Fair Division: for $n > 2$

Question

I apologize if this is a dumb question.

Basically, I'm wondering if the following is equivalent to the last diminisher procedure. If not, I'm curious why this fails. What I have in mind is as follows:

The idea is that the person cutting the cake must give the other players an opportunity to take the slice; however, if none of the other players take the slice, the person who cut the slice must take the slice. So suppose $A, B$, and $C$ are dividing a cake. $A$ gets to cut a slice first. Then $B$ and $C$ each have a kind of right of first refusal. $B$ can take the slice or not. Then $C$ can take the slice or not. If neither $B$ or $C$ take the slice, $A$ must take the slice. Then the same procedure for the remainder of the cake is carried out between $B$ and $C$, which just reduces to the divide and choose procedure.

Answer 1

Alternative approach.

My answer is limited to the case of 3 people sharing a cake, and considers that there may be disagreement among the three people.

A cuts $x$. A is now committed to the idea that $x$ is neither too big nor too small. Then, one of 4 things will happen.

• Option 1: B,C agree to give $x$ to A.
• Option 2: B agrees to give $x$ to A, but C disagrees.
• Option 3: C agrees to give $x$ to A, but B disagrees.
• Option 4: B and C both decide that $x$ is too big to give to A.

The analysis below assumes that when a slice is given to one of the three, A,B,or C, that the 2 other people will share the remainder of the pie in accordance with the standard way of one person making the slice, and then the other person deciding who gets the slice.

$\underline{\text{Option 1:}}$
$x$ is given to A, and B,C divide the remainder.

$\underline{\text{Option 2:}}$
$x$ is given to C, and A,B divide the remainder.

$\underline{\text{Option 3:}}$
$x$ is given to B, and A,C divide the remainder.

$\underline{\text{Option 4:}}$
Here, A is committed to the idea that $x$ is not too large, while B and C are both committed to the idea that $x$ is too large.

B shaves off part of $x$ to create $y$ that B regards as fair. Then, C chooses whether to accept $y$ or give it to B.

A can not complain about getting to share $\displaystyle (1 - y) > (1- x)$ with one other person.

Since B regards $\displaystyle y$ as fair, B can not complain about either accepting $\displaystyle y$ or getting to share $\displaystyle (1 - y)$ with one other person.

Since C is allowed to decide whether to accept $\displaystyle y$ or give it to B, C also can not have any complaints.

Answer 2

Prove it by induction.

$P(n)=$ if there is $X$ amount of cake, and the cutter cuts $y$ of the cake. Then if $y < \frac Xn$ then the cutter will end up with $y< \frac Xn$ of the cake. If $y > \frac Xn$ then the second player will get $y > \frac Xn$ of the cake and all other players including the cutter will get $\frac {X-y}{n-1} < \frac {X -\frac Xn}{n-1}= X\frac {1-\frac 1n}{n-1}=X\frac {\frac {n-1}n}{n-1} = \frac Xn$ of the cake. So it is to the cutters advantange to cut a $\frac Xn$ slice.

Base case: $n= 2$.

If cutter cuts a $y$ slice the other player can take it or refuse it. If she takes the cutter gets $X-y$. If she refuses the cutter gets $y$ and the second player gets $X-y$. The second player will choose the larger option if $y > \frac 12 X$ or $y < \frac 12 X$ and the cutter will end up with the lesser. So it's to the cutter advantage to make as $\frac 12 X$ slice and accept or reject all players get $\frac 12 x$.

Induction step: True for all $n \le k$.

So if there are $k+1$ players and the cutter cuts a slice $y$.

Player $B$ can accept or refuse. If Player $B$ accepts then player $B$ we have $y$ cake. The remaining $k$ players will have $X-y$ cake and this is $P(k)$. And so then it will be to the cutters interest to cut it into $\frac {X-y}n$ and all players but $B$ get that.

If Player $B$ rejects, the someone else will accept and player $B$ will next be involved with $k$ players and $X-y$ cake. So player $B$ will have $\frac {X-y}k$ cake.

$y <; = ; > \frac {X-y}k$ if $y <; = ; > \frac {X}{k+1}$

So if $y < \frac {X}{k+1}$ player $B$ will refuse, and by the same logic so will all other players and the cutter ends with $y < \frac X{k+1}$ cake.

If $y > \frac Xn$ then player $B$ will accept and the cutter ends with $\frac {X-y}k < \frac X{k+1}$ cake.

SO its to the cutters advantage to cut an $\frac X{k+1}$ slice. If so the it doesn't matter whether any player accepts or rejects. If one accepts that player gets $\frac X{k+1}$ cake and all others get $\frac {X -\frac X{k+1}}k = \frac X{k+1}$ cake. And if one rejects someone will take it and that the rejectors will get $\frac {X-\frac X{k+1}}k = \frac X{k+1}$ cake.