The other day I came across a paper that addresses a closely related question:
http://www.unige.ch/math/folks/velenik/Vulg/Paninimania.pdf
If I have understood it correctly, the expected number of packs you would need to buy would be:
$\binom{424}{5}\sum_{j=1}^{424}\left(-1\right)^{j+1}\frac{\binom{424}{j}}{\binom{424}{5}-\binom{424-j}{5}}$
However, as eqperes points out in the comments, the specific question the OP asks is actually covered in detail in another paper that is not open access.
Their final conclusion suggests the following strategy (for an album of 660 stickers):
- Buy a box of 100 packs of 5 stickers (500 stickers, guaranteed to be all different)
- Buy 40 more packs of 5 stickers and swap the duplicates until you have at most 50 missing stickers.
- Purchase the remaining stickers directly from Panini (these cost approx. 1.5 times as much).
This is total of 140 packs + upto 15 extra packs worth of stickers (by cost) purchased in a targeted fashion, equivalent to at most 155 packs.
