TITLE: Probability from randomized dice
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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. 

$i=j$. 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 
$$P( (i,i)) = \frac{ 2\binom{10}{5}}{\binom{12}{6}} \frac{1}{36}=\frac{1}{66}$$ 
$i\ne j$. 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 $7$.