You and your partner treasure hunter have uncovered a box of coins from the ancient civilization of Hyrule. The box contains 2 big coins, 3 silver coins, 4 shiny coins and 5 tiny coins!
The two of you decide to split the contents of the box evenly. You and your partner flip a coin and decide that you will have the hard job of being the divider. Then your partner will pick the pile they value more.
You took a course in Hylian coins in college, but your partner has never heard of them. You know that each big coin is worth $4, each silver coin is worth $2, each shiny coin is worth $3 and each tiny coin is worth $6
1) Divide the coins into two equal piles (pile A and pile B). Your partner will pick one pile and you will get the other pile. In other words, use divide and choose.
What coins will you put in each pile?

2) While you are subdividing the coins, your partner is looking at the coins and informing you of what they think each coin is worth. They think that the big coins are each worth $6, the silver coins are each worth $4, the shiny coins are each worth $5 and the tiny coins are each worth $1
Based on this information, how does your partner value each of the piles in your answer to question #1?
3) You realize that the division you were planning on making is not equal according to your partner. Your partner will be able to get more then what they think their fair share is.
Since you know how your partner values things, you can use that knowledge to make a new (unfair) division so that your partner values Pile C more than Pile D, while you secretly think that Pile D is worth more than Pile C. This way, both of you think that you got more than 50% of the total value.

What coins will you put in the new piles? (Pile C and Pile D)
4) How will your partner value each pile? How do you value each pile?

Answer:

First we find out the TOTAL TRUE VALUE of the coins.

BIG coins 2 nos. @ $4 = $ 4 x 2 = $ 8

SILVER coins 3 nos. @ $2 = $ 2 x 3 = $ 6

SHINY coins 4 nos. @ $3 = $ 3 x 4 = $ 12

TINY coins 5 nos. @ $6 = $ 6 x 5 = $ 30

Hence, Total no. of Coins = (2 + 3 + 4 + 5) = 14 nos. and their TOTAL TRUE value = $ (8 + 6 + 12 + 30) = $ 56

(1) Since I have to make two equal divisions, hence each of Pile-A and Pile-B shall constitute of 7 coins of total value $ 28 each. We can make it in the following way.

Pile-A:

BIG coins 1 no. @ $4 = $ 4 x 1 = $ 4

SILVER coins 3 nos. @ $2 = $ 2 x 3 = $ 6

SHINY coins 0 nos. @ $3 = $ 3 x 0 = $ 0

TINY coins 3 nos. @ $6 = $ 6 x 3 = $ 18

Hence, Total no. of Coins = (1 + 3 + 0 + 3) = 7 nos. and their TOTAL TRUE value = $ (4 + 6 + 0 + 18) = $ 28

Pile-B:

BIG coins 1 no. @ $4 = $ 4 x 1 = $ 4

SILVER coins 0 nos. @ $2 = $ 2 x 0 = $ 0

SHINY coins 4 nos. @ $3 = $ 3 x 4 = $ 12

TINY coins 2 nos. @ $6 = $ 6 x 2 = $ 12

Hence, Total no. of Coins = (1 + 0 + 4 + 2) = 7 nos. and their TOTAL TRUE value = $ (4 + 0 + 12 + 12) = $ 28

Hence, there are two equal Piles, Pile-A and Pile-B each of which constitute of 7 coins of total value $ 28 each........(Answer)

(2) Partner will evaluate the above distribution in following manner.

Pile-A:

BIG coins 1 no. @ $6 = $ 6 x 1 = $ 6

SILVER coins 3 nos. @ $4 = $ 4 x 3 = $ 12

SHINY coins 0 nos. @ $5 = $ 5 x 0 = $ 0

TINY coins 3 nos. @ $1 = $ 1 x 3 = $ 3

Hence, Total no. of Coins = (1 + 3 + 0 + 3) = 7 nos. and their TOTAL value = $ (6 + 12 + 0 + 3) = $ 21

Pile-B:

BIG coins 1 no. @ $6 = $ 6 x 1 = $ 6

SILVER coins 0 nos. @ $4 = $ 4 x 0 = $ 0

SHINY coins 4 nos. @ $5 = $ 5 x 4 = $ 20

TINY coins 2 nos. @ $1 = $ 1 x 2 = $ 2

Hence, Total no. of Coins = (1 + 0 + 4 + 2) = 7 nos. and their TOTAL value = $ (6 + 0 + 20 + 2) = $ 28

Hence, According to Partner, total value of Pile-A is $ 21 and that of Pile-B is $ 28.........(Answer)

(3) Since TOTAL TRUE value of Pile-D should be more than Pile-C, but according to Partner this should appear to be opposte, hence we will make the following distribution.

Pile-C:

BIG coins 1 no. + SILVER coins 2 nos. + SHINY coins 3 nos. + TINY coins 1 no.

Hence, Total no. of Coins = (1 + 2 + 3 + 1) = 7 nos.

Pile-D:

BIG coins 1 no. + SILVER coins 1 no. + SHINY coins 1 no. + TINY coins 4 nos.

Hence, Total no. of Coins = (1 + 1 + 1 + 4) = 7 nos.

(4) Partner will value each Pile in the following manner.

Pile-C:

BIG coins 1 no. @ $ 6 = $ 6 x 1 = $ 6

SILVER coins 2 nos. @ $4 = $ 4 x 2 = $ 8

SHINY coins 3 nos. @ $5 = $ 5 x 3 = $ 15

TINY coins 1 nos. @ $1 = $ 1 x 1 = $ 1

Hence, Total no. of Coins = (1 + 2 + 3 + 1) = 7 nos. and their TOTAL value according to partner = $ (6 + 8 + 15 + 1) = $ 30

Pile-D:

BIG coins 1 no. @ $6 = $ 6 x 1 = $ 6

SILVER coins 1 no. @ $4 = $ 4 x 1 = $ 4

SHINY coin 1 no. @ $5 = $ 5 x 1 = $ 5

TINY coins 4 nos. @ $1 = $ 1 x 4 = $ 4

Hence, Total no. of Coins = (1 + 1 + 1 + 4) = 7 nos. and their TOTAL value according to partner = $ (6 + 4 + 5 + 4) = $ 19

Thus Partner values Pile-C more than Pile-D.

Actually the TRUE value of each Pile is determined in the following manner.

Pile-C:

BIG coins 1 no. @ $ 4 = $ 4 x 1 = $ 4

SILVER coins 2 nos. @ $2 = $ 2 x 2 = $ 4

SHINY coins 3 nos. @ $3 = $ 3 x 3 = $ 9

TINY coins 1 no. @ $6 = $ 6 x 1 = $ 6

Hence, Total no. of Coins = (1 + 2 + 3 + 1) = 7 nos. and their TOTAL TRUE value = $ (4 + 4 + 9 + 6) = $ 23

Pile-D:

BIG coins 1 no. @ $4 = $ 4 x 1 = $ 4

SILVER coins 1 no. @ $2 = $ 2 x 1 = $ 2

SHINY coin 1 no. @ $3 = $ 3 x 1 = $ 3

TINY coins 4 nos. @ $6 = $ 6 x 4 = $ 24

Hence, Total no. of Coins = (1 + 1 + 1 + 4) = 7 nos. and their TOTAL TRUE value = $ (4 + 2 + 3 + 24) = $ 33

Thus actually the TRUE value of Pile-D is more than Pile-C.