My initial hunch was also to calculate the answer as 1 + 4/3 + 2 + 4. Ie, "average tries to get 1st toy, plus average tries to get 2nd toy, plus ...."
However, my memory of probability is very fuzzy and I wasn't sure if my above intuition was mathematically rigorous. Ie, can the expected values of these 4 events be simply summed in order to get the combined expected value?
For anyone else curious about this, here's the mathematical proof that you can indeed do so. Ie, proof that E(X+Y) = E(X) + E(Y). Where X can be defined as the number of tries to get the 1st toy, Y is the number of tries to get the 2nd toy, and so on: https://www.milefoot.com/math/stat/rv-sums.htm
In this specific example, X and Y are independent random variables. But if my reading of the above proof is correct, you can still sum the expected values even if X and Y happen to be correlated.
Linearity of expectation is of course true in general, but in this case summing the expectations of the individual transition probabilities is only valid because the only valid state transitions are going from state i -> i+1 (owning i unique toys to i+1 unique toys) or just staying in state i. If you could magically go from state i to any other state (you lose unique toys or gain more than one unique toy), then this simple calculation would not be correct.
However, my memory of probability is very fuzzy and I wasn't sure if my above intuition was mathematically rigorous. Ie, can the expected values of these 4 events be simply summed in order to get the combined expected value?
For anyone else curious about this, here's the mathematical proof that you can indeed do so. Ie, proof that E(X+Y) = E(X) + E(Y). Where X can be defined as the number of tries to get the 1st toy, Y is the number of tries to get the 2nd toy, and so on: https://www.milefoot.com/math/stat/rv-sums.htm
In this specific example, X and Y are independent random variables. But if my reading of the above proof is correct, you can still sum the expected values even if X and Y happen to be correlated.