Probability of rolling 4 dice and obtaining a sum from 2 dice of 3, 8, or 11?

Question

Problem:

You roll 4 dice. What is the probability of getting 2 of the 4 dice to have a sum of 3, 8, or 11?

Examples:

• 1 2 3 4 -> counts -> as 1 + 2 = 3
• 1 2 3 6 -> counts -> as 1 + 2 = 3 OR 2 + 6 = 8
• 1 3 4 6 -> doesn't count -> since you can't combine any 2 dice to get the sum of 3, 8, or 11
• 1 1 1 1 -> doesn't count -> same as above

Breakdown:

To get a sum of 3, you need the combination (1, 2). To get a sum of 11, you need the combination (5, 6). To get a sum of 8, you can have the combinations (2, 6), (3, 5), or (4, 4).

My take:

For the sums of 3 and 11, there is only one possible combination each. So, the problem can be simplified to "Rolling 4 dice, what is the probability of getting at least one 1 and at least one 2" (or at least one 5 and one 6 for 11).

I used the inclusion/exclusion principle to solve it, which gives an answer of 0.233025.

However, I'm stuck on calculating the probability for a sum of 8 and how to combine all three answers to get the final result.

Note:

I used code to get the final answer, it should be around 0.758488. But I'm still struggling to find a way to solve it using Math.

Answer 1

In what follows, I will assume that the dice are fair so that each outcome is equally likely to occur.

Suppose there is one blue die, one green die, one red die, and one yellow die. Then an outcome can be represented as $(b, g, r, y)$, where $b, g, r, y$ represent, respectively, the numbers on the blue die, green die, red die, and yellow green die. Since there are six equally likely outcomes on each die, there are $6^4$ possible outcomes.

Since the only way to obtain a sum of $3$ with two dice is $3 = 1 + 2$, to obtain a sum of $3$ with two of the four dice, there must be at least one $1$ and at least one $2$ among the four dice.

Since \begin{align*} 8 & = 2 + 6\\ & = 3 + 5\\ & = 4 + 4 \end{align*} we can obtain a sum of $8$ with two of the four dice if there is at least one $2$ and at least one $6$ among the four dice, or there is at least one $3$ and at least one $5$ among the four dice, or there is at least two $4$s among the four dice.

Since the only way to obtain a sum of $11$ with two dice is $11 = 5 + 6$, to obtain a sum of $11$ with a pair of dice, there must be at least one $5$ and at least one $6$ among the four dice.

Since we wish to obtain a sum of $3$ or $8$ or $11$, the favorable cases are the events

• $E_1$: at least one $1$ and at least one $2$ appear on the four dice
• $E_2$: at least one $2$ and at least one $6$ appear on the four dice
• $E_3$: at least one $3$ and at least one $5$ appear on the four dice
• $E_4$: at least two $4$s appear on the four dice
• $E_5$: at least one $5$ and at least one $6$ appear on the four dice

Therefore, we wish to find $\Pr(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_5)$. By the Inclusion-Exclusion Principle, \begin{align*} \Pr(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_5) & = \sum_{i = 1}^{5} \Pr(E_i) - \sum_{1 \leq i < j \leq 5} \Pr(E_i \cap E_j)\\ & \qquad + \sum_{1 \leq i < j < k \leq 5} \Pr(E_i \cap E_j \cap E_k)\\ & \qquad\qquad - \sum_{1 < i < j < k < l \leq 5} \Pr(E_i \cap E_j \cap E_k \cap E_l)\\ & \qquad\qquad\qquad + \Pr(E_1 \cap E_2 \cap E_3 \cap E_4 \cap E_5) \end{align*}

$\Pr(E_1)$: The probability that there is at least one $1$ and at least one $2$ is $1$ minus the probability that there are no $1$s or no $2$s. Since there are $\binom{2}{k}$ ways to exclude $k$ of these two numbers and $(6 - k)^4$ outcomes in which $k$ numbers are excluded, by the Inclusion-Exclusion Principle, $$\Pr(E_1) = 1 - \binom{2}{1}\left(\frac{5}{6}\right)^4 + \binom{2}{2}\left(\frac{4}{6}\right)^4$$

By symmetry, $\Pr(E_1) = \Pr(E_2) = \Pr(E_3) = \Pr(E_5)$.

$\Pr(E_4)$: The probability that at least two $4$s appear on the four dice can be found by subtracting the probability that at most one $4$ appears on the four dice. The probability that no $4$s appear on the four dice is $(5/6)^4$. The probability that exactly one four appears on the four dice is $\binom{4}{1}(1/6)(5/6)^3$ since there are four dice on which the $4$ could appear, the probability of obtaining a $4$ on that die is $1/6$, and the probability of not obtaining a $4$ on each of the other three dice is $(5/6)^3$. Hence, the probability of obtaining at least two $4$s is $$1 - \left(\frac{5}{6}\right)^4 - \binom{4}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^3$$

$\Pr(E_1 \cap E_2)$: We wish to find the probability that at least one $1$, at least one $2$, and at least one $6$ appear on the four dice. The probability that this occurs is $1$ minus the probability that there are no $1$s or no $2$s or no $6$s. Since there are $\binom{3}{k}$ ways to exclude $k$ of these three numbers and $(6 - k)^4$ possible outcomes in which $k$ of the numbers are excluded, there are $$\Pr(E_1 \cap E_2) = 1 - \binom{3}{1}\left(\frac{5}{6}\right)^4 + \binom{3}{2}\left(\frac{4}{6}\right)^2 - \binom{3}{3}\left(\frac{3}{6}\right)^4$$ by the Inclusion-Exclusion Principle.

By symmetry, $\Pr(E_1 \cap E_2) = \Pr(E_2 \cap E_5) = \Pr(E_3 \cap E_5)$.

$\Pr(E_1 \cap E_3)$: We wish to find the probability that the numbers $1, 2, 3, 5$ appear on the four dice. There are $4!$ ways to distribute the four distinct numbers to the four dice. Hence, $$\Pr(E_1 \cap E_3) = \frac{4!}{6^4}$$

By symmetry, $\Pr(E_1 \cap E_3) = \Pr(E_1 \cap E_5) = \Pr(E_2 \cap E_3)$.

$\Pr(E_1 \cap E_4)$: We wish to find the probability that the four dice show two $4$s, a $1$, and a $2$. There are $\binom{4}{2}$ ways to select the dice on which the $4$s appear and $2!$ ways to distribute the $1$ and $2$ to the remaining two dice. Hence, $$\Pr(E_1 \cap E_4) = \frac{\binom{4}{2}2!}{6^4}$$

By symmetry, $\Pr(E_1 \cap E_4) = \Pr(E_2 \cap E_4) = \Pr(E_3 \cap E_4) = \Pr(E_4 \cap E_5)$.

Notice that the only possible intersections of three or more of the five events are $E_1 \cap E_2 \cap E_5$ and $E_2 \cap E_3 \cap E_5$.

$\Pr(E_1 \cap E_2 \cap E_5)$: We showed above that the probability of the numbers $1, 2, 5, 6$ appearing on the four dice is $$\Pr(E_1 \cap E_2 \cap E_5) = \frac{4!}{6^4}$$

By symmetry, $\Pr(E_1 \cap E_2 \cap E_5) = \Pr(E_2 \cap E_3 \cap E_5)$.

Hence, the probability that there are two dice which have $3$, $8$, or $11$ when four fair dice are rolled is \begin{align*} \Pr(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_5) & = 4\left[1 - \binom{2}{1}\left(\frac{5}{6}\right)^4 + \binom{2}{2}\left(\frac{4}{6}\right)^4\right]\\ & \qquad + 1 - \left(\frac{5}{6}\right)^4 - \binom{4}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^3\\ & \qquad\qquad - 3\left[1 - \binom{3}{1}\left(\frac{5}{6}\right)^4 + \binom{3}{2}\left(\frac{4}{6}\right)^2 - \binom{3}{3}\left(\frac{3}{6}\right)^4\right]\\ & \qquad\qquad\qquad - 3 \cdot \frac{4!}{6^4} - 4 \cdot \frac{\binom{4}{2}2!}{6^4}\\ & \qquad\qquad\qquad\qquad + 2 \cdot \frac{4!}{6^4} \end{align*}

Answer 2

Let $E$ be the event of getting two of four thrown dice to have a sum of $3$, $8$, or $11$.

We will calculate the complement of $E$.

Label the four dice outcomes with $D_1$ thru $D_4$.

We partition the sample space with $D_1 = 1,2,3,4,5,6$ to calculate the answer, using 'smart' combinatorial arguments for the $D_1 = 1,2,3,5,6$ blocks, but working with the $4$ block was challenging. However, we found a technique for that block that avoided the necessity of using a pure brute force listing method.

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Let $A_1$ be the event that $D_1 = 1$.
To derive $P(\overline E \cap A_1)$ we have to check which outcomes in

$\; \{1\} \times \{1,3,4,5,6\}^3$

have to be excluded (to wit, the $3 + 5$, $5 + 6$ and $4 +4$ "action" must be rejected).

This is a combinatorial problem and after some thought we can write

$P(\overline E \cap A_1) = \frac{5^3 - (3 \cdot 2 \cdot 4 + 3 \cdot 2 \cdot 4 -3! )- 3 \cdot 4 - 1}{1296}$.

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Let $A_2$ be the event that $D_1 = 2$.
To derive $P(\overline E \cap A_2)$ we have to check which outcomes in

$\; \{2\} \times \{2,3,4,5\}^3$

have to be excluded (to wit, the $3 + 5$ and $4 +4$ "action" must be rejected).

This is a combinatorial problem and after some thought we can write

$P(\overline E \cap A_2) = \frac{4^3 - 3 \cdot 2 \cdot 3 - 3 \cdot 3 - 1}{1296}$.

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Let $A_3$ be the event that $D_1 = 3$.
To derive $P(\overline E \cap A_3)$ we have to check which outcomes in

$\; \{3\} \times \{1,2,3,4,6\}^3$

have to be excluded (to wit, the $1 + 2$, $2 + 6$ and $4 +4$ "action" must be rejected).

This is a combinatorial problem and after some thought we realize this is the same pattern we saw for our $P(\overline E \cap A_1)$ work, so we copy the solution from there and write

$P(\overline E \cap A_3) = \frac{5^3 - (3 \cdot 2 \cdot 4 + 3 \cdot 2 \cdot 4 -3! )- 3 \cdot 4 - 1}{1296}$.

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Let $A_4$ be the event that $D_1 = 4$.
To derive $P(\overline E \cap A_4)$ we have to check which outcomes in

$\; \{4\} \times \{1,2,3,5,6\}^3$

have to be excluded (to wit, the $1 + 2$, $3 + 5$ and $5 + 6$ "action" must be rejected).

While this is a combinatorial problem, we still resort to mixing in some brute force listing technique to solve it.

We can analyze the remaining dice as if they come in as ascending, $D_2 \le D_3 \le D_4$, and then find the appropriate multiplicity factors to cover all possibilities:

$(4,1,1,1) \;(\times 1)$
$(4,1,1,3) \;(\times 3)$
$(4,1,1,5) \;(\times 3)$
$(4,1,1,6) \;(\times 3)$
$(4,1,3,3) \;(\times 3)$
$(4,1,3,6) \;(\times 6)$
$(4,1,5,5) \;(\times 3)$
$(4,1,6,6) \;(\times 3)$
$(4,2,2,2) \;(\times 1)$
$(4,2,2,3) \;(\times 3)$
$(4,2,2,5) \;(\times 3)$
$(4,2,3,3) \;(\times 3)$
$(4,2,5,5) \;(\times 3)$
$(4,3,3,3) \;(\times 1)$
$(4,3,3,6) \;(\times 3)$
$(4,3,6,6) \;(\times 3)$
$(4,5,5,5) \;(\times 1)$
$(4,6,6,6) \;(\times 1)$

This comes in with a total count of $47$ and we can now write

$P(\overline E \cap A_4) = \frac{47}{1296}$.

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Let $A_5$ be the event that $D_1 = 5$.
To derive $P(\overline E \cap A_5)$ we have to check which outcomes in

$\; \{5\} \times \{1,2,4,5\}^3$

have to be excluded (to wit, both $1 + 2$ and $4 + 4$ must be rejected).

This is a combinatorial problem and after some thought we realize this is the same pattern we saw for our $P(\overline E \cap A_2)$ work, so we copy the solution from there and write

$P(\overline E \cap A_5) = \frac{4^3 - 3 \cdot 2 \cdot 3 - 3 \cdot 3 - 1 }{1296}$.

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Let $A_6$ be the event that $D_1 = 6$.
To derive $P(\overline E \cap A_6)$ we have to check which outcomes in

$\; \{6\} \times \{1,3,4,6\}^3$

have to be excluded (to wit, $4 + 4$ must be rejected).

This is a combinatorial problem and after some thought we can write

$P(\overline E \cap A_6) = \frac{4^3 - 3 \cdot 3 - 1}{1296}$.

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So

$P(E) = 1 - $
$\quad \large(\normalsize \, P(\overline E \cap A_1)+P(\overline E \cap A_2)+P(\overline E \cap A_3)+$
$\quad P(\overline E \cap A_4)+P(\overline E \cap A_5)+P(\overline E \cap A_6) \, \large)\normalsize = \frac{983}{1296}$