Jae and Yoon are playing SunCraft. The probability that Jae wins the $n$-th game is $\frac{1}{n+2}.$ What is the probability that Yoon wins the first six games, assuming there are no ties?
1. The probability that Jae wins the $n$-th game is given by:
   \[
   P(\text{Jae wins } n\text{-th game}) = \frac{1}{n+2}
   \]

2. Therefore, the probability that Yoon wins the $n$-th game is:
   \[
   P(\text{Yoon wins } n\text{-th game}) = 1 - \frac{1}{n+2} = \frac{n+1}{n+2}
   \]

3. To find the probability that Yoon wins all of the first six games, we need to multiply the probabilities of Yoon winning each individual game from the 1st to the 6th game:
   \[
   P(\text{Yoon wins all 6 games}) = \left(\frac{2}{3}\right) \cdot \left(\frac{3}{4}\right) \cdot \left(\frac{4}{5}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{6}{7}\right) \cdot \left(\frac{7}{8}\right)
   \]

4. Simplify the product step-by-step:
   \[
   \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8}
   \]

   Notice that the terms in the numerator and denominator cancel out sequentially:
   \[
   = \frac{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}
   \]

   Cancel out the common terms:
   \[
   = \frac{2}{8} = \frac{1}{4}
   \]

5. Therefore, the probability that Yoon wins all of the first six games is:
   \[
   \boxed{\frac{1}{4}}
   \]