[bushindo]

3 people are dividing a cake. Each person wants as much cake as possible for himself, but each person thinks that the other two might be colluding. Lets call the three A, B, and C.

A suggests that he will cut the cake into 3 portions, and then C will pick his piece from the 3, and B then would pick his choice from the 2 remaining. and A takes the last piece. B objects to this scheme, saying that A can cut the cake into 1 large piece, and 2 smaller equal pieces. C then would have the largest piece, and B would have to take the smaller piece. In fact, B claims, If A and C were really colluding, this method would allowing them to get the entire cake to share among them two.

Given this distrusting atmosphere, is there a way to divide the cake so that each person is satisfied? Any proposed method would have to convince each person that he would get his fair share even if the other two were colluding.

I know of two solutions to this, but I wouldn't be surprised if the den can come up with a couple extra solutions.

Edited May 28, 2009 by bushindo

[Guest]

A wud cut a piece for himself and can't claim unless other 2 have taken theirs

similarly, B wud cut a piece for himself and can't claim unless other two have taken theirs

now ...in the order C,B,A.. each can claims their share.

following this order... wud ensure that if anyone cheats, he and his colluder wud lose on their share, since if any two wud cheat 3rd wud automatically get a bigger share and thus any cohort cud be loss making

[Guest]

I know that

rotating the knife

is the correct method.

But as an alternative

A divides the cake into 3 pieces. B and C chooses first and second big cakes, A takes the smallest cake. So A tries to make a fair division so that the smallest slice must be nearly equal to the largest slice.

Than B takes the knife and if he wants makes a slice to any of the remaining slices in order to make these two slices equal, because C will get the larger one and he (B) will get smaller one. Thus he should be fair, if he is not fair, C will get more cake than him.

[Guest]

Randomly decide who cuts the cake, then have the remaining two flip a coin to decide who chooses the next piece.

[Guest]

Use a ruler for a rectangular cake or a protractor for a circular cake to cut the cake evenly.

[Guest]

If A makes one cut, then B makes another cut, then C makes the last cut.

Then you have 6 pieces, so

A choses a piece, the B, then C, then C, then B, then A.

Works out like how a fantasy draft goes, and they've all make a cut, so each person is equally responsible for the size of the pieces (I guess C has more responsibility, but he's chosing 3rd to it's negated.)

[Guest]

If A makes one cut, then B makes another cut, then C makes the last cut.

Then you have 6 pieces, so

A choses a piece, the B, then C, then C, then B, then A.

Works out like how a fantasy draft goes, and they've all make a cut, so each person is equally responsible for the size of the pieces (I guess C has more responsibility, but he's chosing 3rd to it's negated.)

Assume that 5 pieces are equal and one piece is greater. A will get this one and so more than others.

The solution should fit all worst conditions.

Also flipping a coin doesn't assure that all kids get equal pieces, it only prevents a fight.

[Guest]

Have someone else cut the cake into three equal pieces to satisfy them

3 people are dividing a cake. Each person wants as much cake as possible for himself, but each person thinks that the other two might be colluding. Lets call the three A, B, and C.

A suggests that he will cut the cake into 3 portions, and then C will pick his piece from the 3, and B then would pick his choice from the 2 remaining. and A takes the last piece. B objects to this scheme, saying that A can cut the cake into 1 large piece, and 2 smaller equal pieces. C then would have the largest piece, and B would have to take the smaller piece. In fact, B claims, If A and C were really colluding, this method would allowing them to get the entire cake to share among them two.

Given this distrusting atmosphere, is there a way to divide the cake so that each person is satisfied? Any proposed method would have to convince each person that he would get his fair share even if the other two were colluding.

I know of two solutions to this, but I wouldn't be surprised if the den can come up with a couple extra solutions.

[Guest]

Assume that 5 pieces are equal and one piece is greater. A will get this one and so more than others.

The solution should fit all worst conditions.

Also flipping a coin doesn't assure that all kids get equal pieces, it only prevents a fight.

I was suggesting that the cuts be fully through the cake, not partial cuts. This way you couldnt get 1 small piece

[Guest]

I was suggesting that the cuts be fully through the cake, not partial cuts. This way you couldnt get 1 small piece

Also C would have cut the larger piece, and so there would be 2 larger pieces. Both A and B would get a larger piece, and C would be left with the bag. Collusion fails

[plainglazed]

Okay, let's try this

[spoiler=':

']A cuts the cake in what he feels are three equal pieces. B and C place secret ballots as to which piece they want. They reveal their choice at the same time. If they each choose different pieces, problem solved as all is fair. If they choose the same piece, A chooses one of the other pieces. Now there are two remaining pieces. B cuts one in half and C chooses; C cuts the other in half and B chooses.

[Guest]

Ive got it my math teacher did this a week ago

Person A devides it into 3 so that he would be happy with all 3 pieces. B then cuts a part off the 3rd piece and puts it on the second piece so that he would be happy with either of the remaining pieces aside from the piece that A cut and then they Choose in this order: C, B, A

OH!!!!! in ur face

[Guest]

Divide the cake into three portions and give a piece to each one of them. They should cut their individual portions of cakes into pieces further.

So essentially each one is having 3 pieces of the cake with them now. B and C will now go to A, and first B will select the portion from the cake(since B was having problems with the division ).Then A and C will go to B, and C will select the portion first and A second and the remaining is for B. Similarly A and B will go to pick up their pieces of cake cut by C but A will be the one who is choosing his portion first.

I think in this way, everyone will be satisfied. Everyone got to select in an equal way.

[Guest]

ur way makes no sense

[Guest]

ur way makes no sense

your solution is so wrong it makes me laugh

just look at my solution

A cuts first then B cuts the remaining 2 then C chooses then B chooses then A choses

[plainglazed]

Ive got it my math teacher did this a week ago

Person A devides it into 3 so that he would be happy with all 3 pieces. B then cuts a part off the 3rd piece and puts it on the second piece so that he would be happy with either of the remaining pieces aside from the piece that A cut and then they Choose in this order: C, B, A

OH!!!!! in ur face

I dont think your logic works for this particular problem. The OP's second sentence says "each person thinks that the other two might be colluding." and the object is to find a solution that is fair to all. In your scenerio it seems to me that B and C could be in collusion. B cuts all but the tiniest sliver off of one of the pieces, piles it on another, and then B and C split all but that smallest of slivers. Not fair to A

[Guest]

If its a circular cake we have A make 3 cuts across the diameter of the cake. B then chooses two slices for himself with the one rule that he cannot choose the slice directly opposite his own. C then does the same, and A gets the remaining pieces. If the cuts are made across the diameter of the cake, then any given piece of the cake must be the same size as the piece opposite it. This way B and C will always have the same size pieces because even if the slices differ in size, they will each select 1 each of the largest and second largest size slices if there is any difference. A will attempt to cut as fairly as he possibly can because any failure to do so will mean he gets less cake. Neither B nor C would be willing to collude with A because A cannot cut the cake such that B or C will have a larger quantity than the other.

[Guest]

A cuts the cake into two pieces.

B picks a piece.

A and B each cut their piece into thirds.

C picks a piece from both A and B

[Guest]

Okay, let's try this

[spoiler=':

']A cuts the cake in what he feels are three equal pieces. B and C place secret ballots as to which piece they want. They reveal their choice at the same time. If they each choose different pieces, problem solved as all is fair. If they choose the same piece, A chooses one of the other pieces. Now there are two remaining pieces. B cuts one in half and C chooses; C cuts the other in half and B chooses.

I was working on a solution like this, but discarded it because -

if A makes a 49% 48% 3% split, he'd wind up with 48% of the cake, while B and C each get 26%,

-unless B & C collude so that A winds up shafting himself, which is rather poetic

[Guest]

The problem I'm seeing with a lot of solutions is the collusion where Person C cuts the cake 94%, 3%, 3%, A takes 94%, B is stuck with 3% and unhappy going home with less cake but a better waistline, and C takes 3%, B leaves, C and A share their 97% between them.

One true way to split them is to get the blender out, mix with milk, blend thoroughly, and pour the cake into 3 equal measuring glasses. If they can't get along well enough to split a cake without distrust, they don't deserve a real cake.

[Guest]

I assume that each person when cutting cake will cut a tiny sliver and choose the larger piece as their slice, so:

A takes the whole cake.

B chooses from any of A's pieces and cuts off however much he wants to keep.

C chooses from any of A's pieces, and then any of B's pieces and cuts off however much he wants to keep.

This continues, each person in turn choosing one piece from everyone else's pieces, cutting off however much he wants to keep, and adding that to his own group of pieces.

So, after the initial round:

A chooses from B.

A chooses from C.

B chooses from A.

B chooses from C.

C chooses from A.

C chooses from B.

After an infinite number of rounds they would each have an infinite number of inifinitely small slices of cake.

Once all three declare they are satisfied the cutting stops.

[Guest]

Did anybody else think...

measure it? If you cut into even thirds, no one can argue. If a center point is needed, lay to lines on the diameter.

Or, if the cake is square/rectangular, simple thirds can be found even with a piece of string/paper/etc... that can be cut to length and folded in thirds to mark.

If the cake is other geometries, your in a little trouble, but the aid of a kitchen scale should do the trick!

[Guest]

person A cuts the cake into thirds (as closely as possible) persons B & C agree which piece is smallest and give that to A. Now A is done receiving cake. Of the two remaining pieces: the left piece is cut into two equal parts by person B. Now persons A & C must agree which piece is smallest and give that piece to B. C gets the other from that cut. The piece left from the right: Person C cuts that piece into 2 equal parts, and persons A & B agree which piece is smallest and give it to C. C is done receiving cake. The remaining piece goes to B.

Now since these people are soooo greedy, who gets to lick the cakepan?

[Guest]

If A makes one cut, then B makes another cut, then C makes the last cut.

Then you have 6 pieces, so

A choses a piece, the B, then C, then C, then B, then A.

Works out like how a fantasy draft goes, and they've all make a cut, so each person is equally responsible for the size of the pieces (I guess C has more responsibility, but he's chosing 3rd to it's negated.)

B and C could still collude against A. Assuming A makes a slice perfectly through the center dividing the cake in half, B could then make a perfectly perpendicular cut, dividing it into fourths. C then makes a cut dividing two of the fourths such that there are two pieces that are nearly fourths and two pieces that are merely tiny slivers. A and B would then end up with 1 fourth slice and 1 sliver, C would have two near fourth slices. B and C then get together and share their cake, having nearly 3/4 between them.

[Guest]

person A cuts the cake into thirds (as closely as possible) persons B & C agree which piece is smallest and give that to A. Now A is done receiving cake. Of the two remaining pieces: the left piece is cut into two equal parts by person B. Now persons A & C must agree which piece is smallest and give that piece to B. C gets the other from that cut. The piece left from the right: Person C cuts that piece into 2 equal parts, and persons A & B agree which piece is smallest and give it to C. C is done receiving cake. The remaining piece goes to B.

Now since these people are soooo greedy, who gets to lick the cakepan?

If A and C are in cahoots, then A can make one huge slice and two small slices. C can claim that the huge slice is the "smallest" so A can have that cake which I'm sure B will argue. Since A has no say, B and C will argue and it will be a stalemate since its B's word vs C. The cake will get stale due to all the arguing and no one will eat it

[Guest]

My husband cracked me up when he came up with this solution....

The guy with the knife kills the other two and keeps the cake for himself.