2. There are already several red and black balls in the box. After adding some black balls, the red balls account for one quarter of the total number of balls; after adding some red balls, the number of red balls is half of the number of black balls. If the number of black balls and red balls added is the same, then the original ratio of the number of red balls to black balls in the box is $\qquad$ .
【Analysis】In this problem, after the first addition of balls, the change is in the black balls, while the red balls remain unchanged (the original ones), at this point the ratio of red balls to black balls is $1: 3 = 2: 6$; after the second addition of balls, the black balls remain unchanged (the number is the same as after the first change), but the red balls change, at this point the ratio of red balls to black balls is $1: 2 = 3: 6$. By comparing, we can see that the number of black balls is the same, but the red balls have increased by 1 part, which is exactly the number of red balls added in the second addition, and also the number of black balls added in the first addition. From this, we can find the answer to the problem.

【Solution】Solution: $1-\frac{1}{4}=\frac{3}{4}$
$$
\begin{array}{l}
\frac{1}{4}: \frac{3}{4}=\frac{1}{3}=1: 3=2: 6 \\
\frac{1}{2}=1: 2=3: 6 \\
3-2=1 \\
6-1=5
\end{array}
$$

The original number of red balls is 2 parts, and the original number of black balls is 5 parts.
Therefore: The original ratio of red balls to black balls in the box is $2: 5$.