A chi-square test is almost pointless; obviously this is almost impossible to happen just by coincidence, if the probability of equal start rolls is 1/18 = 5.556 %. But I have done it anyway.

Every one of the n = 630 games falls in one of two classes:
class 1: first and second roll equal (expected probability p1 = 1/18)
class 2: first and second roll not equal (p2 = 17/18)

Y1 = 218 is the number of games in class 1.
Y2 = 412 the games in class 2.

now is calculated:
V = (Y1 - n * p1)^2 / (n * p1) + (Y2 - n * p2)^2 / (n * p2)
= (218 - 35)^2 / 35 + (412 - 595)^2 / 595
= 1013.1

If the dice were not biased (probabilities p1 and p2 as expected), then the value V would be lower than 6.635 with probability 99 %, and lower than 10.84 with probability 99.9 %.

(edit:) see this table for the values to compare V with. With 2 classes one must look in the line with DF=1 (degrees of freedom).