Two persons each make a single throw with pair of dice. The probability that the throws are equal, is

$\frac{73}{648}$

The total number of outcomes when each of the two persons throws a pair of dice is

$6^2×6^2=1296$

Throws are equal means that the sum of the numbers on two dice is same. The sum varies from2 to 12. The numbers of ways to get the sum 2, 3, 4, ...., 12 in a single throw of a pair of dice are

Now, if each person can throw the sum i and $a_i$ ways, then both of them will throw the sum i in $a^2_i$ ways

Therefore, the number of ways in which the throws of two persons are equal is

$2(1^2 + 2^2 + 3^2 + 4^2 + 5^2 ) + 6^2 = \frac{2×5 ×(6)(11)}{6}+36 = 146$

Hence, required probability $=\frac{146}{1296}=\frac{73}{648}$