5.2 Divider-Chooser

The first method we will look at is a method for continuously divisible items. This method will be familiar to many parents - it is the “You cut, I choose” method. In this method, one party is designated the divider and the other the chooser, perhaps with a coin toss. The method works as follows:

Notice that the divider-chooser method is specific to a two-party division. Examine why this method guarantees a fair division: since the divider doesn’t know which piece he will receive, the rational action for him to take would be to divide the whole into two pieces he values equally. There is no incentive for the divider to attempt to “cheat” since he doesn’t know which piece he will receive. Since the chooser can pick either piece, she is guaranteed that one of them is worth at least 50% of the whole in her eyes. The chooser is guaranteed a piece she values as at least 50%, and the divider is guaranteed a piece he values at 50%.

Of course, in this example, Fred and Martha probably could have discussed their preferences and reached a mutually agreeable decision. The divider-chooser method is more necessary in cases where the parties are suspicious of each other’s motives, or are unable to communicate effectively, such as two countries drawing a border, or two children splitting a candy bar.

Your Turn

Try it Now 2

Dustin and Quinn were given an apple pie and a chocolate cake, and need to divide them. Dustin values the apple pie at $6 and the chocolate cake at $4. Quinn values the apple pie as $4 and the chocolate cake at $10. Describe a fair division if Quinn is dividing, and specify which “half” Dustin will choose.

# Math in Society 5.2 -- Divider-Chooser ("you cut, I choose")
# Dustin and Quinn were given an apple pie and a chocolate cake to split.
values = {"Quinn":  {"apple pie": 4, "chocolate cake": 10},
          "Dustin": {"apple pie": 6, "chocolate cake": 4}}
DIVIDER, CHOOSER = "Quinn", "Dustin"          # EDIT: swap these and re-run
# Piece 1 is this fraction of each dessert; piece 2 is everything left over.
piece1 = {"apple pie": 0.0, "chocolate cake": 0.7}     # EDIT the cut

items = list(values[DIVIDER])
piece2 = {i: 1 - piece1[i] for i in items}

def worth(who, piece):
    return sum(values[who][i] * piece[i] for i in items)

for who in (DIVIDER, CHOOSER):
    whole = sum(values[who].values())
    print(f"{who:<7} values the whole at ${whole:>6.2f}   fair share ${whole/2:>6.2f}"
          + "".join(f"   {i} ${values[who][i]}" for i in items))

def describe(p):
    return ", ".join(f"{p[i]:.0%} of the {i}" for i in items if p[i] > 0) or "nothing"

print(f"\n{DIVIDER} cuts:")
print(f"   piece 1 = {describe(piece1)}")
print(f"   piece 2 = {describe(piece2)}")
d1, d2 = worth(DIVIDER, piece1), worth(DIVIDER, piece2)
print(f"   to {DIVIDER}: piece 1 = ${d1:.2f}, piece 2 = ${d2:.2f}  "
      + ("  (an equal cut)" if abs(d1 - d2) < 1e-9 else
         "  NOT an equal cut -- a rational divider would not risk this"))

c1, c2 = worth(CHOOSER, piece1), worth(CHOOSER, piece2)
taken, left = ("piece 1", "piece 2") if c1 >= c2 else ("piece 2", "piece 1")
print(f"   to {CHOOSER}: piece 1 = ${c1:.2f}, piece 2 = ${c2:.2f}"
      f"  -> {CHOOSER} takes {taken}")

print("\nFinal split")
for who, got in ((CHOOSER, max(c1, c2)), (DIVIDER, d2 if taken == "piece 1" else d1)):
    whole = sum(values[who].values())
    ok = "fair share or better" if got >= whole / 2 - 1e-9 else "SHORT of a fair share"
    print(f"   {who:<7} ends with ${got:>6.2f} = {got/whole:>5.1%} of the whole"
          f" in their own eyes  ({ok})")

print("\nThe divider can only guarantee themselves exactly half, because they do not")
print("know which piece they will get. The chooser is guaranteed at least half and")
print("usually does better, since the cut was made to someone else's tastes.")
print("Try it: cut 50/50 with piece1 = {'apple pie': 0.5, 'chocolate cake': 0.5}.")
print("Both pieces are worth $5 to Dustin, so the chooser's advantage disappears.")
print("Try it: set DIVIDER, CHOOSER = 'Dustin', 'Quinn' and find a cut Dustin")
print("considers even. Quinn will still come out ahead.")

Answer

There are a lot of possible fair divisions Quinn could make. Since she values the two desserts at $14 together, a fair share in her eyes is $7. Notice since Dustin values the desserts at $10 together, a fair share in his eyes is $5 of value. A couple possible divisions:

Piece 1	Piece 2	Dustin would choose
$\frac{7}{10} chocolate ($7)$	$\frac{3}{10} chocolate ($3)$	Piece 2. Value in his eyes:
$No pie ($0)$	$All pie ($4)$	$$ 6 + 3 / 10 \cdot $ 4 = $ 7.20$
$1/2 chocolate ($5)$	$\frac{1}{2} chocolate ($5)$	Either. Value in his eyes:
$\frac{1}{2} pie ($2)$	$\frac{1}{2} pie ($2)$	$\frac{1}{2} \cdot $ 6 + \frac{1}{2} \cdot $ 4 = $ 5$

Things quickly become more complicated when we have more than two parties involved. We will look at three different approaches. But first, let us look at one that doesn’t work.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.