, 2014

Solution:
Answer: $3$

We instead consider the probability the goon wins. The game clearly must last $n$ days. The probability the goon is not sent to jail on any of these $n$ days is then
$$\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1} \cdots \frac{2}{3}$$
If $n=2$ then the probability the goon wins is $\frac{4}{5} \cdot \frac{2}{3} = \frac{8}{15} > \frac{1}{2}$, but when $n=3$ we have $\frac{6}{7} \cdot \frac{8}{15} = \frac{16}{35} < \frac{1}{2}$, so the answer is $n=3$.

Alternatively, let $p_n$ be the probability that $2n$ townspeople triumph against $1$ goon. There is a $\frac{1}{2n+1}$ chance that the goon is jailed during the first morning and the townspeople win. Otherwise, the goon eliminates one townsperson during the night. We thus have $2n-2$ townspeople and $1$ goon left, so the probability that the town wins is $p_{n-1}$. We obtain the recursion
$$p_n = \frac{1}{2n+1} + \frac{2n}{2n+1} p_{n-1}.$$
By the previous question, we have the initial condition $p_1 = \frac{1}{3}$. We find that $p_2 = \frac{7}{15} < \frac{1}{2}$ and $p_3 = \frac{19}{35} > \frac{1}{2}$, yielding $n=3$ as above.