Modeling Two Dice
When two dice are rolled, an event such as “the sum is 9” is determined by the face shown on each die. The key is to record both results, not just the total: a roll of 3 on the first die and 6 on the second is a different outcome from 6 followed by 3, even though both rolls have sum 9. Keeping the order lets us count every possible result consistently.
The independence assumption means that the probability of a particular pair is the product of the probabilities for its two results: \(P((i,j))=(1/6)(1/6)=1/36\). Thus these 36 outcomes are equally likely. For any event \(A\), its probability is the number of ordered pairs in \(A\) divided by 36. This favorable-count-over-total-count rule applies here because the outcomes in the sample space are equally likely.
For example, \((2,5)\) and \((5,2)\) are separate outcomes. An event that includes both must count both, even if the order does not matter to the wording of the event. The outcomes are ordered because they specify the result of each die; the event itself may care only about the sum or some other feature.
Grouping Outcomes by Their Sum
Let \(S\) denote the sum of the two dice. The smallest possible sum is 2, from \((1,1)\), and the largest is 12, from \((6,6)\). Each sum is produced by a different number of ordered pairs. Listing those pairs makes the changing counts visible.
| Sum | Ordered pairs that give the sum | Number of pairs |
|---|---|---|
| 2 | (1,1) | 1 |
| 3 | (1,2), (2,1) | 2 |
| 4 | (1,3), (2,2), (3,1) | 3 |
| 5 | (1,4), (2,3), (3,2), (4,1) | 4 |
| 6 | (1,5), (2,4), (3,3), (4,2), (5,1) | 5 |
| 7 | (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) | 6 |
| 8 | (2,6), (3,5), (4,4), (5,3), (6,2) | 5 |
| 9 | (3,6), (4,5), (5,4), (6,3) | 4 |
| 10 | (4,6), (5,5), (6,4) | 3 |
| 11 | (5,6), (6,5) | 2 |
| 12 | (6,6) | 1 |
Each row describes an event, and its probability is the row’s count divided by 36. In particular, the table shows that different sums do not have equal probabilities: there are six ways to make 7, but only one way to make 2 or 12. Because the sum events from 2 through 12 cover all 36 outcomes without overlap, their counts must add to 36.
Worked Example: Finding the Probability of a Sum of 9
Find the probability that the sum of two fair dice is 9.
List the favorable ordered pairs. The first die can show 3 while the second shows 6, or the first can show 4 while the second shows 5. Reversing the order gives two more outcomes: \((3,6),(4,5),(5,4),(6,3)\).
There are 4 favorable outcomes among 36 equally likely ordered pairs. Therefore,
The probability is \(1/9\), or about 11.11%. Counting only \((3,6)\) and \((4,5)\) would miss the two outcomes with the dice reversed.
Why the Sum Counts Rise and Fall
The pattern in the table can be derived by counting possible first-die results. For a fixed sum \(s\), each possible first result \(i\) determines exactly one second result, \(j=s-i\). A pair is valid only when both results are between 1 and 6.
For \(2\leq s\leq 7\), the first die can show any integer from 1 through \(s-1\). Each choice leaves a second result from 1 through 6, so there are \(s-1\) pairs.
For \(7\leq s\leq 12\), the first die must be at least \(s-6\) and at most 6. The number of possible first results is \(6-(s-6)+1=13-s\), giving \(13-s\) pairs.
Since every ordered pair has probability \(1/36\), the probability of a sum is its number of pairs divided by 36.
The two counting expressions agree at \(s=7\): each gives 6 pairs. Together, they give the number of outcomes for any possible sum.
This derivation also explains the triangular pattern: the number of ways rises from 1 to 6, then falls from 6 to 1. A useful symmetry check follows by changing every die result \(i\) to \(7-i\). This maps \((i,j)\) to \((7-i,7-j)\), a valid one-to-one mapping of the 36 outcomes. The new sum is \(14-(i+j)\), so a sum \(s\) has the same number of outcomes as the sum \(14-s\). For example, sums 5 and 9 each have 4 outcomes, and sums 2 and 12 each have 1.
Worked Example: Finding the Probability of a Sum at Least 10
Find the probability that the sum is at least 10. The relevant sums are 10, 11, and 12. From the table, they have 3, 2, and 1 ordered pairs, respectively. These sum events are disjoint: one roll cannot have two different sums. Therefore, add their counts.
As a symmetry check, sums of 10, 11, and 12 correspond to sums of 4, 3, and 2 under the mapping above. Those lower sums have \(3+2+1=6\) outcomes as well. The probability is \(1/6\), or about 16.67%.
Counting Events Other Than a Single Sum
The same ordered-pair sample space works for events that involve the individual dice, not just their total. State exactly what qualifies, then count every qualifying pair. When adding counts for overlapping descriptions, account for outcomes that satisfy more than one description; an outcome should be counted only once in the event’s total.
Worked Example: At Least One Die Shows 6
Find the probability that at least one of the two dice shows 6. List the outcomes with a 6 on the first die and those with a 6 on the second die. The first group is \((6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\); the second is \((1,6),(2,6),(3,6),(4,6),(5,6),(6,6)\).
There are 6 outcomes in each group, but \((6,6)\) is in both. Counting it twice would overstate the number of favorable outcomes, so count it once:
Thus,
The result is about 30.56%. The two groups are not disjoint, which is why simply adding their counts, \(6+6\), would be incorrect.
Common Mistakes and Checks
- Counting only unordered pairs. Reasoning that \((3,6)\) and \((6,3)\) are “the same” drops a possible outcome. The dice results are recorded as ordered pairs, and both outcomes belong in the count.
- Assuming all sums are equally likely. There is one outcome with sum 2 but six with sum 7. Equal likelihood applies to the 36 ordered pairs, not to the 11 possible sum values.
- Double-counting an overlap. “A 6 on the first die” and “a 6 on the second die” both include \((6,6)\). When counting their union, subtract the shared outcome once.
- Using 36 without checking the model. The denominator 36 follows from two independent fair six-sided dice. If a die is not fair, or the rolls are not independent, the 36 outcomes need not be equally likely, and favorable counts alone do not determine probabilities.
Check Your Understanding
Try these questions using the ordered-pair model.
- How many ordered pairs produce a sum of 8, and what is the probability of that sum?
- Explain why the probabilities of sums 4 and 10 are equal without listing every pair.
- Find the probability that the sum is less than 5. Show which sum counts you add.
- How many outcomes have the first die showing 2 and the second die showing an even number? What is the probability?
- Why is it incorrect to count 12 outcomes for “at least one die shows 6” by adding two groups of 6?