Stumbled upon this question in a game of dice where a point is dealt for each pair that makes seven. In the game there were a total of six dice that were to be thrown up to three times. Each time a pair of dice made seven 1+6, 2+5 or 4+3), the pair(s) was taken out from the rest, a point was dealt and you got to roll the remaining dice. You can not use dice from the excluded pairs to make up new ones. You win the game by getting the most points/pairs in 3 rounds.


round one [1] [1] [1] [3] [5] [6] - one pair make 7 (1+6)

round two [1] [3] [4] [2] - one pair make 7 (3+4)

round one [2] [6] - no pairs make 7

points awarded: 2

Out of curiosity I have been trying to calculate the probabilities of getting three pairs of sevens in the first round?

  • $\begingroup$ We have six dice and we throw them all together in the first round. Let's say the faces are $3,4,5,2,6,6$. Because there are two pairs that make $7$, we remove the pairs and continue with two dice. Round 2 and 3 continues this way. What is the definition of winning? Is it the situation that we're left with no dice at the end? $\endgroup$
    – gunes
    Nov 10, 2019 at 18:18
  • $\begingroup$ and how is it possible to roll three dice following those rules? $\endgroup$
    – carlo
    Nov 10, 2019 at 18:39
  • 1
    $\begingroup$ Your question is unclear. Your title refers to three dice while your body text refers to six. You describe a game where the number of dice change, altering the probabilities. At which point in that game are you calculating the chance of getting two dice that add to 7? $\endgroup$
    – Glen_b
    Nov 10, 2019 at 21:02
  • $\begingroup$ Sorry if the original edit was confusing, I have tried to clarify. The game description is just for context. The question is in the title and reformulated in the last paragraph. $\endgroup$
    – joho
    Nov 13, 2019 at 14:28

1 Answer 1


Consider a sequence of $2k$ dice each with the possible values $1, 2, \ldots, 2n.$ (The question concerns $k=n=3.$) The possible pairs are $\{1,n\},$ $\{2,2n-1\},$ and so on, through $\{n,n+1\}.$ Denote such a pair by its smallest value $i$ and for each $i$ let $k_i\gt 0$ be the number of such pairs that can be located in the sequence. We need to count the number of equally probable sequences for which $k_1+k_2+\cdots + k_n = k.$

Such a sequence is determined by (a) which pairs occur in it and (b) where in the sequence of $2k$ values each such pair occurs. The permutation group on $2k$ elements acts on the set of such sequences. For all $i,$ the stabilizer of such a sequence permutes the $k_i$ values of $i$ among themselves and the $k_i$ values of $2n+1-i$ among themselves. Thus, applying the method described at https://stats.stackexchange.com/a/415878/919, the number of ways of producing a sequence designated by $\mathrm{k}=(k_1,k_2,\ldots,k_n)$ is

$$p(\mathrm{k}) = \frac{(2k)!}{(k_1!)^2(k_2!)^2\cdots(k_n!)^2}.$$

Thus, the chance is obtained by summing $p(\mathrm{k})$ over all possible $\mathrm{k}$ whose components sum to $k$ and dividing by the total number of sequences, $(2n)^{2k}.$

These possibilities correspond to the weak compositions of $k$ into $n$ parts, which number $\binom{k+n-1}{n-1}.$ However, the amount of calculation is smaller than this, because $p(\mathrm{k})$ does not depend on the order of the $k_i.$ We may therefore do the calculation for all $k_1\ge k_2\ge \cdots \ge k_n \ge 0$ (giving a partition of $k$), multiplying each by the number of distinct re-orderings of $\mathrm k.$ Such sequences correspond to the Ferrers diagrams for $n$ having at most $k$ rows. They are relatively easy to enumerate.

With $k=n=3,$ we have $\binom{3+3-1}{3-1}=10$ possibilities for $\mathrm k,$ but they fall into just three groups corresponding to the partitions $3 = 2+1 = 1+1+1:$

$$p(3,0,0)=p(0,3,0)=p(0,0,3) = \frac{6!}{3!^2} = 20;$$ $$p(2,1,0)=p(2,0,1)=p(1,2,0)=p(1,0,2)=p(0,2,1)=p(0,1,2) = \frac{6!}{2!^21!^2}= 180;$$ $$p(1,1,1) = \frac{6!}{1!^21!^21!^2} = 720.$$

The answer therefore is

$$\Pr(\text{three pairs}) = \frac{3\times 20 + 6\times 180 + 1\times 720}{6^6} = \frac{(5)(31)}{(2^4)(3^5)} \approx 3.9866\%.$$

You have likely figured out by now why I limited this analysis at the outset to dice with even numbers of sides: it eliminates the possibility that some values would be paired with themselves. (For instance, on a five-sided die with values 1 through 5, 3+3=6 forms a pair.) A similar analysis can be carried out for the odd-sided dice. It results in a slightly more complicated formula due to that self-pairing possibility.

  • $\begingroup$ I noticed I got a different answer, 9% vs 4%, but I'm confident enough in my numerical method to post it as an answer. I would appreciate it if you could point out my error. Thanks. $\endgroup$
    – Ron Jensen
    Nov 13, 2019 at 17:49
  • $\begingroup$ @Ron I haven't thoroughly checked my general answer, but I did check the specific answer with a brute force enumeration of all possibilities. At least one mistake in your post is that many sequences without three pairs will sum to 21, such as 5,5,5,4,1,1, which has no pairs at all. Thus you substantially overcount. Here is my implementation: n <- 3; k <- 3; X <- expand.grid(rep(list(seq_len(2*n)), 2*k)); p <- function(x, d=n) { x <- sort(x); sum((x + rev(x))[seq_len(d)] == 2*d+1) }; table(apply(X, 1, p)) $\endgroup$
    – whuber
    Nov 13, 2019 at 19:04
  • $\begingroup$ Ah, that does explain it. My hypothesis that a sum of 21 is sufficient is wrong. I'll delete my answer, then. Thanks. $\endgroup$
    – Ron Jensen
    Nov 13, 2019 at 19:15
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    $\begingroup$ @Ron If you would like to explore the situation in more detail, this code (which follows the creation of data frame X in my previous comment) breaks the population down into all possible pair patterns: pattern <- function(x, d=n) { x <- sort(x); paste0(x[which((x + rev(x))[seq_len(d)] == 2*d + 1)], collapse="") }; table(apply(X, 1, pattern)) $\endgroup$
    – whuber
    Nov 13, 2019 at 19:19

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