To put it in a different way, construct a Markov chain with these states and transitions: (A_0) -> (A_1) -> (A_2) -> (A_3) -> (A_4) State A_i = you have i uni

To put it in a different way, construct a Markov chain with these states and transitions: (A_0) -> (A_1) -> (A_2) -> (A_3) -> (A_4) State A_i = you have i unique toys. Each state from 1-4 also has a transition to itself. The probability of going from A_i -> A_{i+1} = P(A_{i,i+1}) = (1 - i/4). This is pretty obviously: 1, 3/4, 2/4, and 1/4. The transition probability of staying in the same state is 1 - that = i/4, although we don't use those values in our calculations. The expected time to reach A_4 is E[A_0 -> A_1] + E[A_1 -> A_2] + E[A_2 -> A_3] + E[A_3 -> A_4] = 1/P(A_{0,1}) + 1/P(A_{1,2}) + 1/P(A_{2,3}) + 1/P(A_{3,4}) = 1 + 4/3 + 2 + 4 = 8 1/3. This is true due both to linearity of expectation and the fact that each state only either stays the same or advances. If any state could transition to another state besides itself or the subsequent one, you'd have to do a more complex calculation.