Solution:

When two dice are thrown, the total number of outcomes is 36. The possible outcomes of throwing a pair of dice are given below:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Out of these 36 outcomes, there are 4 ways in which the sum is 9: (3,6), (4,5), (5,4), and (6,3). There are 6 ways in which the sum is 7: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).

Case 1: The first roll is 9

The probability of throwing a 9 on the first roll is 4/36 or 1/9. In this case, the person wins the game because he has thrown 9 before 7.

Case 2: The first roll is 7

The probability of throwing a 7 on the first roll is 6/36 or 1/6. In this case, the person loses the game because he has thrown 7 before 9.

Case 3: The first roll is neither 7 nor 9

The probability of throwing a number other than 7 or 9 on the first roll is 1 - 1/9 - 1/6 = 4/9. In this case, the person throws again. If he throws a 9 before a 7, he wins. If he throws a 7 before a 9, he loses. The probability of winning in this case is given by:

P(Win) = (4/36) + (4/9)(1/9) + (4/9)(4/9)(1/9) + ...

This is a geometric series with first term a = 4/36 and common ratio r = (4/9)(1/9) = 4/81. The sum of an infinite geometric series with first term a and common ratio r is given by:

S = a/(1 - r)

Substituting the values of a and r, we get:

P(Win) = (4/36)/(1 - 4/81) = (2/9)/(77/81) = 2/5

Therefore, the chance that the person throws 9 before 7 is 2/5.