Who keeps the ball

Idea. Only the distance from the ball to you matters, and on a circle of five that distance is 0, 1 or 2. Three unknowns, three equations.

Step 1: the unknowns. Let $a_k$ be the probability that you end up keeping the ball when it is $k$ seats away from you, for $k=0,1,2$.

Step 2: the equations.

• At distance 0 (you hold it): you keep it with probability $\frac{1}{3}$; otherwise it moves to distance 1.
• At distance 1: the holder keeps it (you lose), passes it to you (distance 0), or passes it away (distance 2).
• At distance 2: the holder keeps it, passes it to distance 1, or passes it to the other person at distance 2 (on a circle of five, the two people at distance 2 sit next to each other).

$$\begin{array}{rl}{a}_0& =\frac{1}{3}+\frac{2}{3}{a}_1,\\ a_1& =\frac{1}{3}{a}_0+\frac{1}{3}{a}_2,\\ a_2& =\frac{1}{3}{a}_1+\frac{1}{3}{a}_2.\end{array}$$

Step 3: solve. The last equation gives $a_2=\frac{1}{2}{a}_1$. Then the second gives $a_1=\frac{1}{3}{a}_0+\frac{1}{6}{a}_1$, so $a_1=\frac{2}{5}{a}_0$. The first becomes $$a_0=\frac{1}{3}+\frac{4}{15}{a}_0,a_0=\frac{5}{11}.$$

So you keep the ball with probability $\frac{5}{11}\approx 0.45$. By the same symmetry, a neighbour of yours keeps it with probability $a_1=\frac{2}{11}$ (you are at distance 1 from them), and each of the two people opposite with probability $a_2=\frac{1}{11}$. As a check, $\frac{5}{11}+2\cdot \frac{2}{11}+2\cdot \frac{1}{11}=1$.