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| | |-+  In the event of no EC majority, 58 House members could decide the election
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Author Topic: In the event of no EC majority, 58 House members could decide the election  (Read 515 times)
Del Tachi
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« on: December 06, 2012, 09:42:42 am »
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In the event that no candidate wins a majority of electoral votes, the election is decided by the House of Representatives.  In the House, members vote as a delegation from the state they are representing.  While it has never been seen before for obvious reasons, one could assume that a candidate would only have to receive a majority of any particular state's House delegation in order to "win" the entire state.  Whichever candidate wins the majority (or 26) of state delegations is elected President.

Assuming that all of the above is correct, it would only take 58 members of the House of Representatives to deliver a majority of state delegations in the event of no Electoral College majority. 

These 58 members would be:

1) The whole delegations of the states of Alaska, Idaho, Wyoming, Montana, North Dakota, South Dakota, Vermont, New Hampshire, Maine, Rhode Island, and Delaware; 

2) A simple majority of the delegations of the states of Hawaii, Oregon, Nevada, Utah, New Mexico, Nebraska, Kansas, Oklahoma, Iowa, Arkansas, Mississippi, Kentucky, West Virginia, South Carolina, and Connecticut. 

While only comprising 13.3% of House membership, these representatives compromise the 26-state majority needed to elect a President.

A scenario like the one mentioned above is certainly very unlikely to happen, but it could make for good fiction.  For example, in the event of a three-way tie in the Electoral College with an independent of third party candidate costing the two major an Electoral College majority, the third-place candidate could attempt to consolidate support from the 58 members that could (if they coalesced) determine the outcome of the election. 

This would be very undemocratic, however, and I seriously doubt that an attempt to do this would not be countered by severe civil disorder. 
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Ernest
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« Reply #1 on: December 06, 2012, 02:08:56 pm »

Your premise is correct, but your list of States and the number of members needed is not.

It requires 60 members, including all from the green states below, all but one from the blue, and all but two from the green.



Of course the idea that those 60 would be able to get together despite complete opposition from everyone else is pure fantasy.
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DreamTheater
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« Reply #2 on: December 20, 2012, 01:06:09 am »
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This would be hilarious Wink  I think I saw a youtube video that talked about this possibility once.
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Gary J
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« Reply #3 on: December 22, 2012, 08:03:46 am »
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Presumably the theoretical lowest possible number of members of the House, needed to elect a President, is 26 (since only one member of a state delegation needs to participate to decide the state's vote, if no other member of the delegation casts a ballot).
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Ernest
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« Reply #4 on: December 22, 2012, 02:00:04 pm »

The absolute theoretical minimum is 34.  The quorum needed for the vote is a minimum of one Member from two-thirds of the States.
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Quote from: Ignatius of Antioch
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Gary J
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« Reply #5 on: December 23, 2012, 11:56:00 am »
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You have to have at least one House member from two-thirds of the states participating in the contingent election, to meet the quorum requirement. Only 26 of them need to vote for the winning candidate.
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Ernest
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« Reply #6 on: December 23, 2012, 12:36:29 pm »

And if the other 8 needed for a quorum don't show up, there is no election, and if the result would be different if the full Congress weren't there, you can bet that those 8 wouldn't show up unless they supported the same person as the 26.
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Quote from: Ignatius of Antioch
He that possesses the word of Jesus, is truly able to bear his very silence. Epistle to the Ephesians 3:21a
The one thing everyone can agree on is that the media is biased against them.
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