How about this system? In order, each player has the option to CUT a slice, TAKE a slice already cut, or GIVE a cut slice to the person who cut it. Once a person has taken or been given a slice, that person is out. Repeat until everyone has a slice.
I think this works, assuming rational players intending to maximize their own portion, and no collusion between players.
Player 1 must cut, since no slices exist already. If they cut a bigger-than-fair slice, player 2 will take it. If they cut a smaller-than-fair slice, player 2 will give it to them. So it seems that cutting a fair slice is the only reasonable move.
Of course, with non-rational players, suboptimal play may result in someone other than oneself being punished, so that still may not be ideal.
Interesting. Can we assume that each person should be satisfied with the piece which they themselves cut? (I am making that assumption). In your scenario, then person 3 is still left with a piece which they neither chose nor cut.
What if, coming to the end, the choice of piece then moves backwards (from 4, to 3, to 2, then 1)? In your scenario, each person would then have a slice which they either cut or chose.
This seems to workâŚnice! (and assuming no collusion is clearly required for this problem, although it might still work is someone wants less than an equal shareâŚ)
For n=2, itâs easily solved, as there is a divider and a selector. Two roles, such that unfair division only hurts the divider. For n=2*m, it can similarly be solved: divide the cake in two according to the procedure above â replacing the divider with a representative of m-person team A, and the selector with a representative or other voting process of m-person team B â and them let each m-person team divide up their own cake from thereâŚ
So really, we have to solve the problem for reduced version of the problem, n=p, an arbitrary odd prime. But the two-role method canât be extended any further. You canât manufacture an arbitrary number of roles.
The only fair distribution method I know of that can satisfy that many people is the âdodgeballâ method. (In a two person example of picking teams for dodgeball, captain Alice picks, then Bob, then Bob again, then Alice, then Alice again, then Bob, and so on.)
Divide the cake into considerably more than n pieces, say, 2*n to be simple. Put the people in order, then let them choose a piece in order from 1 to n, then go down the order in reverse (from person n to person 1). For example, in a three-person division Alice might get the first pick and last pick, Bob the second pick and penultimate pick, and Charlie gets third and fourth picks. Higher multiples of n lead to âmore fairâ results as there should be less difference between each piece, and because you are forced to make more passes in this process.
In the end, every person should be satisfied that the sum of the pieces she chose are close to 1/n of the full cake.
What goes wrong here is that person 3 should have next choice after person 4, rather than person 1. He will choose the other 30% piece, then 2 will have to take his piece (which he canât complain about) and then 1 takes his own piece (which he canât complain about). Basically when i+1 chooses to cut, then i gets the piece that they themselves cut.
So this way there is no incentive for anyone to intentionally make a small piece in hopes of gaining advantage, because if others believe that the piece you cut was too small, you will be stuck with it. Whereas if you cut a piece that you believe is fair then you shouldnât mind getting that piece even if the next guy doesnât want it.
The flipside is if you intentionally cut a too large piece. In this case we have to assume everyone is 100% self interested for it to work; otherwise i can grant i+1 an extra large piece of cake at the expense of himself and the rest of the group. e.g:
1 cuts 30%
2 takes 1âs cut
at this point it is 3âs turn and heâs working with only 70% of the cake. the best he can hope for, barring othersâ altruism, is 1/3 of 70% because if he cuts more then 4 will take that leaving 1 to cut the remaining in half and now heâs guaranteed less. And as before, cutting less will leave him stuck with his own slice.
But if we stipulate that everyone is trying to get at least their fare share and not trying to help out another at their own expense, then neither the cut too big nor the cut too small strategy will be employed because both result in the person attempting such a thing to get less than their fair share by their own reckoning.
this is not actually different than the first proposed solution, where 1 cuts, 2 chooses to take that piece or cut, and so forth, because if 1âs cut is considered too small he ends up with it anyway. It is not necessary to give 2 the option of assigning 1âs cut to 1, itâll happen anyway.
Cakewrecks.com used to rail against the aesthetic deficiencies of cupcake-cakes on a regular basis. Though I would respond that if one wanted something aesthetic, one would not proceed to eat it.
Itâs been a while since Iâve checked in at http://www.cakewrecks.com/ . Iâm a little surprised to see that itâs still going.
also by taking out the GIVE option we donât actually have to rule out all collusion, only collusion wherin a player sacrifices everyone to some degree (including himself) to enrich a particular player. There is no way for a conspiracy of 2 or more to enrich themselves at the expense only of those outside the conspiracy
You are right of course. I think the mistake then in my algorithm was passing the cutter role forward always. If instead of 4 cutting after 3 takes, cutter role stays with 2, then it still works out, because that way the person at the end of the line is always a chooser, never a cutter. So in your example, it would be 2 cutting (his second) (and third, I guess) 26.6% piece, 4 picks one, 2 gets the other, and 1 ends up with his own piece.
HOWEVER having given this now entirely too much thought, I think there are some issues not yet considered here. What if someone makes a cut which they think is fair, but at least two others think is on the large side? one of those two or more will now believe there is not enough to go around so how can they guarantee that they donât wind up with the short end of the stick? After all the original question centers on belief.
1 cuts 25% according to himself
2 and 4 both think that itâs more like 26%
2 takes the piece
1 cuts another 25% that everyone agrees is 25%
3 takes that piece
at this point, 1 is happy to split the last into 2 equal slices and consider himself fairly treated, but 4, who until this point has not even gotten a chance to pick, considers himself unfairly treated because he has to choose between two 24.5% pieces! It is OK of course that 2 feels he got a good deal because that fits the requirement but what about 4?
In this solution, 1 and 2 can collude to get more cake as long as 1 trusts 2:
First, 1 cuts a much larger than fair slice (letâs say 45% of the cake even though there are 10 people), then 2 accepts it. After everyone else (including 1) gets a small piece, 2 shares some of the very large piece with 1 so the two of them come out ahead of everyone else.
