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. 

If I'm interpreting this correctly...
There are a total of $36$ combinations you can get when rolling a pair of $6$-sided dice.  There is $1$ way to get a $2$, $2$ ways to get a $3$, $3$ ways to get a $4$, etc.
You have a total of $36\cdot 36= 1296$ events.  Now, we do a bit of counting.  There is $1$ way both people can get a $2$, $2 \cdot 2$ ways both people can get a $3$, $3 \cdot 3$ ways both people can get a $4$, etc.
Thus, the number of successes is (by symmetry)
$2(1+4+9+16+25)+36 = 146$.
The answer you are looking for is $\frac{146}{1296} = \frac{73}{648}$.
