Candy store with eggs

Question

The owner of a candy shop has 11 hollow chocolate eggs in his display, all of the same size but different weights of 1 lb, 2 lbs,… 11 lbs respectively. Each of them is marked with a different sticker, so that they are distinguishable. A customer, intending to buy one egg, claims that he knows all the individual weights. The owner, being rather suspicious, asks him to guess the egg that weighs 1 lb. For this reason, he gave him an empty plastic bag that can hold up to exactly 11 lbs, otherwise it will be torn and can’t be used. Please describe the strategy of the customer in order to demonstrate which one is the egg of 1 lb. What is the minimum number of uses of the plastic bag and which eggs will he put inside it in each use?

With any 5 eggs the bag will be torn.

With any 4 eggs, the bag will also be torn except from the cases 1,2,3,4 and 1,2,3,5.

With any 3 eggs, the bag will be torn except from 16 out of 165 cases, of which only 4 do not contain the egg of 1 lb.

With any 2 eggs, the bag will be torn except from 25 out of 55 cases (only 9 containing the egg of 1 lb) and obviously with only 1 egg it will not be torn.

But how do we identify the egg of 1 lb in the least number of uses of the bag?

Answer 1

UPDATE: here's a two-use strategy that doesn't involve breaking the bag: The customer bags together α/β/γ/δ and shows that the bag is intact. They then bag together α/ε/ζ and show that the bag is intact.

Now, let's set Ω=α+β+γ+δ+ε+ζ. On the one hand, Ω is the sum of six different numbers, so it must be $\geq 21 (=1+2+3+4+5+6)$. On the other hand, since α+β+γ+δ $\leq11$ (from the first bagging) and ε+ζ+α $\leq11$ (from the second bagging), we have α+β+γ+δ+ε+ζ+α = Ω+α $\leq22$. But Ω$\geq21$ and Ω+α$\leq22$ together imply α $=1$. (Many thanks to Joffan in the comments for this easier argument!)

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(For posterity, here's my initial four-use strategy: our customer bags together α/β/γ/δ and shows that the bag is intact, then bags together α/β/γ/ε and shows that the bag is intact. Since the only two four-egg bags that don't break the bag are 1/2/3/4 and 1/2/3/5, we now have two known sets: we know that {α, β, γ} are {1,2,3} in some order, and we know that {δ, ε} are {4,5} in some order (though we haven't differentiated members of those sets).

Next, the customer bags together δ/ζ and shows that the bag remains intact; since δ is either 4 or 5, then ζ must be either 6 or 7.

Finally, the customer tries to bag β/γ/ζ and shows that the bag breaks. Any combination of two eggs from the set {1,2,3} with 6 would be 'good' and not break the bag; only the combination {2,3,7} can break it. This proves that ζ is 7 and that {β, γ} are {2,3} in some order, and therefore that α is egg 1.)