Let $K{}$ be an equilateral triangle of unit area, and choose $n{}$ independent random points uniformly from $K{}$. Let $K_n$ be the intersection of all translations of $K{}$ that contain all the selected points. Determine the expected value of the area of $K_n.$
1. **Rescaling the Problem:**
   Let \( K \) be an equilateral triangle with unit area. The height \( h \) of \( K \) is given by:
   \[
   h = \sqrt{\frac{4 \cdot \text{Area}}{\sqrt{3}}} = \sqrt{\frac{4 \cdot 1}{\sqrt{3}}} = \sqrt{\frac{4}{\sqrt{3}}} = \frac{2}{\sqrt[4]{3}}
   \]
   For simplicity, we rescale \( K \) to have height 1. Let \( H_n \) denote the height of \( K_n \), the intersection of all translations of \( K \) that contain all \( n \) points.

2. **Height Calculation:**
   Let \( X_1, X_2, X_3 \) be the distances from the sides of \( K \) to the respective parallel sides of \( K_n \). Then:
   \[
   H_n = 1 - X_1 - X_2 - X_3
   \]
   We need to compute \( E(H_n^2) \).

3. **Joint Distribution of \( X_1, X_2, X_3 \):**
   The probability that all \( n \) points lie within a subtriangle of \( K \) with height \( 1 - t_1 - t_2 - t_3 \) is:
   \[
   P(X_i > t_i \forall i) = \left( (1 - t_1 - t_2 - t_3)^2 \right)^n
   \]
   Taking partial derivatives, the joint density function of \( X_1, X_2, X_3 \) is:
   \[
   f(t_1, t_2, t_3) = 2n(2n-1)(2n-2) (1 - t_1 - t_2 - t_3)^{2n-3}
   \]
   where \( t_i > 0 \) and \( t_1 + t_2 + t_3 < 1 \).

4. **Transformation to \( S_1, S_2, S_3 \):**
   Let \( S_1 = X_1 \), \( S_2 = X_1 + X_2 \), \( S_3 = X_1 + X_2 + X_3 \). By the Lebesgue transformation theorem, the joint density of \( S_1, S_2, S_3 \) is:
   \[
   g(s_1, s_2, s_3) = 2n(2n-1)(2n-2) (1 - s_3)^{2n-3}
   \]
   where \( 0 < s_1 < s_2 < s_3 < 1 \).

5. **Distribution of \( S_3 \):**
   The density function of \( S_3 \) is obtained by integrating out \( s_1 \) and \( s_2 \):
   \[
   h(s) = 2n(2n-1)(2n-2) (1 - s)^{2n-3} \frac{1}{2} s^2
   \]

6. **Expected Value Calculation:**
   We need to compute \( E(H_n^2) = E((1 - S_3)^2) \):
   \[
   E((1 - S_3)^2) = \int_0^1 (1 - s)^2 h(s) \, ds
   \]
   Substituting \( h(s) \):
   \[
   E((1 - S_3)^2) = \int_0^1 2n(2n-1)(2n-2) (1 - s)^{2n-1} \frac{1}{2} s^2 \, ds
   \]
   Simplifying:
   \[
   E((1 - S_3)^2) = n(2n-1)(2n-2) \int_0^1 (1 - s)^{2n-1} s^2 \, ds
   \]

7. **Integration:**
   Expanding \( (1 - s)^2 = 1 - 2s + s^2 \):
   \[
   \int_0^1 (1 - s)^{2n-1} s^2 \, ds = \int_0^1 (1 - s)^{2n-1} s^2 \, ds
   \]
   Using the Beta function \( B(x, y) = \int_0^1 t^{x-1} (1-t)^{y-1} \, dt \):
   \[
   \int_0^1 (1 - s)^{2n-1} s^2 \, ds = B(3, 2n)
   \]
   \[
   B(3, 2n) = \frac{\Gamma(3) \Gamma(2n)}{\Gamma(3 + 2n)} = \frac{2! (2n-1)!}{(2n+2)!} = \frac{2 (2n-1)!}{(2n+2)(2n+1)(2n)!} = \frac{2}{(2n+2)(2n+1)(2n)}
   \]

8. **Final Calculation:**
   \[
   E((1 - S_3)^2) = n(2n-1)(2n-2) \cdot \frac{2}{(2n+2)(2n+1)(2n)} = \frac{(2n-1)(2n-2)}{(2n+1)(2n+2)}
   \]

The final answer is \(\boxed{\frac{(2n-1)(2n-2)}{(2n+1)(2n+2)}}\).