Probability of drawing 4 unique prizes in 5 draws?

Question

Suppose there are 4 prizes. You get one prize for every box of Crackerjack you buy. Assume the chance for each prize is 25%. Let n be the number of boxes bought until you collect all 4 prizes. What's the chance that n=4? What's the chance that n=5?

It seems like N=4 would be 4!/4^4 (24/256) = 9.375%. Is this correct? And what about N=5?

Answer 1

P(4 prizes | n = 4) = 1*(3/4)(1/2)(1/4) = 3/32 = 0.09375

P(4 prizes | n = 5)
= 1 - [(4C1)(3/4)^5 - (4C2)(1/2)^5 + (4C3)(1/4)^5]

= 1 - [4*3^5/4^5 - 6/2^5 + 4/4^5]

= 1 - [3^5/4^4 - 3/2^4 + 1/4^4]

= 1 - [243/256 - 3/16 + 1/256] = 1 - (243 - 48 + 1)/256

= 1 - 196/256 = 1 - 49/64 = 15/64 = 0.234375

This is

1
less
4 times the probability that one prize was not received
There are four possible ways for this to occur.

(But if two prizes were not received there was some double counting.)

plus (add back in)
6 times the probability that two prizes were not received
There are six possible ways for this to occur.

(But if three prizes were not received there was some double counting. These possibilities were added back in too many times.)

less
4 times the probability that three prizes were not received
There are four possible ways for this to occur.

Answer 2

that sounds correct for 5 it would be 4^5 on the bottom with 4*4! on top because your first two draws could be anything
96/1024 so the percentage is once again 9.375%