Polarity Under a “Democratic” Assumption

Posted on February 19, 2011 by jingjinyu

The first article in the book A Mathematician Reads the Newspaper by John Allen Paulos is about voting schemes. From the article I came to know the Banzhaf power index, which is a measure of how much power does a party have in deciding the final outcome of a vote. Basically the Banzhaf power index of a party is defined as the number of times the party’s vote matters among different combinations of minimal passing votes. As an example, assuming that parties {A, B, C}, and {D} are the share holders of a company with {34\%, 34\%, 17\%}, and {15\%} voting power, respectively. Assume that any proposal requires {51\%} vote to pass. In this scnarios, party {D} has no “effective” power in the sense that its decision is never going to change the outcome. Such a {D} is called a “dummy”. On the other hand, parties {A, B}, and {C} have equal power in the sense that any two can pass an item and no single party can pass an item. The power indices for the parties in this case are {2, 2, 2}, and {0}. A slight redistribution of shares, for example {30\%, 30\%, 20\%}, and {20\%} will give parties {C, D} the same (non-zero) power.

As I am reading the essay, it seems that the inevitable power imbalance under seemingly “democratic” assumptions could lead to quite extreme polarity in wealth (and other limited resources) as well. Consider the example in which {n = 2^k + 1} parties are to divide a unit of wealth. Let us denote the set of all parties as {P_0}. Assume that they agree to vote on any dividing proposal and a majority vote ({\ge 2^{k-1} + 1} vote-fors) will pass the proposal. How polar can the dividing become? It is clear that all we need is {2^{k-1} + 1} votes. Hence, the measure may leave {2^{k-1}} parties penniless; let us assume so and denote this set of parties {P_0^-}. Let {P_1 := P_0\backslash P_0^-}, which has {2^{k-1} + 1} parties. To get all parties in {P_1} to vote for the proposal, each party should be allocated at least {\frac{1}{2^k +1}} unit of wealth. Let us give {\frac{1}{2^k +1} + \epsilon_1} wealth to {2^{k-2}} parties among parties in {P_1} to buy their votes; denote this subset {P_1^-} and {P_2 := P_1 \backslash P_1^-}. Let us assume that parties in {P_1^-} are happy, which is a reasonable assumption: After all, they only need to cast a vote to get more than a fair share. We are now left with 1 - 2^{k-2}(\frac{1}{2^k +1} + \epsilon_1) \frac{2^k + 1 - 2^{k-2}}{2^k + 1} - 2^{k-2}\epsilon_1 \approx \frac{3}{4} unit of wealth to divide among {2^{k-2} + 1} parties in {P_2}, which is about {\frac{1}{4}} of {P_0}. Some “democracy”, is it?

What is a “fair” division of the left wealth among parties in {P_2}? Let us make the simplifying assumption that for a subset of {2^{k-3}} parties in {P_2} (denoted {P_2^-}), we allocate {(1 + 0.5)(\frac{1}{2^k + 1} + \epsilon_1)} unit of wealth to them and in exchange, each member of {P_2^-} will need to recruit two parties in the set {P_1^-}. Translating into English, each member of {P_2^-} will get a {25\%} commission for recruiting a member of {P_1^-}. In total, {P_2^-} gets {2^{k-3}(1 + 0.5)(\frac{1}{2^k + 1} + \epsilon_1) \approx \frac{3}{16}} unit of wealth. That is, these is about {1 - \frac{1}{4} - \frac{3}{16} = \frac{9}{16}} unit of wealth left. On the other hand, there are {2^{k-3} + 1} parties left to share this wealth, which is about {\frac{1}{8}} of {P_0}. Clearly, this is more divided than one round ealier; but the proposal seems plausible.

As one can expect, the plot goes on from here and the party at the top will get a very big share of that unit of wealth. Depending on how the recursion is defined, the growth of share for the “elites” may be exponential or may stop at some {k}. More or less, this mirrors what happens in reality with the difference being that the real world may not be as dramatic, at times. What a world.

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