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Let $A$ be the expected area. Then: $$\lim_{n \rightarrow \infty} \frac{n}{\ln n} (1 - A) = \frac{8}{3} \;.$$ This can be found in many places, e.g., this MathWorld article.

[AddedUpdated with comparisons between the above formula (Asymp) and the exact formula (Exact) found by quid.] Some evaluations: $$ \begin{array}{lcc} begin{array}{lcccc} n & & \mathrm{Asymp} & &\mathrm{Exact} \\ n=10 & : & A = 0.39\0.39 & : & 0.44 \\ n=100 & : & A = 0.89\0.89 & : & 0.88 \\ n=1000 & : & A = 0.98 & : & 0.98 \end{array} $$
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Let $A$ be the expected area. Then: $$\lim_{n \rightarrow \infty} \frac{n}{\ln n} (1 - A) = \frac{8}{3} \;.$$ This can be found in many places, e.g., this MathWorld article.

[Added] Some evaluations: $$ \begin{array}{lcc} n=10 & : & A = 0.39\\ n=100 & : & A = 0.89\\ n=1000 & : & A = 0.98 \end{array} $$
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Let $A$ be the expected area. Then: $$\lim_{n \rightarrow \infty} \frac{n}{\ln n} (1 - A) = \frac{8}{3} \;.$$ This can be found in many placeplaces, e.g., this MathWorld article.
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