<s>[INST] Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal? Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal?

Since the same throws can result if either both of them get 1 or both of them get 2 or both of them get 3 or both of them get 4 or both of them 5 or both of them get 6.
So I calculated probability as $\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}=\frac{1}{6}$
But my answer is wrong and the correct answer is $\frac{73}{648}$.Please help me with the correct approach to solve it.Thanks. 
 [/INST] Deducing, like Sherlock Holmes, that the question is:
Two people throw two dice each. What is the probability that both get the same sum ?
Ways of throwing sums of $2 - 12$ with $2$ dice follows the pattern
$1-2-3-4-5-6-5-4-3-2-1$
So P(both get the same sum) = $\dfrac{(1^2+2^2+3^2+ ....+6^2+5^2+4^2... +1^2)}{36^2}= \dfrac{146}{1296}=\dfrac{73}{648}$
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