Dice probability calculator

How to use the calculator

At the top you pick three things: how many dice, which type (d4, d6, d8, d10, d12 or d20) and what you want to work out. The notation NdX means N dice with X faces: 2d6 is two ordinary six-sided dice and 1d20 is a single twenty-sided die. There are four targets:

The result is the exact probability, worked out from every possible combination, plus a chart of the whole distribution of totals so you can see which values are common and which are rare. If you want to check it by rolling, press Roll or drag the dice and drop them in the tray: the 3D dice use real physics, so each roll is independent of the last. Bear in mind that over ten or twenty rolls the results can stray a long way from the theoretical percentage; only over hundreds of rolls does the observed frequency settle close to the probability.

The basics: sample space, events and equally likely outcomes

Rolling a die is a random experiment: you know what can happen, but not what will happen. The set of all possible outcomes is the sample space. For one six-sided die it is {1, 2, 3, 4, 5, 6}. With two dice it helps to tell them apart (picture one red and one blue), and the sample space then has 6 × 6 = 36 ordered pairs, from (1,1) to (6,6). This is the grid you draw as a sample space diagram at school. With three dice there are 6³ = 216 outcomes, and in general NdX has Xᴺ.

An event is any set of outcomes you care about: "rolling an even number" is {2, 4, 6}; "two dice totalling 4" is {(1,3), (2,2), (3,1)}. When every outcome is equally likely, as with a fair die, the classical definition (often credited to Laplace) applies:

P(A) = number of favourable outcomes / total number of outcomes

So the probability of an even number on one die is 3/6 = 1/2, and the probability of two dice totalling 4 is 3/36 = 1/12. Two more ideas solve nearly every dice question:

• The complement: P(not A) = 1 − P(A). It is the short route for any "at least one" question.
• Independent events: what one die shows has no effect on another die or on the next roll, so probabilities multiply. Two sixes in a row: 1/6 × 1/6 = 1/36.

The most common slip when starting out is counting combinations without order. A total of 4 with two dice can come from 1+3 or 2+2, but these are not equally likely: 1+3 happens in two ways (1 on red and 3 on blue, or the other way round) and 2+2 in only one. That is why the probability is 3/36, not 2/21.

Probability of rolling at least one 6 in n rolls

Working out "at least one 6" directly means adding up the cases with one 6, two sixes, three sixes and so on. Going through the complement is much quicker. The chance of no 6 on one roll is 5/6, and over n independent rolls it is (5/6)ⁿ. So:

P(at least one 6 in n rolls) = 1 − (5/6)ⁿ

It makes no difference whether you roll one die n times or n dice at once: the probability is the same.

The table shows three things. Four rolls is the first point where the odds pass 50%. Six rolls do not guarantee a 6, even though "one in six" seems to suggest it: the probability is 66.51%. And it never reaches 100%: you need 13 rolls to pass 90% and 26 to pass 99%.

The same formula works for any value on any die: on a d20, the chance of at least one natural 20 in n rolls is 1 − (19/20)ⁿ, and on a d4 the chance of at least one 4 is 1 − (3/4)ⁿ. In the calculator this is the "at least one V" target.

Two dice: probability table for each total

With two six-sided dice there are 36 equally likely outcomes and the total runs from 2 to 12. Each total has a different number of combinations: 7 comes up in six ways (1+6, 2+5, 3+4, 4+3, 5+2 and 6+1), whereas 2 and 12 come up in only one. The formula is P(S) = (6 − |S − 7|) / 36.

The cumulative columns are the ones you use in games. Rolling 8 or more has a probability of 15/36 = 41.67%; 10 or more, 6/36 = 16.67%. A few other two-dice figures worth knowing: a double (both the same) is 6/36 = 1/6; at least one 6 is 11/36 = 30.56%; double six is 1/36 = 2.78%.

Because of this triangular shape, games that hand out resources on the total of two dice revolve around 7. In Catan, for instance, hexes numbered 6 and 8 produce more often than those numbered 2 or 12, and a 7 produces nothing and moves the robber.

Three dice and the question put to Galileo

With three dice the sample space has 216 outcomes and totals run from 3 to 18. The shape is no longer a triangle but a bell: the middle values, 10 and 11, are the most likely, with 27 combinations each (12.5%), and 3 and 18 can only happen one way (0.46%). The distribution is symmetrical about 10.5, so a total of 3 is as likely as 18, 4 as likely as 17, and so on.

Why 10 turns up more often than 9

Dice players had noticed that with three dice 10 appeared slightly more often than 9, even though each can be made from six different sets of numbers: 9 as 1+2+6, 1+3+5, 1+4+4, 2+2+5, 2+3+4 and 3+3+3, and 10 as 1+3+6, 1+4+5, 2+2+6, 2+3+5, 2+4+4 and 3+3+4. Galileo Galilei explained it in a short paper known as Sopra le scoperte dei dadi ("On a discovery concerning dice"). The question is usually said to have come from the court of the Grand Duke of Tuscany, where Galileo had been mathematician since 1610; the text itself only says he is writing to oblige whoever asked him.

The key is order. A set with three different numbers, such as 1+2+6, can appear in 6 ways depending on which die shows which number; one with a pair, such as 1+4+4, in 3 ways; and three of a kind, such as 3+3+3, in just one. Adding up: 9 has 6+6+3+3+6+1 = 25 outcomes and 10 has 6+6+3+6+3+3 = 27. The gap is 2 cases out of 216 (11.57% against 12.5%), less than one percentage point, and players had still picked it up from experience.

The Chevalier de Méré, Pascal and Fermat (1654)

Antoine Gombaud, known as the Chevalier de Méré, was a French writer with a taste for gambling and mathematics. He put two questions to Blaise Pascal. The better known is the problem of points: how to divide the stakes fairly when a game of several rounds is broken off before the end. Pascal discussed it in letters with Pierre de Fermat during the summer of 1654, and that correspondence is regarded as the starting point of modern probability theory.

The other question was about dice. In his letter to Fermat of 29 July 1654, Pascal reports that Méré saw a contradiction in the numbers:

• Betting on at least one 6 in 4 rolls of one die is favourable, by 671 to 625.
• Betting on at least one double six in 24 rolls of two dice is unfavourable.

Méré reasoned by proportion: one die has 6 faces and 4 rolls are enough; two dice have 36 combinations, so 24 rolls should give the same edge, since 4 is to 6 as 24 is to 36. Proportion does not work here, and the complement shows why:

The figures in the letter match exactly: of the 6⁴ = 1296 outcomes of four rolls, 5⁴ = 625 contain no 6 and the remaining 671 contain at least one. For the double-six bet to pass 50% you need 25 rolls. The gap at 24 is small, under one point below 50%, but it is enough to make the bet a loser in the long run.

The first written study of dice probabilities is older still: Gerolamo Cardano's Liber de ludo aleae, finished around 1563 but not published until 1663, almost a century later.

Distribution of NdX, mean and variance

For N dice with X faces, the probability of each total comes from counting how many of the Xᴺ combinations give that total. The general method is called convolution: start with one die (each face with probability 1/X) and add dice one at a time. The probability of total S with one more die is the average, over each face c of the new die, of the probability of having reached S − c with the earlier dice. That is how the tables above are built, and how the calculator gives exact figures rather than estimates from simulation. There is also a closed formula using binomial coefficients, but by hand it is easier to keep adding columns.

Each extra die smooths the shape: one die gives a flat distribution, two a triangle, three a bell with almost straight sides, and from four or five dice the curve looks very like the normal (Gaussian) bell curve. This is the central limit theorem: the sum of many independent variables with the same distribution tends towards a normal distribution, whatever the shape of each one. With 4d6, for example, the most likely total is 14 (146 of 1296 combinations, 11.27%).

Mean and variance of one die

The mean, or expected value, of a six-sided die is (1+2+3+4+5+6)/6 = 3.5. It is not a value that can be rolled but the average that many rolls tend towards. The variance is the mean of the squares minus the square of the mean: 91/6 − 3.5² = 35/12 ≈ 2.92, and the standard deviation is √(35/12) ≈ 1.71. For a die with X faces numbered 1 to X, the mean is (X+1)/2 and the variance (X² − 1)/12.

With N dice, the mean and the variance are multiplied by N (variances add because the dice are independent). So 10d6 has a mean of 35 and a standard deviation of √(10 × 35/12) ≈ 5.40. The exact probability of a total between 30 and 40 is 68.70%, very close to the 68% the normal distribution predicts for one standard deviation either side of the mean.

Dice probability worked examples and common mistakes

These are the question types that come up most often in GCSE and A level probability. The dice are always fair and six-sided.

1. A total of 7 with two dice

Possible outcomes: 36. Favourable: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), which is 6. P = 6/36 = 1/6 ≈ 16.67%.

2. At least one 6 when rolling two dice

Complement: no 6 at all, which happens in 5 × 5 = 25 cases. P = 1 − 25/36 = 11/36 ≈ 30.56%. The classic mistake is to add 1/6 + 1/6 = 12/36, which counts (6,6) twice.

3. A total of 10 or more with two dice

Total 10: 3 cases; total 11: 2 cases; total 12: 1 case. Altogether 6. P = 6/36 = 1/6.

4. The product of two dice is even

The product is odd only if both dice are odd: 3 × 3 = 9 cases. P(even) = 1 − 9/36 = 3/4 = 75%.

5. An even number or a number greater than 4 on one die

Even: {2, 4, 6}. Greater than 4: {5, 6}. The union is {2, 4, 5, 6}, 4 cases. P = 4/6 = 2/3. Adding 3/6 + 2/6 = 5/6 counts the 6 twice; you have to subtract the intersection: 3/6 + 2/6 − 1/6 = 4/6.

6. Three dice all showing different numbers

The first die can show anything (6 options), the second something different (5) and the third something different from both (4): 6 × 5 × 4 = 120 out of 216. P = 5/9 ≈ 55.56%. All three the same: 6/216 = 1/36.

7. Exactly two sixes with four dice

This is a binomial distribution. Ways to choose which two dice are sixes: C(4,2) = 6. Each has probability (1/6)² × (5/6)² = 25/1296. P = 6 × 25/1296 = 25/216 ≈ 11.57%.

8. How many rolls for at least a 90% chance of a 6

Find the smallest n with 1 − (5/6)ⁿ ≥ 0.9, that is, (5/6)ⁿ ≤ 0.1. Using logarithms: n ≥ log 0.1 / log (5/6) ≈ 12.63. Answer: 13 rolls, giving 90.65%.

Common mistakes