AMC 12 Mock Exam 2 Problem 21

Only the number $k$ of switches that are on matters. From state $k$, the next flip moves to $k+1$ with probability $\frac{4-k}{4}$ and to $k-1$ with probability $\frac k4$.

Let $t_k$ be the expected number of minutes to go from $k$ switches on to $k+1$ switches on for the first time. After one minute we either succeed, or (with probability $\frac k4$) we drop to $k-1$, from which we need $t_{k-1}$ minutes to return to $k$ and then $t_k$ more. So $$t_k=1+\frac k4\left(t_{k-1}+t_k\right),\qquad\text{i.e.,}\qquad t_k=\frac{1+\frac k4t_{k-1}}{1-\frac k4}.$$ Starting from $t_0=1$: $t_1=\frac{1+\frac14}{\frac34}=\frac53$, $t_2=\frac{1+\frac12\cdot\frac53}{\frac12}=\frac{11}{3}$, $t_3=\frac{1+\frac34\cdot\frac{11}3}{\frac14}=15$.

The expected total is $1+\frac53+\frac{11}3+15=\boxed{\tfrac{64}{3}}$.

(Ignoring the backward moves, as if each minute's flip were always useful until it hit a new switch, gives the coupon-collector value $\frac44+\frac43+\frac42+\frac41=\frac{25}{3}$, which is far too small.)