see his slip: 5 + 6 and 6 + 5 are two outcomes, so 11 is twice as likely.

Jean d’Alembert wrote in 1754 that if you toss a coin twice, the chance of at least one head is 2/3: a head first (and you needn’t toss again), a tail then a head, or two tails, and two of the three have a head. But “a head first” is half of all games by itself. List both tosses, like the red and blue dice: HH, HT, TH, TT, four equally likely outcomes, three with a head. The answer is 3/4.

When the dice aren’t fair

Everything so far assumes fair dice, each face exactly one time in six. Real dice are never quite perfect. In 1894 the English biologist Walter Weldon threw 12 dice at once 26,306 times and counted the 5s and 6s. For fair dice, 5 or 6 should come up one time in three, 33.33%; his gave 33.77%. That sounds like nothing, but over 315,672 dice it was far too much to be luck: his dice favored 5 and 6 a little. In 2009 Zacariah Labby built a machine that rolled and photographed 12 dice 26,306 times. His dice showed 1 and 6 slightly more often (16.86% and 16.88% instead of 16.67%), and when he measured them, they were about 0.2% shorter between the 1 and the 6 than across the other faces.

So 1/36 for a double six is a model of perfect dice. It fits ordinary dice so well that it takes tens of thousands of throws to catch them out; when rolls and model disagree by more than luck can explain, as Weldon’s did, the dice are not quite fair.

More dice, a rounder hill

One die gives six equal bars. Two dice give a triangle, peaked at 7. Three give a hill with a rounded top at 10 and 11; four, a rounder hill peaked at 14 (146 of the 1,296 outcomes). The more dice you add, the more the shape becomes a smooth bell, as at Science Playground’s Galton Board, where each ball bounces left or right at random at every peg: there are far more ways to end up in the middle than at the edges.

• Is This Game Fair? A fairground game that looks generous, played a thousand times on paper to see who really wins.
• How Many Ways Can You Walk to School? Counting routes, and a triangle of numbers that also says where the Galton Board’s balls land.
• Should You Switch Doors? (Grades 9–12) A game-show puzzle where counting the equally likely outcomes beats almost everyone’s instinct.
• What Is a Game Worth on Average? (Grades 9–12) Expected numbers turned into prizes.
• How Can Throwing Rice Find π? (Computing, Going Further 6–8) Chance put to work: rice thrown at random on a square drawing finds a famous number.
• In the grade books: Grade 10’s probability, with the addition and multiplication laws behind “or” and the red-and-blue table.