5.3 How not to divide with 3 parties

When first approaching the question of 3-party fair division, it is very tempting to propose this method: Randomly designate one participant to be the divider, and designate the rest choosers. Proceed as follows:

# Math in Society 5.3 -- Why the obvious 3-party method is NOT fair
# "Divider cuts three pieces, chooser 1 picks, chooser 2 picks, divider takes the
# rest." It looks fair. Run it and check the last chooser.
table = {"Chooser 1": [40, 30, 30],       # each row is that person's valuation,
         "Chooser 2": [45, 30, 25],       # as a percent of the whole. EDIT any row.
         "Divider":   [100/3, 100/3, 100/3]}
PICK_ORDER = ["Chooser 1", "Chooser 2"]   # EDIT: reverse this and re-run

people = list(table)
n = len(people)
fair = 100 / n
divider = [p for p in people if p not in PICK_ORDER][0]

print(f"{n} parties, so a fair share is {fair:.1f}% of the whole in your own eyes.")
print(f"   {'':<11}" + "".join(f"{'Piece '+str(i+1):>10}" for i in range(n)))
for p in people:
    print(f"   {p:<11}" + "".join(f"{v:>9.1f}%" for v in table[p]))

left = list(range(n))
got, offered = {}, {}
print()
for p in PICK_ORDER:
    offered[p] = list(left)
    best = max(left, key=lambda i: table[p][i])
    got[p] = best
    left.remove(best)
    print(f"   {p} picks the best piece still on the table: Piece {best+1}"
          f", worth {table[p][best]:.1f}%")
got[divider] = left[0]
print(f"   {divider} is left with Piece {left[0]+1}, worth {table[divider][left[0]]:.1f}%")

print()
short = []
for p in people:
    v = table[p][got[p]]
    ok = v >= fair - 1e-9
    if not ok:
        short.append(p)
    print(f"   {p:<11} ends with {v:>5.1f}%   "
          + ("at least a fair share" if ok else "SHORT -- this division is NOT fair"))

if short:
    p = short[0]
    menu = ", ".join(f"Piece {i+1} at {table[p][i]:.1f}%" for i in offered[p])
    print(f"\n{p} picked last and never had a fair share on offer: the choice was")
    print(f"   {menu}")
    print(f"and nothing there reaches {fair:.1f}%. A method is only fair if a fair")
    print("share is guaranteed to be available no matter what the others do first.")
else:
    print("\nEveryone did fine THIS time -- but a method is only fair if it works for")
    print("every valuation table, not the friendly ones.")

print("\nTry it: change Chooser 2's row to [40, 35, 25] and re-run. Now the method")
print("happens to work. Change it back and reverse PICK_ORDER: the unfairness just")
print("moves to whoever ends up last. The fix is in Section 5.4, Lone Divider.")

1) Have the divider divide the item into 3 pieces

2) Have the first chooser select any of the three pieces they feel is worth a fair share

3) Have the second chooser select either of the remaining pieces

4) The divider gets the piece left.

Example 1 Don’t do this – it is bad!

Suppose we have three people splitting a cake. We can immediately see that the divider will receive a fair share as long as they cut the cake fairly at the beginning. The first chooser certainly will also receive a fair share. What about the second chooser? Suppose each person values the three pieces like this:

	Piece 1	Piece 2	Piece 3
Chooser 1	$40 %$	$30 %$	$30 %$
Chooser 2	$45 %$	$30 %$	$25 %$
Divider	$33.3 %$	$33.3 %$	$33.3 %$

Show solution

Since the first chooser will clearly select Piece 1, the second chooser is left to select between Piece 2 and Piece 3, neither of which she values as a fair share (1/3 or about 33.3%). This example shows that this method does not guarantee a fair division.

To handle division with 3 or more parties, we’ll have to take a more clever approach.

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.