Question

This question illustrates how bidding dishonestly can end up hurting the cheater.

Answer 1

I've had this same question before, here's my answer. ( If you're wondering why I did this so quickly is becuase I did mine digitally, not on paper.)

AND WE APPLY THE LONE DIVIDER METHOD, STEP ONE,

THE DIVIDER DIVIDES THE ITEM INTO N PIECES,

WHICH WE'LL LABEL S SUB 1 THROUGH S SUB N,

WHERE EACH PIECE HAS THE SAME VALUE BASED

UPON THE VALUE SYSTEM OF THE DIVIDER.

TWO, EACH OF THE CHOOSERS WILL SEPARATELY LIST THEIR PIECES

THEY CONSIDER TO BE A FAIR SHARE.

THIS IS CALLED THE DECLARATION, OR BID.

THREE, THE LISTS ARE EXAMINED, AND THERE ARE TWO POSSIBILITIES:

"A", IF IT IS POSSIBLE TO GIVE EACH PARTY A PIECE

THEY DECLARED, THEN DO SO,

AND THE DIVIDER GETS THE REMAINING PIECE.

B, IF TWO OR MORE PARTIES WANT THE SAME PIECES AND NO OTHERS,

MEANING WE HAVE A STANDOFF,

THEN GIVE A NON-CONTESTED PIECE TO THE DIVIDER,

RECOMBINE THE REMAINING PIECES,

AND REPEAT THE ENTIRE PROCEDURE WITH THE REMAINING PARTIES.

IF THERE ARE TWO PARTIES,

THEN WE USE THE DIVIDER-CHOOSER METHOD.

SO THIS QUESTION WILL ILLUSTRATE HOW BIDDING DISHONESTLY,

OR BEING GREEDY, CAN END UP HURTING THE CHEATER.

WE'LL FIRST TAKE A LOOK AT THIS PROBLEM WHERE EVERYONE

IS HONEST, AND THEN WHEN GREEDY IS DISHONEST.

SO FOUR PARTNERS ARE DIVIDING A $1 MILLION PROPERTY

USING THE LONE DIVIDER METHOD.

USING A MAP, DANNY DIVIDES THE PROPERTY INTO 4 PARCELS,

S SUB 1 THROUGH S SUB 4, AS WE SEE IN THIS FIRST ROW.

THE TABLE SHOWS THE VALUE OF THE 4 PARCELS

IN THE EYES OF EACH PARTNER,

AND THE UNITS ARE THOUSANDS OF DOLLARS.

SO AGAIN, NOTICE BECAUSE DANNY IS THE DIVIDER,

EACH PIECE HAS THE SAME VALUE,

WHICH IS 1 MILLION DIVIDED BY 4, OR 250,000.

THIS WOULD BE THE FAIR SHARE FOR EACH PLAYER, OR PARTNER.

FIRST, ASSUMING ALL PLAYERS BID HONESTLY,

WE WANT TO DETERMINE WHICH PIECE GREEDY WILL RECEIVE,

BUT WE'LL HAVE TO GO THROUGH THE PROCESS

TO DETERMINE WHICH PIECE EACH PARTNER RECEIVES.

SO NOW THAT DANNY HAS DIVIDED UP THE PROPERTIES,

EACH PARTNER NOW MAKES THEIR DECLARATION, OR BID,

WHICH WOULD BE THE PROPERTIES THEY VALUE AT $250,000 OR MORE.

SO NOTICE THAT BREANNA WOULD ONLY BID S SUB 1,

CARLOS WOULD BID S SUB 1 AND S SUB 2, AND GREEDY,

WHEN HE IS HONEST, WOULD BID S SUB 1, S SUB 2, AND S SUB 3.

SO NOW NOTICE THAT S SUB 3 AND S SUB 4 ARE NON-CONTESTED,

SO DANNY CAN RECEIVE S SUB 4, AND GREEDY CAN RECEIVE S SUB 3.

NOTICE GREEDY RECEIVES S SUB 3,

WHICH HE OR SHE VALUES AT $320,000.

SO AGAIN, GREEDY RECEIVES S SUB 3, DANNY RECEIVES S SUB 4,

SO THAT LEAVES S SUB 1 AND S SUB 2 FOR BREANNA AND CARLOS.

WE'LL NOTICE NOW CARLOS CAN RECEIVE S SUB 2,

AND BREANNA CAN RECEIVE S SUB 1.

NOTICE HERE EACH PERSON RECEIVES THEIR FAIR SHARE.

SO AGAIN, BREANNA RECEIVED S SUB 1,

AND CARLOS RECEIVED S SUB 2.

NOW LET'S LOOK AT THE SAME PROBLEM,

BUT WE'LL ASSUME THAT BREANNA AND CARLOS BID HONESTLY,

BUT GREEDY DECIDES TO ONLY BID FOR S SUB 1,

FIGURING THAT DOING SO WILL GET HIM S SUB 1.

SO AGAIN, DANNY IS THE DIVIDER,

BREANNA AND CARLOS' DECLARATIONS OR BIDS WILL STILL BE HONEST.

SO AGAIN, BREANNA WILL ONLY BID FOR S SUB 1,

CARLOS WILL BID FOR S SUB 1 AND S SUB 2,

AND NOW GREEDY, BEING DISHONEST, IS ONLY GOING TO BID

FOR S SUB 1, NOT FOR S SUB 2 AND S SUB 3.

AND BY DOING SO, HE THINKS HE WILL GET S SUB 1.

BUT NOTICE IN THIS CASE, IF CARLOS GETS S SUB 2,

THERE'S GOING TO BE A STANDOFF BETWEEN BREANNA

AND GREEDY FOR S SUB 1.

BEFORE WE FIGURE THAT OUT THOUGH, AGAIN,

BECAUSE DANNY AND CARLOS ARE NOT PART OF THE STANDOFF,

THEY CAN RECEIVE THEIR FAIR SHARES.

LET'S SAY THAT DANNY RECEIVES S SUB 3,

AND CARLOS RECEIVES S SUB 2.

NOTICE HOW THAT LEAVES S SUB 1 AND S SUB 4 FOR TWO PLAYERS,

WHICH MEANS NOW WE WOULD APPLY THE DIVIDER-CHOOSER METHOD

BETWEEN BREANNA AND GREEDY.

SO ONCE AGAIN, THE REMAINING PIECES S SUB 1 AND S SUB 4

ARE PUT BACK TOGETHER,

AND BREANNA AND GREEDY WILL SPLIT THEM

USING THE BASIC DIVIDER-CHOOSER METHOD.

IF GREEDY IS SELECTED TO BE THE DIVIDER,

WHAT WILL BE THE VALUE OF THE PIECE THAT HE RECEIVES?

WELL, GREEDY HAS TO DIVIDE S SUB 1 AND S SUB 4

INTO TWO PIECES OF EQUAL VALUE.

NOTICE THE VALUE OF EACH PIECE WOULD BE 340,000 + 80,000

DIVIDED BY 2, OR $210,000.

BUT REMEMBER WHEN GREEDY BID HONESTLY, HE RECEIVED S SUB 3,

WHICH HE VALUED AT $320,000.

SO NOTICE HOW BY BEING DISHONEST, GREEDY LOST 110,000,

THE DIFFERENCE BETWEEN 320,000 AND 210,000.

THIS IS THE REASON WHY IT OFTEN DOESN'T PAY

TO BE GREEDY OR DISHONEST WHEN USING THE LONE DIVIDER METHOD.

I HOPE YOU FOUND THIS HELPFUL.