TL;DR:
When outcomes are equally likely, probability is favourable outcomes divided by total outcomes.
Two dice have 36 equally likely outcomes, and 6 of them sum to 7, so the chance is
6/36 = 1/6.For "at least one", it is often easier to find the chance of none and subtract from 1.
Roll two dice and add them. Which total is most likely? Many people say 6, or 7, or "they are all about the same". You can settle the question with a list, no guessing required. Dice are a good place to learn probability because every outcome can be written down.
Outcomes and equally likely events
The sample space is the list of everything that can happen. For one fair die it is 1, 2, 3, 4, 5, 6, each equally likely. When all outcomes are equally likely, the probability of an event is favourable outcomes / total outcomes. The chance of an even number on one die is 3/6 = 1/2, since 2, 4 and 6 are even.
Probabilities are always between 0 (impossible) and 1 (certain). The probabilities of all the separate outcomes must add to 1.
Worked example: the chance of rolling a sum of 7 with two dice
Make the sample space for two dice as ordered pairs: first die, second die. There are 6 × 6 = 36 pairs, because each of the 6 faces on the first die pairs with 6 faces on the second. Treat the dice as different, say one red and one blue. That is what makes the 36 outcomes equally likely.
List the pairs adding to 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). That is 6 pairs. So P(sum = 7) = 6/36 = 1/6.
The whole table of sums
Sum | Ways | Chance |
|---|---|---|
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 |
Check that the ways add to 36: 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1 = 36. Sum 7 is the most likely, with 6 ways. The sums 2 and 12 each need both dice to show a particular face, so they have just 1 way.
A common error is to list the 11 possible sums, 2 to 12, and say each has chance 1/11. But the sums are not equally likely. A sum of 7 can happen in 6 ways and a sum of 12 in 1. Probability by counting only works when you count outcomes that are equally likely, which here means the pairs.
Complements: the easy way to do "at least one"
What is the chance of at least one 6 when rolling two dice? Counting every pair with a 6 is possible, but there is a shortcut. The opposite event, no sixes, is easier: each die has 5 non-6 faces, so there are 5 × 5 = 25 pairs with no 6. Then P(at least one 6) = 1 − 25/36 = 11/36.
Check by counting directly: pairs where the first is 6 are 6 in number, pairs where the second is 6 are 6, but the pair (6, 6) was counted twice, so the total is 6 + 6 − 1 = 11. The two routes both give 11/36.
Independent events multiply
The result on one die does not affect the other, so the events are independent. The chance of 6 on the first die and 6 on the second is 1/6 × 1/6 = 1/36, matching the single pair (6, 6). The chance of a double of any kind is 6/36 = 1/6, since the doubles are (1,1), (2,2), …, (6,6).
What probability does and does not tell you
A probability of 1/6 does not mean that exactly one roll in six gives a 7. It describes the long-run pattern. In 60 rolls you would expect about 10 sevens, but 8 or 12 would not be surprising. A die has no memory either: after five rolls with no 6, the chance of a 6 on the next is still 1/6.
Try two more yourself
What is the chance that the total is 10 or more? The sums 10, 11 and 12 have 3, 2 and 1 ways, so P = (3 + 2 + 1)/36 = 6/36 = 1/6. What is the chance of a total of 7 or 11? These cannot happen together, so add: (6 + 2)/36 = 8/36 = 2/9. Adding works here because the two events are mutually exclusive, a sum cannot be both 7 and 11.
How to check your answer
Range. Is your answer between 0 and 1?
Total. Do all your cases add to 1 or 36?
Complement. Does
P(event) + P(not event) = 1?Two methods. Count directly and by the complement.
Common mistakes
Treating sums as equally likely.
Forgetting that (1, 6) and (6, 1) are different outcomes.
Adding probabilities for "and" instead of multiplying.
Using the gambler's fallacy: believing a 6 is "due".
Leaving the answer as a count rather than a fraction of the total.
Where you will use it next
Once outcomes depend on earlier information, you need conditional probability with a screening test. For summarising many outcomes, see mean, median and mode. Counting methods for bigger lists are the subject of permutations and combinations.
Frequently asked questions
Is a sum of 7 really the most common?
Yes. Seven has 6 ways out of 36. No other sum has more than 5.
What about a loaded die?
Then the faces are not equally likely, and counting outcomes no longer works. You would need the measured chance of each face.
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