Expected Steps Between Circle Intersections
Two unit circles are centered at $(0,0)$ and $(1,0)$. They overlap and share exactly two intersection points -- call them $A$ (the upper one) and $B$ (the lower one).
A particle starts at $A$. At each step, it picks one of the two circles uniformly at random (prob $\frac{1}{2}$ each), then jumps to a uniformly random point on that circle's boundary. The process stops when the particle lands at $B$.
Under the natural discrete interpretation (each step can only land at one of the two intersection points), what is the expected number of steps to reach $B$?
Hints
- First find the two intersection points explicitly -- they share an $x$-coordinate of $\frac{1}{2}$ and are symmetric about the $x$-axis. Then think about what 'uniformly random point on the boundary' actually means for the state space.
- Note that both circles pass through both intersection points $A$ and $B$. So whichever circle the particle picks, it has equal probability of landing at either intersection. This reduces to a two-state problem.
- Let $E$ be the expected steps to reach $B$ from $A$. At each step, $P(\text{reach } B) = \frac{1}{2}$, giving the first-step equation $E = \frac{1}{2}(1) + \frac{1}{2}(1 + E)$.
Worked Solution
How to Think About It: The first thing to notice is that this problem is degenerate under a purely continuous interpretation -- on a continuous circle, the probability of landing on any single exact point is zero, so the particle would never reach $B$ in finite expected time. The problem only makes sense if we interpret each step as jumping between the discrete states: the two intersection points. Both circles pass through both $A$ and $B$, so when you pick a circle and jump to a uniformly random point on it, the relevant outcome is which of the two shared points you land on. That reduces the problem to a simple geometric waiting time.
Quick Estimate: At each step you pick a circle (each with prob $\frac{1}{2}$) and then land at either intersection point with equal probability. So regardless of which circle you pick, you jump to $A$ or $B$ with prob $\frac{1}{2}$ each. The probability of reaching $B$ on any given step from $A$ is $\frac{1}{2}$. That is a Geometric$(\frac{1}{2})$ distribution, so the expected number of steps is $\frac{1}{1/2} = 2$. The answer feels right -- you expect to flip a fair coin twice before getting heads.
Approach: Set up a first-step equation using the state structure, then solve.
Formal Solution:
First, find the intersection points. Solve $x^2 + y^2 = 1$ and $(x-1)^2 + y^2 = 1$ simultaneously: $$x^2 = (x-1)^2 \implies 2x = 1 \implies x = \frac{1}{2}$$ $$y = \pm\frac{\sqrt{3}}{2}$$
So $A = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $B = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$.
Under the discrete interpretation, at each step the particle chooses one of the two circles (each with prob $\frac{1}{2}$) and lands at one of the two intersection points of that circle with equal probability $\frac{1}{2}$. Since both circles pass through both $A$ and $B$, the choice of circle does not affect the outcome distribution: the particle moves to $A$ with prob $\frac{1}{2}$ and to $B$ with prob $\frac{1}{2}$, independently of where it currently is.
Let $E$ be the expected number of steps to reach $B$ starting from $A$. By the first-step equation: $$E = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot (1 + E)$$ $$E = \frac{1}{2} + \frac{1}{2} + \frac{1}{2}E$$ $$\frac{1}{2}E = 1$$ $$E = 2$$
Alternatively, the number of steps is Geometric$\left(\frac{1}{2}\right)$, so $E[T] = \frac{1}{p} = 2$ directly.
Answer: The expected number of steps is $\boxed{2}$.
Intuition
The deeper lesson here is about recognizing when a problem is degenerate. Under a truly continuous uniform distribution on the circle, the probability of hitting any specific point is exactly zero -- the particle wanders forever without ever reaching $B$. This is not a trick; it is just measure theory. When a problem seems to ask for the expected time to hit a zero-measure set, you have to ask: what was the problem setter really intending? The discrete interpretation -- where the state space is just the two intersection points -- is the only version that gives a finite, interesting answer. Spotting this early in an interview saves you from going down a dead end.
Once you commit to the discrete interpretation, the problem collapses to something almost trivially simple: both circles share both intersection points, so the choice of circle is irrelevant. Every step is an independent fair coin flip between $A$ and $B$. The expected first passage time for a fair coin is 2. The broader principle: when a random walk has a symmetric structure that makes the transition probabilities identical regardless of which 'mechanism' you pick, the labels on those mechanisms are noise -- you can ignore them and treat the walk directly.