# Two person fair division

## You cut, I choose

The standard method for dividing a piece of cake: you cut I choose, is fair in the sense that it is proportionate (i.e., each person gets at least 1/2 of the cake by their own evaluation) and it is envy free (i.e., neither person would prefer the other person's portion). Yet it is not fair, I would much prefer to have you cut, which gives me the choice, rather than cut the cake and give you the choice. The reason is that cake is not homogeneous, your and my relative preferences for cake, frosting, and aesthetic appearance may be different. The cutter divides the cake into two portions which he perceives as equal, but the chooser may not perceive them as equal. Hence the cutter gets half the cake (by his perception), but the chooser may get more than half the cake (by his perception). For example, if a cake were half chocolate and half white, and the cutter was indifferent to chocolate versus white cake, the cutter might cut the cake down the middle producing one chocolate and one white piece; in that case, if the chooser preferred chocolate cake, he would be much happier with the chocolate piece than with the white piece.

Conversely, if the preferences of the chooser (or both parties) are known, the advantage is shifted to the cutter. For example, if the chooser prefers frosting, while the cutter likes both frosting and cake equally well, the cutter will put enough excess frosting in the smaller portion to entice the chooser to take it. Hence the chooser will get just over half the value of the cake (by the chooser's values) (the cutter must make sure the chooser chooses the proper portion), while the cutter will get substantially over half the value of the cake (by the cutter's values). Using the example of the cake which is half chocolate and half white, if the cutter is indifferent between chocolate and white cake but the chooser is allergic to chocolate (and the cutter knows that fact); the cutter could cut the cake into 1/3 white (i.e., 2/3 of the white half) versus 1/6 white with 1/2 chocolate. The chosser would choose the 1/3 white, and the cutter would end up with 2/3 of the original cake.

## Quantified preferences and trading to pareto optimality

Assume you like chocolate cake twice as much as white (i.e., you have equal preference for one square of chocolate or two squares of white). Assume also that your sibling likes white cake thrice as much as chocolate (i.e., your sibling has equal preference for one square of white or three squares of chocolate). The Pareto optimal divisions are as described above (either you have no white or your sibling has no chocolate). But what apportionment would ensue from trading after your mother's "fair" division?

Since you like chocolate equal to twice as much white cake, you would be willing to give up your 1/4 white cake for one-eighth chocolate cake, resulting in your sibling having all the white cake (one-half cake) and one-eighth chocolate cake. But your sibling would be willing to give up chocolate cake for one-third as much white cake, so if you offered one-twelfth white cake (1/3 of the 1/4 your mother gave you), you could get all your sibling's chocolate cake (one-quarter cake) with the ultimate result you would have all the chocolate cake (one-half cake) plus 1/6 white cake, while your sibling had only 1/3 white cake. Hence even with an initial division, there will in general not be a unique Pareto optimal division which ensues from it. How much you can get will depend on whether you know your siblings preferences, and how good you are at bargaining. Any division between the two extremes calculated is possible.

Exercise: How would Jack Sprat divide a roast if he did not know his wife's preferences? How would he divide the roast if he knew his wife's preferences?
If you knew the chooser wanted exactly equal amounts of cake and frosting, how would you divide a cake?
If you preferred chocolate cake thrice as much as white, and your sibling preferred white cake quatrice as much as chocolate, what are the Pareto optimal divisions of a cake which is 1/2 chocolate and 1/2 white? Which of those partitions could ensue from an initial division giving each of you 1/4 chocolate and 1/4 white cake?

## The case of discrete objects

If the matter to be divided is discrete rather than a continuum, the chooser has the obvious advantage. For example, if there are 5 marbles to be divided, the fairest division is 2 and 3, and the chooser gets three. If the objects are not the same, the advantage may not be so clear. For example, if there is a red shirt, a green shirt, red pants, and green pants, the "cutter" decides according to his pleasure whether the wardrobes will be mixed or matched; if the "cutter" decides on mixed wardrobes, the chooser canot get green pants with a green shirt.

A more common implementation is alternating choices, in which case the first choice is a definite advantage (unless compensated by the next two choices going to the second individual, or some other caompensatory mechanism).

## If opponent's preferences known

If you know your opponents preferences, you may have a significant advantage under alternate choices. Assume a family estate consists of A tailcoat (T), a wedding gown (W), a pair of cufflinks ((C), a necklace (N), a samovar (S), and a mirror (M).Annabelle has the first choice, and the preference order WNSMTC, Alfred has the second choice and the preference TCSMNW. Aternating choices should provide Annabelle with WNS and Alfred with TCM. But if Alfred knows Annabelle does not want cufflinks, he can put them at the bottom of his preference and end up with TSC while Annabelle gets WNM. (Of course if Annabelle knew Alfred's preferences, she could put the wedding gown at the bottom of her list ... .)

1) I read my web page on two person fair division (cake cutting) (above) and thought it could be improved. Hence I have drafted the following (below) which I hope clarifies some of it.

Cake cutting:

The key to many of the problems entailing cutting a cake with two flavors (or dividing a pizza which is half mushroom and half sausage) is to evaluate the cake in terms of a single flavor (like converting Swiss Francs and Euros to U. S. Dollars). [You could have a similar problem dividing up nickels and dimes where your younger brother prefers nickels because they are larger, but you prefer dimes because they are worth more.]

For example, if you like chocolate cake three times as much as vanilla, then you are indifferent between one square of chocolate or three squares of vanilla. If you have three squares of chocolate and three squares of vanilla, to you it would be the same as three squares of chocolate and one square of chocolate (i.e., four squares of chocolate) or nine squares of vanilla and three squares of vanilla (i.e., twelve squares of vanilla). If you had 1/4 chocolate cake and 1/4 vanilla cake, it would be the same to you as having 1/4 chocolate cake and (1/3)(1/4)=1/12 chocolate cake (i.e., 1/4 + 1/12 = 1/3 chocolate cake) or 3/4 vanilla cake and 1/4 vanilla cake (i.e., 1 vanilla cake).

Assuming the 3:1 preference ratio above, this lets you determine whether you prefer 1/4 chocolate and 1/4 vanilla to 1/6 chocolate and 2/5 vanilla. Using chocolate to evaluate the portions, 1/4 chocolate and 1/4 vanilla is equivalent to 1/3 chocolate as calculated above. 1/6 chocolate and 2/5 vanilla is equivalent to 1/6 chocolate and (1/3)(2/5) chocolate equals 1/6 + 2/15 = 9/30 = 3/10 chocolate. Since .3 < .33 (= 1/3), you prefer 1/4 chocolate and 1/4 vanilla to 1/6 chocolate and 2/5 vanilla.

If you were offered 1/6 chocolate and 3/5 vanilla, since 1/6 + (1/3)(3/5) = 11/30 =.37 > .33, you would prefer 1/6 chocolate and 3/5 vanilla to 1/4 chocolate and 1/4 vanilla.

This also lets you evaluate the most you (or your opponent) could get by trading. If you start with 1/4 chocolate and 1/4 vanilla, you will not accept anything you value at less than 1/3 chocolate (assuming the 3:1 conversion ratio above). Therefore your opponent (who prefers vanilla) could never get more than the entire cake less 1/3 chocolate (i.e., 1/2 chocolate and 1/2 vanilla less 1/3 chocolate equals 1/6 chocolate and 1/2 vanilla)