# Probability involving n dice

• I
• goraemon
This means that after throwing the first six dice, the probability of a case that has at least one die that shows 3 is\begin{pmatrix}1/6 & 1/6 & 1/6 & 1/6\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ \vdots & \vdots & 1 & 0.\end{pmatrixf

#### goraemon

I'm studying probability and am currently stuck on this question:

Let's say we have n distinct dice, each of which is fair and 6-sided. If all of these dice are rolled, what is the probability that there is at least one pair that sums up to 7?

I interpreted the above as being equivalent to the following:

1 - (Probability that there is no pair that sums up to 7)

So if I were to consider just one pair of dice, then the probability that the pair adds up to 7 is 1/6, I think?

So Pr(one pair doesn't add up to 7) = 5/6.

But then I'm kind of stuck on how to proceed. Because there are lots of possible pairs amongst the n die, and some of these pairs overlap...for example, (die1, die2) is a pair, (die1, die3) is a pair, and so on. So I don't know how to account for these overlaps.

I tried breaking down the problem into a number of cases where there is no way for any pair to add up to 7:

(1) All of the dice show exactly one number.
(2) All of the dice show exactly two numbers which do not add up to 7 -- e.g. All the dice show either 3 or 6. Or all the dice show 2 or 4. And so on...
(3) All of the dice show exactly three numbers e.g. (1, 2, 4), no two of which can possibly add up to 7.
(4) If all of the dice show 4 or more numbers, then there MUST exist a pair that adds up to 7, so I don't consider any of these cases.

I suppose that I could add up the probabilities for all of the above cases, then I'd have the total probability that no two dice add up to 7? But then, how do I compute these?

In fact, is there a better / easier approach than the one I have thought up? Thanks.

I might do it the way you suggest. I'd look at using induction on n.

Because there are lots of possible pairs amongst the n die,
Do you know a function which can tell you exactly how many pairs there are?

I definitely agree that the way to begin this problem is to find the probability that no pairs sum to 7, which is of course 1 - the probability you are ultimately seeking.

It might help organize the calculation if you specify what the sample space is: the set {1, 2, 3, 4, 5, 6} raised to the cartesian power of n, just a fancy way of saying the set of "all ordered n-tuples consisting of any of those 6 numbers". Then specify what the probability is for each point of this space, and what subset of the space you are finding the probability of. Finally, break that subset into disjoint pieces, each of which you can find the probability of.

You have virtually done this already, but it doesn't hurt to conceive of what you have done systematically, since that helps to see how various pieces that you have broken your target event into are disjoint from each other.

It will help if you figure out how many possible pairs of these numbers do not sum to 7. And then how many triples of these numbers involve no pair that sums to 7.

Let's do an example:

(2) You know that two numbers are shown from the ##n## dice. We have the following choices, and all these choices have the same probabilities:
a) 1 and 2
b) 2 and 3
c) 1 and 3
...
(How many choices do we have for two numbers to show?)

Let's analyse case (a). The other cases are similar. So we know that all dice show either ##1## or ##2##. Since the dice are all distinct, this means there are ##2^n## ways of ##n## dice to show either ##1## or ##2##. However, ##2## of these ways are to show only ##1## and ##2##. So there are ##2^n - 2## ways to satisfy ##(a)##.

...
I tried breaking down the problem into a number of cases where there is no way for any pair to add up to 7:

(1) All of the dice show exactly one number.
(2) All of the dice show exactly two numbers which do not add up to 7 -- e.g. All the dice show either 3 or 6. Or all the dice show 2 or 4. And so on...
(3) All of the dice show exactly three numbers e.g. (1, 2, 4), no two of which can possibly add up to 7.
...

Let $p_k(n)$ be the probability that case k has occurred after n dice have been thrown. It is not hard to show that these probabilities satisfy a recurrence relation
$$\begin{pmatrix}p_1(n+1) \\ p_2(n+1) \\ p_3(n+1) \end{pmatrix} = \begin{pmatrix} \frac{1}{6} & 0 & 0 \\ \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & \frac{1}{2} \end{pmatrix} \begin{pmatrix}p_1(n) \\ p_2(n) \\ p_3(n) \end{pmatrix}$$
with $p_1(1)=1, p_2(1)=0=p_3(1)$. Since the matrix is lower triangular and the eigenvalues (along the diagonal) are distinct, it's not overly difficult to find the matrix powers to solve the recurrence relation. After a bit of algebra we find that the probability of obtaining at least one pair adding to 7 after n throws of the dice is
$$1-(p_1(n)+p_2(n)+p_3(n)) = 1-\frac{4}{2^{n-1}}+\frac{4}{3^{n-1}}-\frac{1}{6^{n-1}}$$

micromass