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:
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.
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.
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.