You have two dice, what is the probability of rolling a sum of 3 BEFOR

First of all please note that this question is far beyond the GMAT scope.

\(P(sum=3)=\frac{2}{36}=\frac{1}{18}\): either (1,2) or (2,1) out of total of 36 different combinations of two dice;

\(P(sum=7)=\frac{6}{36}=\frac{1}{6}\): (1,6), (2,5), (3,4), (4,3), (5,2), or (6,1);

\(P(other \ sum)=1-(\frac{1}{18}+\frac{1}{6})=\frac{7}{9}\), probability of sums: 4, 5, 6, 8, 9, 10, 11, and 12;

Winning scenarios:
{sum=3} - we have sum of 3 on the first roll of two dice - \(P_1=\frac{1}{18}\);
{other sum; sum=3} - on the first roll we have other sum and sum of 3 on the second roll - \(P_2=\frac{7}{9}*\frac{1}{18}\);
{other sum; other sum; sum=3} - \(P_3=(\frac{7}{9})^2*\frac{1}{18}\);
{other sum; other sum; other sum; sum=3} - \(P_4=(\frac{7}{9})^3*\frac{1}{18}\);
...

So probability of rolling a sum of 3 BEFORE rolling a sum of 7 would be the sum of the above infinite series:
\(P=\frac{1}{18}+\frac{7}{9}*\frac{1}{18}+(\frac{7}{9})^2*\frac{1}{18}+(\frac{7}{9})^3*\frac{1}{18}+...=\frac{1}{18}(1+\frac{7}{9}+(\frac{7}{9})^2+(\frac{7}{9})^3+...)=\frac{1}{18}(1+\frac{\frac{7}{9}}{1-\frac{7}{9}})=\frac{1}{18}(1+\frac{7}{2})=\frac{1}{4}\) (for geometric progression with common ratio \(|q|<1\), the sum of the progression: \(b_1, b_2, ...\) is \(Sum=\frac{b_1}{1-q}\).).

OR:

As \(P(sum=3)=\frac{2}{36}\) and \(P(sum=7)=\frac{6}{36}\) then getting the sum of 7 is 3 times more likely than getting the sum of 3, so the sum of 3 has 1 chance out of 4 to get first out of any number of tries, so \(P=\frac{1}{4}\) or \(P=\frac{\frac{2}{36}}{\frac{2}{36}+\frac{6}{36}}=\frac{1}{4}\).

Hope it's clear.