Prisoners, apples & oranges

Question

You are in a group of sixty prisoners, and the warden has a game to play with you.

In a room there are sixty boxes. Each box can contain either two apples, two oranges or one of each, but you don't know which contains how many. When the game starts, each prisoner can go in the room and pick one fruit each from two of the boxes (including labelled ones). Then he is allowed to put '2 Apples', '2 Oranges' or 'Apple & Orange' labels on one or two of them (if they are unlabeled). EDITED: He is also allowed to change the labels of up to two boxes he didn’t look in. When all 60 prisoners have visited the room and there is at least one unlabeled box, the process starts again from the first prisoner.

After that, they will be put in a soundproof room, with there is no communication, one for every prisoner. When all the boxes are labeled, the prisoners are freed if they are all placed correctly, or exceuted otherwise.

You can simply have every box checked, then label each one with the fruit you get, but you only have a $(\frac{2}{3})^{60} = 2.7197216 \times 10^{-11}$ chance of getting freed.

What is the optimal plan to maximise the chances of the prisoners being freed?

Answer 1

As I understand the problem, the answer is:

There is no strategy to maximize the probability of freedom.

Because

You can never be sure someone saw both fruits of a given box. So there will always remain some uncertainty for some boxes.
If nothing is reset between rounds, you can always add a round by not labelling uncertain boxes, to learn more and reduce the probability to miss a fruit and mislabel a box.
But that means that to maximize the probability to be free the prisoners have to play forever.
To actually be free the prisoners must compromize between the risk of death and the time to conclude.

Answer 2

Here is the process that all prisoners need to follow:

1. all prisoners are assigned numbers from 1 to 60.

2. Randomly, any prisoner will go into room and will examine Box one = 2 * ((prisoner number) mod 30)- 1 Box two = 2 * ((prisoner number) mod 30)

   like Prisoner_1 and prisoner_31 both will examine Box_1 and Box_2, P_2 and P_32 both will examine B_3 and B_2,
   P_30 and P_60 both will examine B_59 & B_60

3. if selected box has no label, he will pick fruit from left side of the box and label it as per following table:

box	fruit	old label	new label
1	orange		2-orange
1	apple		2-apple
2	orange		2-orange
2	apple		2-apple

4. if selected box has label, he will pick fruit from right side of the box and label it as per following table:

box	fruit	old label	new label
1	orange	2-orange	2-orange
1	orange	2-apple	apple and orange
1	apple	2-orange	orange and apple
1	apple	2-apple	2-apple
2	orange	2-orange	2-orange
2	orange	2-apple	apple and orange
2	apple	2-orange	orange and apple
2	apple	2-apple	2-apple

Once, all 60 prisoners have visited room, all the 60 boxes will have right labels.

This process will work even if all boxes have just one fruit in all of them OR any permutations/combinations of orange and apple in all the boxes.

PS: It has been mentioned in comment that one fruit may not be picked even after all prisoners looks into the box. My point is: It is a possibility. not a constraint. It is like Everyone tosses a coin and it is possible that no one gets a head. There s no restriction or constraint that prisoners can not decide from which side of the box they can pick, right? Prisoners are at liberty to pick any one fruit from the box whichever way they want. Also, it was mentioned that boxes are left untouched, unshuffled and unchanged throughout the process. So if all prisoners agree beforehand to follow the same above strategy, they all can be freed with 100% guaranteed result

Edited: Edited to make it more clear.. pick left side of fruit was changed to pick fruit from left side of the box and likewise for right side. also, added PS: phrase.