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:

$C={36}^{2}\times {}_{5}{\mathrm{P}}_{3}\times 2\times 3\times 5-D$

Where D is duplicates. But if this is correct, not sure what pattern the duplicates follow.

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:

$C={36}^{2}\times {}_{5}{\mathrm{P}}_{3}\times 2\times 3\times 5-D$

Where D is duplicates. But if this is correct, not sure what pattern the duplicates follow.

Answer & Explanation

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 $|A\cap B\cap C|={36}^{5}-|{A}^{\prime}\cup {B}^{\prime}\cup {C}^{\prime}|$

Step 2

By the Inclusion-Exclusion Principle,

$|{A}^{\prime}\cup {B}^{\prime}\cup {C}^{\prime}|=|{A}^{\prime}|+|{B}^{\prime}|+|{C}^{\prime}|-|{A}^{\prime}\cap {B}^{\prime}|-|{A}^{\prime}\cap {C}^{\prime}|-|{B}^{\prime}\cap {C}^{\prime}|+|{A}^{\prime}\cap {B}^{\prime}\cap {C}^{\prime}|$

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, $|{A}^{\prime}|={34}^{5}$.

|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, $|{B}^{\prime}|={33}^{5}$.

|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, $|{B}^{\prime}|={31}^{5}$.

$|{A}^{\prime}\cap {C}^{\prime}|$ Excluding sums of 3 and 8 eliminates $2+5=7$ possible outcomes for each of the five rolls. Hence, $|{A}^{\prime}\cap {C}^{\prime}|=(36-7{)}^{5}={29}^{5}$.

$|{B}^{\prime}\cap {C}^{\prime}|$: Excluding sums of 4 and 8 eliminates $3+5=8$ possible outcomes for each of the five rolls. Hence, $|{A}^{\prime}\cap {B}^{\prime}|=(36-8{)}^{5}={28}^{5}$.

$|{A}^{\prime}\cap {B}^{\prime}\cap {C}^{\prime}|$: Excluding sums of 3, 4, and 8 eliminates $2+3+5=10$ possible outcomes for each of the five rolls. Hence, $|{A}^{\prime}\cap {B}^{\prime}\cap {C}^{\prime}|=(36-10{)}^{5}={26}^{5}$.

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 ${36}^{5}-{34}^{5}-{33}^{5}-{31}^{5}+{31}^{5}+{29}^{5}+{28}^{5}-{26}^{5}$.

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 $|A\cap B\cap C|={36}^{5}-|{A}^{\prime}\cup {B}^{\prime}\cup {C}^{\prime}|$

Step 2

By the Inclusion-Exclusion Principle,

$|{A}^{\prime}\cup {B}^{\prime}\cup {C}^{\prime}|=|{A}^{\prime}|+|{B}^{\prime}|+|{C}^{\prime}|-|{A}^{\prime}\cap {B}^{\prime}|-|{A}^{\prime}\cap {C}^{\prime}|-|{B}^{\prime}\cap {C}^{\prime}|+|{A}^{\prime}\cap {B}^{\prime}\cap {C}^{\prime}|$

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, $|{A}^{\prime}|={34}^{5}$.

|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, $|{B}^{\prime}|={33}^{5}$.

|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, $|{B}^{\prime}|={31}^{5}$.

$|{A}^{\prime}\cap {C}^{\prime}|$ Excluding sums of 3 and 8 eliminates $2+5=7$ possible outcomes for each of the five rolls. Hence, $|{A}^{\prime}\cap {C}^{\prime}|=(36-7{)}^{5}={29}^{5}$.

$|{B}^{\prime}\cap {C}^{\prime}|$: Excluding sums of 4 and 8 eliminates $3+5=8$ possible outcomes for each of the five rolls. Hence, $|{A}^{\prime}\cap {B}^{\prime}|=(36-8{)}^{5}={28}^{5}$.

$|{A}^{\prime}\cap {B}^{\prime}\cap {C}^{\prime}|$: Excluding sums of 3, 4, and 8 eliminates $2+3+5=10$ possible outcomes for each of the five rolls. Hence, $|{A}^{\prime}\cap {B}^{\prime}\cap {C}^{\prime}|=(36-10{)}^{5}={26}^{5}$.

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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