Coin tossing, again

Let $h$ and $t$ denote the number of heads and tail flipped, respectively.

• $P(B) = P(t=0) + P(t=1) = \dfrac{1}{2^n} + \dfrac{n}{2^n}$
• $P(A) = 1 - (P(t=0) + P(h=0)) = 1 -\left(\dfrac{1}{2^n} + \dfrac{1}{2^n}\right) = \dfrac{2^n-2}{2^n}$
• $P(B\mid A) = \dfrac{P(B\cap A)}{P(A)} = \dfrac{P(t=1)}{P(A)} = \dfrac{^n\!/\!_{2^n}}{^{2^n-2}\!/\!_{2^n}} = \dfrac{n}{2^n-2}$

By definition, if $A$ and $B$ are independent then $P(B) = P(B\mid A)$ . Thus we have $\begin{aligned} \frac{1}{2^n} + \frac{n}{2^n} &= \frac{n}{2^n-2} \\ \frac{2^n-2}{2^n} + \frac{n(2^n-2)}{2^n} &= n \\ 1 + n - \frac{2n + 2}{2^n} &= n \\ \frac{2n + 2}{2^n} &= 1 \\ 2n + 2 &= 2^n \end{aligned}$

By inspection, $n=\boxed{3}$ is a solution. Since $y=2x+2$ intersects $y=2^x$ at only one point in the first quadrant, this is the unique solution.