Expected Steps Between Circle Intersections
Two unit circles are centered at $(0,0)$ and $(1,0)$. They meet at exactly two 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 (probability $\frac{1}{2}$ each), then jumps to one of that circle's two intersection points with the other circle, choosing each with probability $\frac{1}{2}$ (it may land back where it is). The process stops when the particle lands at $B$.
What is the expected number of steps to reach $B$?
Hints
- Both circles pass through both $A$ and $B$. Does the choice of circle change where the particle can land, or with what probability?
- 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 heuristic is strip away the irrelevant mechanism. The circle choice looks important, but both circles pass through both $A$ and $B$, so whichever circle is picked, the particle lands on $A$ or $B$ with probability $\frac{1}{2}$ each. Once you see the circle label is noise, each step is a fair coin flip for "reached $B$", and the question becomes a geometric waiting time.
(Why the jump is restricted to the intersection points: if the particle instead jumped to a uniformly random point anywhere on a circle, the chance of landing exactly on the single point $B$ would be $0$ on every step, and the particle would never stop. The intersection-point version is the meaningful one.)
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 $B = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$.
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 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 lesson is to spot when a piece of the setup is pure decoration. Two circles, a random circle choice, explicit coordinates for $A$ and $B$: none of it matters, because both circles pass through both points, so every step is the same fair coin flip between $A$ and $B$ regardless of which circle is chosen. The expected first-passage time for a fair coin is 2.
A related trap is worth knowing: if the jump were to a uniformly random point anywhere on the circle, the probability of hitting the single point $B$ would be zero and the expected time would be infinite. Checking whether the target has positive probability is the first thing to do in any hitting-time problem. The broader principle: when the transition probabilities are identical whichever 'mechanism' fires, the mechanism labels are noise and you can model the walk directly.