I set it up as a Markov chain pretty quickly, which felt right.
Use the linearity of expectation by conditioning on the first spin's outcome. Define the expected additional spins needed from each possible first region, then solve the resulting system of equations or use the formula for the expected number of trials to get a different outcome.
Pro tip: After computing the answer, verify it by checking that the expected value lies between the minimum and maximum possible values (2 and 1 + 1/min(p_i) respectively). This sanity check catches arithmetic errors and shows rigor.
Let X be the number of spins until at least two different regions have been landed on. Note that X is at least 2.
After the first spin, we have seen one region. The expected additional spins depends on which region was seen first, so define E_i as the expected additional spins given that the first spin landed in region i.
For each region i, if the next spin is in a different region, we stop; if it is the same region, we need 1 + E_i more spins. Thus E_i = 1 + p_i * E_i, where p_i is the probability of landing in region i.
Solve E_i = 1/(1 - p_i). Then the overall expected number of spins is 1 + sum_i p_i * E_i, since the first spin is always needed.
Plug in the given probabilities: p1=1/6, p2=1/3, p3=1/2. Compute the sum and simplify. Check that the result is reasonable (between 2 and 3).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.