Entropy of two dice

First, let's figure out probability of each even occurring. Dice has numbers 1,2,3,4,5 and 6. The sum of two dice therefore takes values from 2 to 12. To get sum = 2, dice $a$ must show 1 and dice $b$ must show 1 too. The probability of such event occurring is $\frac{1}{6}*\frac{1}{6} = \frac{1}{36}$ . To get sum = 3, dice $a$ must show 1 while dice $b$ must show 2. However, the sum = 3 also when dice $a$ is 2 while dice $b$ is 1. Therefore, probability of getting the sum = 3 is $\frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} = \frac{2}{36}$ . If we continue with this logic, the sum with highest probability will be 7: there are 6 possible combinations of two numbers from 1 to 6 to give the sum = 7. After 7, probability of next sums will decline. For example, to get sum = 12, both dice must show 6, which is an event with probability $\frac{1}{6}*\frac{1}{6} = \frac{1}{36}$ . The final formula to calculate entropy is $\frac{1}{36}log_2(\frac{36}{1}) + \frac{2}{36}log_2(\frac{36}{2}) + \frac{3}{36}log_2(\frac{36}{3}) + ... + \frac{6}{36}log_2(\frac{36}{6}) + \frac{5}{36}log_2(\frac{36}{5}) + \frac{4}{36}log_2(\frac{36}{4}) + ... + \frac{1}{36}log_2(\frac{36}{1}) = 3.274402$

Solving the problem using Mathematica:

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pmf[x_] = (6 - Abs[x - (6 + 1)]) / 36;
-Sum[pmf[x] * Log2[pmf[x]], {x, 2, 12}] // N