Number of combinations with given outcomes for a set number of dice rollsHow to find...

glycleWogry

glycleWogry

Answered

2022-06-26

Number of combinations with given outcomes for a set number of dice rolls
How to find the number of combinations for n rolls of a pair of 6-sided dice with all of the specified sums rolled at least once?
Example. A pair of 6-sided dice is rolled 5 times. What is the number of combinations containing each of: 3, 4, 8 at least once? So that (3, 7, 8, 5, 4) is counted and (3, 4, 2, 2, 12) is not.
Thoughts on the example. It's possible to roll 3 in 2 ways, 4 in 3 and 8 in 5 and the total number of outcomes is 36. So I think the number of combinations containing 3, 4, 8 is:
C = 36 2 × 5 P 3 × 2 × 3 × 5 D
Where D is duplicates. But if this is correct, not sure what pattern the duplicates follow.

Answer & Explanation

iceniessyoy

iceniessyoy

Expert

2022-06-27Added 27 answers

One way to address your problem is to use the Inclusion-Exclusion Principle, which will allow us to avoid doing case work. Let's look at your example.
Since there are 36 possible outcomes for each roll of a pair of six-sided dice, there are 36 5 possible outcomes for five rolls of a pair of dice. From these, we must exclude those outcomes in which a sum of 3, a sum of 4, or a sum of 8 is missing. We want to find | A B C | = 36 5 | A B C |
Step 2
By the Inclusion-Exclusion Principle,
| A B C | = | A | + | B | + | C | | A B | | A C | | B C | + | A B C |
You observed that there are two ways to obtain a sum of 3, three ways to obtain a sum of 4, and five ways to obtain a sum of 8.
|A′|: Since there are two ways to obtain a sum of 3, there are 36 2 = 34 ways to not obtain a sum of 3 on each of the five rolls. Hence, | A | = 34 5 .
|B′|: Since there are three ways to obtain a sum of 4, there are 36 3 = 33 ways to not obtain a sum of 4 on each of the five rolls. Hence, | B | = 33 5 .
|C′|: Since there are five ways to obtain a sum of 8, there are 36 5 = 31 ways to not obtain a sum of 8 on each of the five rolls. Hence, | B | = 31 5 .
| A C | Excluding sums of 3 and 8 eliminates 2 + 5 = 7 possible outcomes for each of the five rolls. Hence, | A C | = ( 36 7 ) 5 = 29 5 .
| B C | : Excluding sums of 4 and 8 eliminates 3 + 5 = 8 possible outcomes for each of the five rolls. Hence, | A B | = ( 36 8 ) 5 = 28 5 .
| A B C | : Excluding sums of 3, 4, and 8 eliminates 2 + 3 + 5 = 10 possible outcomes for each of the five rolls. Hence, | A B C | = ( 36 10 ) 5 = 26 5 .
By the Inclusion-Exclusion Principle, the number of outcomes in which sums of 3, 4, and 8 all appear when five dice are rolled is 36 5 34 5 33 5 31 5 + 31 5 + 29 5 + 28 5 26 5 .

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