Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal?

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 $\frac{146}{1296} = \frac{73}{648}$.

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) = $\dfrac{(1^2+2^2+3^2+ ....+6^2+5^2+4^2... +1^2)}{36^2}= \dfrac{146}{1296}=\dfrac{73}{648}$