, 2014

Solution:
Answer: $6$

By a similar inductive argument, the probability for a given $n$ is
$$p_{n} = \frac{n!}{(2n+1)!!}.$$
Clearly this is decreasing in $n$. It is easy to see that
$$p_{5} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}{3 \cdot 5 \cdot 7 \cdot 9 \cdot 11} = \frac{8}{693} > 0.01$$
and
$$p_{6} = \frac{6}{13} p_{5} = \frac{48}{693 \cdot 13} < 0.01$$
Hence the answer is $n=6$. Heuristically, $p_{n+1} \approx \frac{1}{2} p_{n}$ for each $n$, so arriving at these estimates for the correct answer of $n$ is not difficult.