How to divide their loot? The thieves' dilemma

Question

Three thieves rob a jewelry store at gunpoint and end up with the following loot.

10 necklaces

8 bangles

6 rings

It so happened that the jewelry was antique and valuable. They asked the scared jeweler the value of those items. “Individually, the necklaces are worth 4000 dollars each, the bangles are worth 3000 dollars each and the rings are worth 2000 dollars each" The jeweler said.

"What do you mean individually?"

"Well, the sets are worth more!" The jeweler said.

"What do you mean?"

"A set of all three items together is worth $15000. A set of one necklace and one bangle is worth 12000 dolllars, a set of one necklace and one ring is worth 8000 dollars and a set of a bangle and a ring is worth 10000 dollars." The jeweler said.

Now the three thieves have a dilemma. How to divide the loot?

They had agreed that the loot must be divided equally (money wise). They had also decided to take off in three different directions after the robbery and never contact each other for any reason.

So they must come up with a strategy to divide the loot so that all three shares are equal (money wise) and with the highest possible money value.

What should be their strategy? The answer should show the final division of loot with explanation. Each one's loot will have equal money value but they can be different items. Like one thief can have different number of items than the others.

Remember they can only get higher value if they have sets. No programming please.

Answer 1

The final division should be

$2 \times RBN + 1 \times RB$ for two of them and the rest $2 \times NB + 4 \times N$ for the third. Where, of course, $N$ stands for necklace, $R$ for ring and $B$ for bangle.

Reasoning

The maths become easy if we introduce the paired of $R,N,B$ as $pR = R+1k$, $pN = N+1k$ and $pB = B+4k$. We can check that this gives the right totals for any pair and for the triple.

The best total ignoring the splitting constraint is $6\times pR + 8 \times pB + 10 \times pN = 124k$. This is, however, not divisible by 3. The nearest, $3\times 41k$, can be ruled out by a mild amount of case bashing (using the fact that the triple is the only odd$-k$ valued combination). Showing that our solution, $3\times 40k$ is optimal.

Answer 2

Observations:

1 A triple and a single is worth more if rearranged into two pairs
2 Pairs always yield an even number (of thousand dollars)
3 The problem can be reformulated as: Necklaces are worth 5K, bangles 7K, and rings 3K, but you have to pay a fee if you want to sell a single piece (1K,3K and 1K resp.)

Solution:

From 3 we conclude that the max worth is 124K i.e. 41333/thief.
with 1+2 we can generate possible distributions
41K with triplet: 15+10+8+8 (only solution for a single thief)
41K without triplets: needs single bangle(s) to make odd; costs more than 1K fee
Conclusion: Since we require exactly 1K fee: There is no 41K solution
40K with triplets: This requires 2 triples (to get to an even result) Two triplets can be rearrange as three pairs with the same value; this option can be handled there.
40K options with only pairs:
RB,RB,RB,RB
NB,NB,NR,NR
NB,NR,RB,RB
NR,NR,NR,NR,NR
40K options with singles (and no more than 4K fee):
N,NB,NB,NB (1K fee)
N,N,NR,NR,NR,NR (2K fee)
N,N,N,N,NB,NB (4K fee)
R,R,R,R, NR,NR,NR,NR (4K fee)
After a little trying: Solution 6N2B / 4N2B2R / 4B4R gives every thief 40K loot
This is maximal considering the reasoning above

Answer 3

First, we have the following underlying values (I will be using 1/1000 values for simplicity):

Necklace ($x$) = 4, Bangle ($y$) = 3, Ring ($z$) = 2. I will refer to these with $x$, $y$, and $z$ respectively. Sets are always worth more than the individual pieces:
$xy$ = 12, which is worth 5 more than $x+y$
$xz$ = 8, which is worth 2 more than $x+z$
$yz$ = 10, which is worth 5 more than $y+z$
$xyz$ = 15, which is worth 1 more than both $xy+z$ or $yz+x$, and 4 more than $xz+y$

Interestingly, 2 full sets ($xyz$) are worth the same as three 2-part sets with the same pieces ($xy+xz+yz$).

A maximised divisible worth of all objects:

If we use 6 full sets, we can get 122 ($6xyz+2xy+2x$), but if we split the full sets into 2-part, it becomes clear that with 5 sets we can get of 123:
$5xyz + 3xy + 1xz + 1x$ (= 5 * 15 + 3 * 12 + 8 + 4 = 123)

This maximum is unobtainable if split in three, since each thief would need to receive 41, which does not work with the available sets. If we reduce this maximum by 3, each thief would gain 40, which seems more plausible.
One way of reducing the total amount by 3, is by splitting up 3 full sets into one partial set and 3 individual items. We have two possibilities: $xyz = yz+x$ or $xyz = xy+z$
This gives us two possible sets of loot:
A) $2xyz + 3xy + 1xz + 1x + 3xy + 3z = 2xyz + 6xy + 1xz + 1x + 3z$
B) $2xyz + 3xy + 1xz + 1x + 3yz + 3x = 2xyz + 3xy + 1xz + 3yz + 4x$

Dividing the loot

We have two possible sets to divide into 40 for each thief. Both full sets $xyz$ should go to one person, as it is the only uneven value in both set A and set B. This leaves us with 10 additional for that person.

Set A does not work in this case, since the only way of obtaining 10 would be to combine one $x$ and 3 $z$, but those could then be recombined for more value into $xz + 2z$.
Set B does work:
Thief 1 receives $2xyz$ (30) + $1yz$ (10)
Thief 2 receives $2yz$ (20) + $1xy$ (8) + $1xz$ (12)
Thief 3 receives $2xz$ (24) + $4x$ (16)

Final Answer:

Thief 1 receives 2 necklaces, 3 bangles, and 3 rings. Thief 2 receives 2 necklaces, 3 bangles, and 3 rings. Thief 3 receives 6 necklaces and 2 rings.
As such, each thief could fence their goods for $40 000.