Caption: The probability distribution for a pair of thrown dice.

This is the probability distribution for true dice---ideal perfect dice---which real dice approximate closely---except when they don't---loaded dice.

The horizontal axis coordinate (i.e., the independent variable) S is the random variable and probability p(S)is the vertical axis coordinate (i.e., the dependent variable).

There are 11 outcomes (S values) and the probability distribution is symmetric about the maximum probability outcome S = 7---lucky number 7.

We can solve for the probability distribution:

1. As always, let's tackle the problem with generality. You have two dice: white and black to distinguish them, but otherwise idential. Let there be I faces with dot count running i=1,2,3,...,I. Let the probability of each face be p_i.

2. The probability of a distinct throw turning up faces i and j is p_i*p_j.

3. But what we what is the probability p_k of getting value k=i+j. The k values run 2,3,...,2I. So we have to sum the probabilities for the throws that yield this value. We get

     p_k = ∑ p_i*p_j = ∑ p_i*p_(k-i)  .

   The real hard part is setting the limits on the summation. If you organize, mentally at least, the distinct throws into an I X I table, the diagonals that bridge the axes have the distinct throws with constant k=i+j. A little thought and cursing shows that

     p_k = ∑_(i=1)^(i=k-1) p_i*p_(k-i)  for k∈[2,I+1] with k-1 terms
   
     p_k = ∑_(i=k-I)^(i=I) p_i*p_(k-i)  for k∈[I+1,2I] with 2I-(k-1) terms, 

   where the formulae agree for k=I+1 which has I terms.

4. Now specializing to the case of all p_i = 1/I, we get

     p_k = (k-1)*[1/(I**2)]  for k∈[2,I+1] 
   
     p_k = [2I-(k-1)]*[1/(I**2)]  for k∈[I+1,2I]  .  

   If we define k' = k - (I+1), we get k = k' + (I+1) and

     p_k' = (I-|k'|)*[1/(I**2)]  for k'∈[-(I-1),I-1] 

   which shows that p_k' is an even function of k'. Note that p_(k') from k'=0 to k'=±(I-1) runs 1/I, (I-1)/I**2, (I-2)/I**2, ..., 1/I**2.

5. Now specializing to ordinary dice with I=6 and p_i=1/6, we get

     p_k = (k-1)*[1/36]  for k∈[2,7]
   
     p_k = [12-(k-1)]*[1/36)]  for k∈[7,12]  
   
     p_k' = (6-|k'|)*[1/(36)]  for k'∈[-5,5]
   
     p_2 =1/36, p_3 =1/18, p_4 =1/12, p_5=1/9, p_6=5/36, p_7=1/6
   
     p_12=1/36, p_11=1/18, p_10=1/12, p_9=1/9, p_8=5/36, p_7=1/6 . 

   The results agree with the image---which is jolly good.