Amazon 4-for-3 deal:
Buy 4 item and have the one with the least price free. 4-for-3 items are under $10.
I believe if I buy 4n books, then it will be made into a 4n-for-3n deal that's not in writing, where n items with the least price is free. I did not test it yet, hope someone try that for me. Maxium saving is 25%, not bad.

How to maximize the amount of saving, suppose there is no such thing as shipping fee(If you are in your Amazon prime trial period).

1. Assume one have to buy n items.
2. If n is divisible by 4, the best strategy is to sort the items by price, forms a series of prices , where . This also forms set B.(uhh.. this set allows repetition of elements...)
3. Buy a vector of them each time for , which will maximize saving.
Proof: The total amount of money paid for this method is

Make each of the number removed into a set, S
Each time find one element for S, the 4 element of B is removed.
A non-rigorous proof
Create n blocks out of 4n elements in B, name it
S is set containing the minimal for each .
we can see has to be part of S, and .
suppose is not in , but in ,
and is the smallest number in other than .
If one switch with , we can find all the smallest number in is larger than . All the rest blocks doesn't change at all.
Which shows that must be in the same block with to allow the sum of S be maximized.
use the same reasoning for and .
and use the same reasoning for all blocks.
we get the original way have to be the best way.
If the count of elements in B is not divisible by 4, the best way is still the same, group items of 4 from most expensive to least expensive, remain a few least expensive ones in one group, buy each group separately.

This can be extended to any n-for-n-1 deals and should be able to extend to any n-for-k deals, where k lesser than n.