## Power on the United Nations Security Council

The Security Council of the United Nations consists of 15 members. Five of these members, China, France, Russia, the United Kingdom, and the United States, are permanent members. The other ten members are elected to two-year terms. A resolution in the Security Council is adopted when nine of the 15 members vote for the resolution, and none of the permanent members veto the resolution.

### An Equivalent Weighted Voting System

Consider the weighted voting system   $[39:7,7,7,7,7,1,1,1,1,1,1,1,1,1,1]$.   In this system, there are 15 players, just as the Security Council has 15 members. Five of those players are alike in that they each possess seven votes, and the other ten players are alike in their single votes. Therefore, these 15 players have a total of   $5\times 7+10\times 1=45$   votes.

We shall call each player having seven votes a dominant player. If one of the dominant players did not favor a motion, then at most 38 votes would be available for that motion.   Since the quota is 39 votes, a motion could not pass without every dominant player's support.  That is, each dominant player has a veto, just as each permanent member of the Security Council has a veto.

Having established that a successful motion requires the support of every dominant player, what about the other players? The dominant players will cast 35 votes in favor of a motion, and therefore another 4 votes are required to meet the quota. That is, four of the ten remaining players must also support a successful motion. In other words, a successful motion will be supported by at least nine players (the five dominant players, and four others) out of the 15 players in the voting system. This is the same standard as the requirement for a successful resolution in the Security Council.

### Computing the Banzhaf Power Indices

With 15 players, there are actually coalitions, but the winning coalitions come only in seven types.  The characteristics and count of each type is given in the following table.

 Winning Coalitions Number of Coalitions Votes Extra Votes Critical Players All permanent members,four non-permanent members ${}_{10} C_4=210$ 39 0 All members of coalition All permanent members,five non-permanent members ${}_{10} C_5=252$ 40 1 All permanent members All permanent members,six non-permanent members ${}_{10} C_6=210$ 41 2 All permanent members All permanent members,seven non-permanent members ${}_{10} C_7=120$ 42 3 All permanent members All permanent members,eight non-permanent members ${}_{10} C_8=45$ 43 4 All permanent members All permanent members,nine non-permanent members ${}_{10} C_9=10$ 44 5 All permanent members All permanent members,ten non-permanent members ${}_{10} C_{10}=1$ 45 6 All permanent members Total:  $848$

For any particular type of winning coalition, the number of coalitions is a combination of the number of ways to select the necessary number of players from the ten available.  Thus, there are 848 winning coalitions.  Since the permanent members always have veto power, they are critical players in every coalition.  The non-permanent members are critical only in the coalitions where the number of available votes is exactly equal to the quota.

Each permanent member is a member of every winning coalition.  Therefore, each permanent member is critical 848 times.

Each non-permanent member is critical only when they are a member of a coalition having exactly 39 votes.  There are 210 such coalitions, each with four non-permanent members.  Thus the ten non-permanent members are critical players a total of   $210\times 4=840$   times.  Since these ten non-permanent members are completely identical with regards to their power, they must be critical   $\dfrac{840}{10}=84$   times each.

Summarizing, the total number of times any player is critical is   $5\times 848+10\times 84=5080$   times.  Therefore, each permanent member has   $\dfrac{848}{5080}=16.69\%$   of the power, and each non-permanent member has   $\dfrac{84}{5080}=1.65\%$   of the power.

### Computing the Shapley-Shubik Power Indices

With 15 players, there are   $15!=1307674368000$   sequential coalitions.  For each sequential coalition, we must identify the pivotal player.  When the computation for the number of sequential coalitions contains four factors, the first factor is for the choice of the pivotal player, the second factor is for the partitioning of the non-permanent supporters, the third factor is for the arrangement of the preceding supporters, and the last factor is for the arrangement of the remaining players.

 Preceding supporters Preceding vote total(and votes needed) Pivotal Player Remaining players Number of sequential coalitions Four permanent,four non-permanent 32 (7 needed) Permanent 6 non-permanent $5\times{}_{10}C_4\times 8!\times 6!=30481920000$ Four permanent,five non-permanent 33 (6 needed) Permanent 5 non-permanent $5\times{}_{10}C_5\times 9!\times 5!=54867456000$ Four permanent,six non-permanent 34 (5 needed) Permanent 4 non-permanent $5\times{}_{10}C_6\times 10!\times 4!=91445760000$ Four permanent,seven non-permanent 35 (4 needed) Permanent 3 non-permanent $5\times{}_{10}C_7\times 11!\times 3!=143700480000$ Four permanent,eight non-permanent 36 (3 needed) Permanent 2 non-permanent $5\times{}_{10}C_8\times 12!\times 2=215550720000$ Four permanent,nine non-permanent 37 (2 needed) Permanent 1 non-permanent $5\times{}_{10}C_9\times 13!=311351040000$ Four permanent,ten non-permanent 38 (1 needed) Permanent none $5\times 14!=435891456000$ All permanent,three non-permanent 38 (1 needed) Non-permanent 6 non-permanent $5\times{}_9C_3\times 8!\times 6!=24385536000$ Total:  $15!=1307674368000$

We see that a non-permanent member is a pivotal player in just $24,385,536,000$ coalitions.  Each non-permanent member is pivotal in one-tenth of those coalitions, or exactly   $\dfrac{24385536000}{10}=2438553600$   times.  Therefore, each non-permanent member has   $\dfrac{2438553600}{1307674368000}=0.1865\%$   of the power.

A permanent member will be the pivotal player in the other $1,283,288,832,000$ coalitions.  With five permanent members, each is pivotal in one-fifth of those coalitions, or exactly   $\dfrac{1283288832000}{5}=256657766400$   times.  Therefore, each permanent member has   $\dfrac{256657766400}{1307674368000}=19.6270\%$   of the power.