Four Corners is a popular children's game, in which a person who is "it" is blindfolded in a four-corner room, with everyone else attempting to avoid being found. Whoever is "it" spins around, and randomly chooses a corner. The other players choose a corner to stand in, and if their corner gets chosen by whoever is "it," they are out. The game ends when the last person is found.
Imagine only two people playing the game, one person who is "it" and one who is trying not to lose. If the corner is chosen at random for both the "it" person and the person running, how many turns is the game expected to last?
Here's what I have:
Expected value calculation: $n_1p_1+n_2p_2+n_3p_3+...$
$n_k$ is the number of turns, so $n_k=k$.
$p_1=\frac{1}{4}\cdot(\frac{3}{4})^0$
$p_2=\frac{1}{4}\cdot(\frac{3}{4})^1$
$p_k=\frac{1}{4}\cdot(\frac{3}{4})^{k-1}$
Now multiply and sum up everything:
$\lim_{n\rightarrow\infty}\sum ^{n}_{k=0}\frac{1}{4}k(\frac{3}{4})^{k-1}$
Now I'm stuck. What's next?
If the expected number of turns is $n$, on the first turn you find the opponent with probability $\frac 14$ and otherwise you are at the same place you started, so it will take $n$ turns. This gives $$n=\frac 14(1) + \frac 34 n\\n=4$$ Otherwise, you can sum the geometric series and get the same result.