Roll $4$ dice and probability that there exists at least one pair of dice such that the sum of faces add up to $7$.

Question

Disclaimer: This isn't a homework question. I was independently studying probability and this question randomly came up in my mind.

I was thinking of subtracting "the probability of having no pair of $(n,m)$ that adds up to $7$" from $1$.

I shall denote the probability in quotation by $P$. If four numbers are all different, then P trivially becomes 1 by pigeonhole principle. (for simplicity, let's say three holes are $\{1,6\}$, $\{2,5\}$, $\{3,4\}$) Hence, I assume that there exist at least two duplicated numbers.

Total number of ways of choosing $4$ numbers is: $T = \binom{6}{4} + \binom{6}{3}\binom{3}{1} + \binom{6}{2}[\binom{2}{1}+1] + \binom{6}{1}$. Then, I break up into three cases to calculate $P$, where

1. only $2$ numbers are the same: $ P_{1} = \binom{6}{1}\binom{2}{1}\binom{2}{1}/T$ (choose $1$ number of duplicate, choose two numbers from each hole.)

2. $2$ duplicates: $ P_{2} = \binom{6}{1}\binom{4}{1}/T$

3. only $3$ numbers are the same: $ P_3 = \binom{6}{1}\binom{4}{1}/T$ (choose $1$ number of duplicate, then choose $1$ last number.

4. all $4$ numbers are the same $ P_4 = \binom{6}{1}/T$

$P = P_1 + P_2 + P_3 + P_4$, and the solution I am looking for is, $1-P$. Am I missing anything, or overcounting anything?

P.S.

1. How should I solve this problem if if we have a large number instead of 4 dice, say $n$?

2. Is there a more elegant way to solve this problem using inclusion and exclusion? I can't really think of a way to apply inclusion and exclusion principle. Answer to this question shall easily suffice as a solution to the previous question.

Answer 1

Throw $n$ dice and for $A\subseteq\left\{ 1,2,3,4,5,6\right\} $ let $E_{A}$ denote the event that all faces of the dice take a value in $A$.

Let $E:=E_{\left\{ 1,2,3\right\} }\cup E_{\left\{ 1,2,4\right\} }\cup E_{\left\{ 1,3,5\right\} }\cup E_{\left\{ 1,4,5\right\} }\cup E_{\left\{ 2,3,6\right\} }\cup E_{\left\{ 2,4,6\right\} }\cup E_{\left\{ 3,5,6\right\} }\cup E_{\left\{ 4,5,6\right\} }$

Then to be found is $1-P\left(E\right)$.

With inclusion/exclusion and symmetry we find:

$$\begin{aligned}P\left(E\right) & =8P\left(E_{\left\{ 1,2,3\right\} }\right)-12P\left(E_{\left\{ 1,2\right\} }\right)-12P\left(E_{\left\{ 1\right\} }\right)+24P\left(E_{\left\{ 1\right\} }\right)-6P\left(E_{\left\{ 1\right\} }\right)\\ & =8\cdot2^{-n}-12\cdot3^{-n}+6\cdot6^{-n}\\ & =2^{3-n}-4\cdot3^{1-n}+6^{1-n} \end{aligned} $$

So we conclude that: $$1-P\left(E\right)=1-2^{3-n}+4\cdot3^{1-n}-6^{1-n}$$

In special case $n=4$ we find: $$1-P(E)=\frac{139}{216}$$