## Power on the Nassau County Board of Supervisors

In 1964, the Nassau County Board of Supervisors had 6 members, each member representing one district.  Four of the districts were urban, and two were rural.  To provide some sort of representation equal to the population of each district, the Board used a weighted voting system.

 District Votes Hempstead 1 31 Hempstead 2 31 Oyster Bay 28 North Hempstead 21 Long Beach 2 Glen Cove 2

The six members had a total of 115 votes.  In order to pass a motion, a simple majority of 58 votes was needed.  In other words, the Nassau County Board of Supervisors of 1964 is an example of the weighted voting system   $[58:31,31,28,21,2,2]$.

### Three Districts Are Powerless

Clearly, no supervisor can pass a motion on their own, since no supervisor has more than 31 votes, which is far short of the 58 votes needed.

Two supervisors could agree and pass a motion.  There are three different ways this could happen.

 Coalition Votes Hempstead 1 & Hempstead 2 31 + 31 = 62 Hempstead 1 & Oyster Bay 31 + 28 = 59 Hempstead 2 & Oyster Bay 31 + 28 = 59

No other pair of supervisors could pass a motion, since they would have at most 52 votes (using North Hempstead and Hempstead 1, for example).  Notice that three of the districts were large enough to be able to form a two-district coalition.  We shall call these three districts (Hempstead 1, Hempstead 2, Oyster Bay) the large districts.  In other words, any two of the three large districts can pass a motion.

If all three small districts (North Hempstead, Long Beach, Glen Cove) were to propose a motion, they could not pass it without help.  The three small districts only have a total of 25 votes.  And even if one of the large districts helped the three small districts with their motion, it would still not pass, as they would have at most 56 votes.  In other words, the three small districts need the help of at least two of the large districts.  But the two large districts could pass the motion without the help of the small districts.  Therefore, the wishes of the small districts are actually irrelevant.  The small districts have no power to affect the passage of any motion.  (It is quite amazing that North Hempstead, with 21 votes, is actually a "small" district in this analysis.)

Identical results would occur if the weighted voting system had been   $[2:1,1,1,0,0,0]$.   That is to say, the system   $[58:31,31,28,21,2,2]$   is equivalent to the system   $[2:1,1,1,0,0,0]$.   For the computation of the power indices, we shall use the latter system.

### Computing the Banzhaf Power Indices

With 6 players, there are actually   $2^6=64$   coalitions, but there are only 32 winning coalitions which come in eight types.  The characteristics and count of each type is given in the following table.

 Districts in Winning Coalitions Number of Coalitions Votes Extra Votes Critical Players Two large, no small 3 2 0 Two large Two large, one small 9 2 0 Two large Two large, two small 9 2 0 Two large Two large, three small 3 2 0 Two large Three large, no small 1 3 1 None Three large, one small 3 3 1 None Three large, two small 3 3 1 None Three large, three small 1 3 1 None Total: 32

The small districts are never critical.  In each of the 24 coalitions having just two large districts, both of the large districts are critical players.  Thus, the three large districts are critical a total of 48 times, or 16 times each. Therefore, each large district has   $\dfrac{16}{48}=33.33\%$   of the power, and each small district has zero power.

### Computing the Shapley-Shubik Power Indices

With 6 players, there are   $6!=720$   sequential coalitions.  For each sequential coalition, we must identify the pivotal player.  Since the second large district is always the pivotal player, there are only four cases.  In the computation of the number of sequential coalitions, the first factor is for the partitioning of the large districts, the second factor for the partitioning of the small districts, the third factor is the arrangement of the preceding supporters, and the fourth factor is the arrangement of the remaining players.

 Preceding supporters Preceding vote total(and votes needed) Pivotal Player Remaining players Number of sequential coalitions One large, no small 1 (1 needed) Large One large, three small $3!\times 1\times 1\times 4!=144$ One large, one small 1 (1 needed) Large One large, two small $3!\times 3\times 2\times 3!=216$ One large, two small 1 (1 needed) Large One large, one small $3!\times 3\times 3!\times 2=216$ One large, three small 1 (1 needed) Large One large, no small $3!\times 1\times 4!\times 1=144$ Total:  $720$

In each case, one of the large districts is the pivotal player.  Since each large district has the same number of votes, each must be pivotal the same number of times.  Therefore, each large district is pivotal   $\dfrac{720}{3}=240$   times, and has   $\dfrac{240}{720}=33.33\%$   of the power.  Of course, the small districts were never pivotal, and therefore have no power.