I find this a quite terrible explanation of a fairly simple concept, but perhaps I haven't had enough coffee. Here's an alternative try:

First, for the computer scientists out there, this is the Coupon Collector's problem - so you probably know already it's Θ(N log N). That doesn't give us a precise answer but it should guide our intuition - it should take somewhere like 4 * log(4) ~= 8 draws to get what we want. [1]

Being exact:

When you have already have 3 of the 4 toys, what's the chance you get the 4th? It's 1 in 4. The expected number of draws to get an item with P=1/4 is 4.

Generalizing, the expected number of tries to get an item with probability p is 1/p. When you have 2 toys, you have a 1/2 probability of getting one of the two you haven't seen already, etc. When you have no toys, you always get something you don't have.

So the only slightly tricky case is when you have 1 toy, where you have a 3 in 4 chance of getting something new. 1/(3/4) = 1 1/3.

So the answer is very simply 4 + 2 + 1 1/3 + 1 = 8.3333.

More on the Coupon Collector's Problem: https://en.wikipedia.org/wiki/Coupon_collector%27s_problem

[1] The actual bounds use ln, not log2, but from a "getting intuition" perspective log2 gets us into the right range.