Suppose you have a pair of dice that have removable stickers for numbers on each of their 6 sides. Suppose that you unstick all 12 of the stickers from the dice and reapply them randomly to the 2 dice. You will still have 2 occurrences of each number 1 through 6. However, they may both occur on the same die. (For instance: after rearranging the stickers, you may have dice $d_1$ and $d_2$ with sides $d_1 = [1,2,2,4,4,6]$ and $d_2 = [1,3,3,5,5,6]$.)

Suppose now that you roll this randomized pair of dice. Is there a concise way to calculate the probability of each outcome? What is the probability of each possible outcome?

Just by working out some of the possible arrangements, it seems like $p(2)$ should be $\frac{1}{72}$ (which might not be correct), but the other probabilities are more difficult to compute this way.

  • $\begingroup$ How did you calculated $p(2)=\frac1{72}$ ? $\endgroup$ Jun 6, 2017 at 18:03
  • $\begingroup$ I made an error. Starting with the fact that $p(2)=p(2|1$s on same die$)+p(2|1$s on different dice$)$, the first term is zero, so it reduces to $p(2)=p(2|1$s on different dice$)=\frac{1}{36}*p(1$s on different die$)$. My mistake was in error in the combinatorics $p(1$s on different die$)=\frac{6}{11}$ (not $\frac{1}{2}$). $\endgroup$
    – DavidWayne
    Jun 6, 2017 at 18:42

3 Answers 3


Here's a formula that generalizes to more digits.

Let's label the dice A and B, and write $(i,j)$ for the getting $i$ on A and $j$ on $B$. Consider the following two cases.

  1. $i=j$. This can only happen if we have exactly one $i$ on each die. The probability of this is $\frac{ 2\binom{10}{5}}{\binom{12}{6}}$ (chose one of the two $i$'s on A and the other on B, then choose $5$ more numbers from the remaining $10$ for die A). Once we have exactly one $i$ on each die, the probability of both landing $i$, is $\frac 16 \times\frac 16$. Therefore $$P( (i,i)) = \frac{ 2\binom{10}{5}}{\binom{12}{6}} \frac{1}{36}=\frac{1}{66}$$
  2. $i\ne j$. Since this is independent of the choice of $i$ and $j$, and there are exactly $6*5=30$ such pairs, the probability is equal to $$ \frac {1}{30} (1- \sum_{i} P( (i,i) )=\frac{1}{30}(1-\frac {1}{11})=\frac {1}{33}.$$

Now it remains to find the probability of a sum equal to $k$.

a. If $k\le 7$, there are exactly $k-1$ ways to write it: $ (1,k-1),(2,k-2),\dots, (k-1,1)$. Now if $k$ is odd, then in all of these $i\ne j$. Therefore the probability is $(k-1)/33$. If $k$ is even, then exactly one of these is of the form $i=j$, therefore the answer is $(k-2)/33 + 1/66$.

b. Finally if $k \ge 8$, and $(i,j)$ is such that $i+j=k$. Then $(7-i,7-j)$ has exactly the same probability, and has sum between $\{2,\dots,6\}$. Therefore the probability to get $k$ is the same as the probability to get $14-k$.

In other words, the distribution of the sum is symmetric about $7$.


The problem can be viewed as picking $2$ numbers from list of twelve numbers $1$,$1$,$2$,$2$,$3$,$3$,$4$,$4$,$5$,$5$,$6$,$6$ and looking at their sum.

Now there are $12 \cdot 11 = 132$ ways to pick two numbers from the list. In how many ways we can pick two same (specific) numbers? There are apparently exactly $2$ ways to do it (depends on which of the same two numbers you pick first). In how many ways we can pick two different numbers in given order (e.g. $1,2$)? There are exactly $2\cdot2 = 4$ ways to do it ($2$ two choices for first pick, $2$ choices for second pick).

Now for the sums, in how many ways we can get sum $2$? That is possible only as $1+1$, which by reasoning above can happen $2$ times (we have to pick two same numbers), so probability is $$p(2)=\frac{2}{132}=\frac{1}{66}$$

In how many ways can we get sum $3$? We have $3=1+2=2+1$ and again by reasoning above there is overall $4+4=8$ possibilities. This is because to get $2+1$ we need to pick two different numbers (and we know that is possible in $4$ ways), same for $1+2$. So probability is $$p(3)=\frac{8}{132}=\frac{2}{33}$$

Similarly $4=1+3=2+2=3+1$ with $4+2+4=10$ possibilities, so probability is $$p(4)=\frac{10}{132}=\frac{5}{66}$$

And so on...

  • $\begingroup$ Very nice answer; a very clever way of reducing it to a simpler problem. I have to give the best answer to @Fnacool, however, for providing a closed form solution for computing the probabilities. $\endgroup$
    – DavidWayne
    Jun 6, 2017 at 20:33
  • $\begingroup$ @DavidWayne That is alright, I would prefer closed form solution as well. $\endgroup$
    – Sil
    Jun 7, 2017 at 17:51

I wrote a basic Python program to find the distribution of the different values. There are ${12 \choose 6} = 924$ ways to select six values out of twelve, so the number of possible value distributions equals $\frac{924}{2} = 462$. For each of these combinations, I increased the probability of the 36 combinations of the two dice, resulting in the following probabilities:

 2 - 0.0152 -  1/66
 3 - 0.0606 -  4/66
 4 - 0.0758 -  5/66
 5 - 0.1212 -  8/66
 6 - 0.1364 -  9/66
 7 - 0.1818 - 12/66
 8 - 0.1364 -  9/66
 9 - 0.1212 -  8/66
10 - 0.0758 -  5/66
11 - 0.0606 -  4/66
12 - 0.0152 -  1/66

For instance, to get a 2 we need the two 1's to be on two different dice (probability $\frac{6}{11}$), in which case the probability of hitting a 2 equals $\frac{1}{36}$. The total probability of getting a 2 thus equals $\frac{5}{11} \cdot 0 + \frac{6}{11} \cdot \frac{1}{36}=\frac{1}{66} \approx 0.0152$. The Python program:

import itertools
import math
from collections import defaultdict

values = [x for x in range(1, 13)]
d = defaultdict(int)

for v in itertools.combinations(values, 6):
    w = [x for x in values if x not in v]
    for i in range(6):
        for j in range(6):
            d[math.ceil(v[i]/2) + math.ceil(w[j]/2)] += 1/36/924

for k in d:
    print(k, d[k])
  • 1
    $\begingroup$ It looks like all your probabilities are divisible by $1/66$ (alternatively, d[k] is always a multiple of 14.) It's not obvious why this should be. $\endgroup$ Jun 6, 2017 at 18:14
  • $\begingroup$ @MichaelLugo That is correct. I updated the table to make this clear. $\endgroup$
    – jvdhooft
    Jun 6, 2017 at 18:20

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