Define a coalition as a subset of the set of the 51 states (counting DC as a state) that make the USA. Define a coalition as winning if the total number of electoral votes of the state in that coalition is 270 or more (let's ignore at first that two small states make things more complicated by allowing a mixed elector group). There are 2^51 total coalitions and each state belongs to 2^50 (about 1000 trillion) of them. For every state we can define its power as the number of winning coalition to which it belongs.

Has the power of every state been computed?

I am pretty sure it has been computed, it is just about 2000 trillion additions of less than 50 terms, but where can I find this. Also I am pretty sure that what I call power here has a more specific name, but I don't know it which prevents me probably to find what I am looking for using google.

  • Wouldn't there be 50! total coalitions? Or is my combinatorics too rusty? – David says Reinstate Monica Feb 3 '17 at 2:58
  • @DavidGrinberg A state/district can either be in a coalition or not. There are 51 of them. That gives 2^51 coalition splits: one for each binary number with 50 digits. There are 51! orderings of the states. 51 because we're include the District of Columbia. This is actually wrong, because Maine and Nebraska vote by district. So it's actually 2^56. But if we assume all congressional districts in ME and NE vote as their state does, 2^51. – Brythan Feb 3 '17 at 4:16
  • FiveThirtyEight, probably. – user4012 Feb 3 '17 at 12:19
  • 1
    This would be a great question for math.SE since it's connection to politics is minimal. However, it might be NP hard to solve. Basically, you're looking at combinations of electoral votes excluding a given state that add up to (270-state's electoral votes) or more. – user2565 Feb 3 '17 at 16:02
  • 2
    Also, since all states not in the coalition are part of the other coalition, we've counted all coalitions double and can divide the number of 2^51 coalitions by 2 to yield a possible 2^50 coalitions possible. – SQB Feb 3 '17 at 16:34

I happened to still have a copy of a computer program I wrote back in college to compute the Banzhaf Power Index of the Electoral College. Running it on the 2010 census apportionment (used for the 2012, 2016, and 2020 presidential elections) gives the following “power” values as a function of electoral votes:

 3: 0.022622 WY DC VT ND AK SD DE MT
 4: 0.030169 RI NH ME HI ID
 5: 0.037720 NE WV NM
 6: 0.045277 NV UT KS AR MS IA
 7: 0.052842 CT OK OR
 8: 0.060416 KY LA
 9: 0.067999 SC AL CO
10: 0.075594 MN WI MD MO
11: 0.083202 TN AZ IN MA
12: 0.090823 WA
13: 0.098460 VA
14: 0.106113 NJ
15: 0.113784 NC
16: 0.121475 GA MI
18: 0.136921 OH
20: 0.152464 PA IL
29: 0.223975 FL NY
38: 0.298862 TX
55: 0.471147 CA
| improve this answer | |
  • Good answer, although you may want to explain a bit a the Banzhaf power index. – SQB Feb 5 '17 at 11:39
  • Good answer, thank you. I believe I have seen your paper somewhere on the web. There is a subtlety however in the way the Banzhaf power index is defined. If I believe the paper I have seen, there is the Banzhaf power index of a state, define by counting for a given state the number of winning coalition it is a part of, and which would become losing if the state switched sides. This is a (probably better) variant of what I asked. But then, there is the Banzhaf Power index of a typical voter in a state, which is different, and is what is reported in your post if I am right... – Joël Feb 5 '17 at 15:46
  • It is estimated (not precisely computed) as the power index of the state divided by the *square root of its population, for reason I have not yet taken the time to understand, being rather busy this week. But am I right about the interpretation of your figures? – Joël Feb 5 '17 at 15:47
  • 1
    @Joël: My figures on consider the states, not the population within them. As for the square root thing, the reason is that, if voters are considered as independent random Bernoulli variables, the probability of an exact tie (which can be broken by a single voter) is O(1/√n), where n is population. – dan04 Feb 5 '17 at 16:23

Without performing the actual calculation, it's easy to see that the more electoral votes a state has, the more power (as per your definition) it will have.

Let's suppose a smaller country, with 4 states, conveniently named A, B, C, and D, having 1, 2, 3, and 4 votes in that country's electoral college each respectively. Those votes add up to 10, so a coalition needs at least 6 votes to be a winning coalition.

There are 8 divisions possible, yielding two coalitions each:

  1. A + B + C + D vs. no-one
  2. A + B + C vs D
  3. A + B + D vs C
  4. A + C + D vs B
  5. B + C + D vs A
  6. A + B vs C + D
  7. A + C vs B + D
  8. A + D vs B + C

For state D, with its 4 votes in the electoral college, there are two ways of deciding the coalitions that will not see it win:

  1. State D by itself (option 2 above)
  2. States A and D together (a tie; option 8 above).

All other 6 ways of dividing the states, will see the coalition D is in, win.

State A on the other hand, with its single vote, has only 4 ways winning (options 1, 2, 3, and 4 above).

For state B there are 5 ways of winning (1, 2, 3, 5, 7) and for state C there are 5 as well (1, 2, 4, 5, and 6).

The site 270 to Win has an interactive map that you may peruse to experiment with this for yourself.

| improve this answer | |
  • 1
    Note that "more electoral votes = more power" is not absolute. It's possible to construct electoral-vote allocations that give smaller states more power than their vote counts would seem to indicate. (Source: "Mathematical Recreations" in an old edition of Scientific American; setting up such an arrangement is harder with 51 states than with the small numbers used in the article.) – Mark Feb 6 '17 at 1:54
  • @Mark: Though the opposite, an electoral vote allocation that gives smaller states less power than their vote counts would seem to indicate, is trivial. Have 61.6% of the national population move to one state, so that it gets 268 of the 435 House seats (and thus, 270 of the 538 electoral votes). Then that one super-state would have total control of presidential elections. – dan04 Feb 6 '17 at 2:14

Let me start by saying that I'm still working on figuring out the problem for the most recent election, but I think I'm making some progress and wanted to share the progress while the real problem is computing. There are a lot of possible coalition sets. But there are a number of ways to strategically reduce the number of sets in the solution space. The first cut we can make is defining the minimum number of states required to obtain an electoral college majority. For 2012, the minimum number of states to reach 270 is 12.

California - 55
Texas - 38
Florida, New York - 29
Illinois, Pennsylvania - 20
Ohio,  18
Georgia, Michigan - 16
North Carolina - 15
New Jersey - 14
Virginia - 13

This reduces the number of possible combinations from 2251799813685247 to 2251735594475336. That is not a big reduction, but I'm working on some methods to make further cuts.

I thought it would be interesting to look at a more manageable problem, the first Presidential election in 1788. In 1788 there were 69 total electoral votes from 10 states.

Connecticut - 7
Delaware - 3
Georgia - 5
Maryland - 6
Massachusetts - 10
New Hampshire - 5
New Jersey - 6
Pennsylvania - 10
South Carolina - 7
Virginia - 10

In this case a coalition needed a majority of the 69 total votes, 35 votes, to be on the winning side. This translates to a minimum coalition size of four states. There are 848 total coalitions that contain at least four states. Now testing each possible coalition

{Connecticut, Delaware, Georgia, Maryland}
{Connecticut, Delaware, Georgia, Massachusetts}
{Connecticut, Delaware, Georgia, New Hampshire}
{Connecticut, Delaware, Georgia, Maryland, Massachusetts, New Hampshire, New Jersey, Pennsylvania, South Carolina, Virginia}

it is possible to determine the probability that a particular state is part of the coalition that selects the president. During the first election, the probabilities are:

Connecticut - 0.353774    
Delaware - 0.316038
Georgia - 0.334906
Maryland - 0.346698
Massachusetts - 0.39033
New Hampshire - 0.334906
New Jersey - 0.346698
Pennsylvania - 0.39033
South Carolina - 0.353774
Virginia - 0.39033

I'll be sure to update if I can make any progress on the current election cycle.

| improve this answer | |
  • 1
    In reality, it isn't a simple question of mathematical combinations. Based on demographics, you have states that are solidly one party or the other, states that are normally one party or the other but can vote differently in unusual or extreme circumstances, and swing states. The vast majority of combinations are theoretical noise. – user11810 Feb 4 '17 at 20:17
  • @fixer1234 It depends on what question you have really in mind. If the question is "does the electoral college advantages a state or an other, over the long run", then it is not really relevant that right now, some states are surely in one party's column and only a few are swing states. Over a hundred year, I am sure that there are not many states, if any, that have never changed their vote (democratic or republican) in presidential elections. – Joël Feb 5 '17 at 15:39
  • @Marchi Thanks, great answer. It is really surprising how Delaware with just 3 electoral votes, has almost the same power than Virgina which has 10. Are you sure of your figures? – Joël Feb 5 '17 at 15:49
  • @fixer1234 In fact, no state has voted constitently for one party since 1964, as I (easily) checked. In the not so long run, all states are swing states. However, you're right that there exist correlation between states, states who tend to vote together. The old south is one such block that jump to the eyes when looking at the electoral maps, but there doesn't seem to be many others. – Joël Feb 5 '17 at 18:21
  • 1
    @fixer1234: You are absolutely correct in stating that in real elections states do not have discrete probabilities of voting for one candidate/party over another. In practice, each state has a distribution of probabilities associated with the electoral selection. To then find the probabilities of a particular candidate to win, and if a certain state participated in the winning coalition, is a much more involved statistical model. You most likely would have to do some bootstrapping to get the answer. – Marchi Feb 5 '17 at 21:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .