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Rerunning the electoral corner case with new population stats?
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Phil Carmody
Sep 13, 2012, 7:19:54 AM9/13/12
to
Can someone come up with a figure for the theoretical lowest
proportion of votes cast that would elect a president, assuming no
"faithless electors"?
I think it was one of the regulars (possibly MSB?) who did it in the
past. I presume that all that would be required would be sticking new
population figures in the top and applying the same handle-turning.
As population increases, the minimal proportion should diminish, so
there should be a more shocking figure than last time.
Phil
--
Regarding TSA regulations:
How are four small bottles of liquid different from one large bottle?
Because four bottles can hold the components of a binary liquid explosive,
whereas one big bottle can't. -- camperdave responding to MacAndrew on /.
proportion of votes cast that would elect a president, assuming no
"faithless electors"?
I think it was one of the regulars (possibly MSB?) who did it in the
past. I presume that all that would be required would be sticking new
population figures in the top and applying the same handle-turning.
As population increases, the minimal proportion should diminish, so
there should be a more shocking figure than last time.
Phil
--
Regarding TSA regulations:
How are four small bottles of liquid different from one large bottle?
Because four bottles can hold the components of a binary liquid explosive,
whereas one big bottle can't. -- camperdave responding to MacAndrew on /.
Mark Brader
Sep 13, 2012, 2:34:13 PM9/13/12
to
Phil Carmody:
a repeat of an earlier posting that's the one Phil is thinking of.
I'm not interest in redoing any computations myself, but note that
I did not produce a definitive answer even in 2004.
> As population increases, the minimal proportion should diminish, so
> there should be a more shocking figure than last time.
The critical number is not the country's total population but how
much of it is concentrated into a few states, because the less
populous states have a higher ratio of electoral votes to voters.
############################## 2004 ##############################
D. Gates writes:
| What's the smallest number of people that could pick the U.S.
| president?
1. This has occurred in 1865, 1881, 1901, 1963 (although some
dispute this), and by different means in 1974.
The next smallest is 5. This occurred in 2000 (although some dispute
this interpretation). A smaller number would be possible in this mode
of picking if there were vacancies on the Supreme Court, or some of
the justices were to recuse themselves.
The next smallest number that has actually occurred is 8, in 1877.
If you meant the smallest number of people to *elect* a president,
the current theoretical minimum is 26, and the practical minimum
is 56. This is based on the 12th Amendment:
# ...The person having the greatest number of votes for President, shall
# be the President, if such number be a majority of the whole number of
# Electors appointed; and if no person have such majority, then from the
# persons having the highest numbers not exceeding three on the list of
# those voted for as President, the House of Representatives shall choose
# immediately, by ballot, the President. But in choosing the President,
# the votes shall be taken by states, the representation from each state
# having one vote; a quorum for this purpose shall consist of a member or
# members from two-thirds of the states, and a majority of all the states
# shall be necessary to a choice...
If at least one member from each of at least 33 states is present in
the House when the question falls to it, and there are 26 states for
which only a single member is present, then those 26 members would
form a majority.
But in practice we can be sure that all members would be present and
voting on such an occasion; and we can assume that a state would cast
a vote only if a majority of its House members supported that candidate.
In that case the practical minimum is 1 vote from the 7 states that
only *have* one representative; 2 votes each from the 10 states that
have 2 or 3; 3 votes each from the 7 states with 4 or 5; and 4 votes
each the 2 states with 6 representatives, making 56 votes from 26 states.
(See the Electoral College table below to see which states these are,
subtracting 2 from the number of EC votes in each case, and ignoring
DC, which is represented on the EC but not in the House.)
If there were more than two candidates (quite likely if the vote was
going to the House in the first place) and a plurality of members from
a particular state, rather than a majority, was considered sufficient
for that state to cast its vote, then 2 votes from one state could be
a plurality, and the above total of 56 would be theoretically reduced
to 45.
| I'm thinking that the answer would lie in finding the least populated
| states, figuring out how many of those states you'd need to get enough
| electoral votes, then adding up the number needed to get 51% in each
| of those states.
Oh, *popular* votes. For popular votes the answer is 0. This has
never happened, but imagine a case where both the President-elect
and the VP-elect die before the Electoral College votes. In practice,
most likely people in their party would select a substitute candidate
and the Electors pledged to the original candidate would vote for
him. Or if no majority in the Electoral College was achieved that
way, the election would fall to the House of Representatives. In
any case, the person elected would not have received any popular
votes for President or VP.
The total of 0 could also theoretically be achieved if there were
270 "faithless electors", i.e. electors who voted for someone other
than the candidate they were pledged to.
Rather than continuing, I'll now quote the related question that
Leroy Quet asked here in March:
> What is the maximum percentage of total votes cast for any
> presidential candidate, in a two-person race, that a candidate can
> receive in the nation-wide POPULAR vote, yet still lose the electoral
> count?
I answered this as follows (having covered the "faithless elector"
case in a separate posting). All text from here on was previously
posted at that time.
*
... I make the answer to be about 99.999988% (using year 2000 data
for the sake of definiteness).
> I wonder if this is really a straight-forward question, since I
> believe that some states do not simply carry out a straight popular
> vote so as to determine who wins those states' electoral votes, but
> instead tally up only the number of counties/districts won by each
> candidate.
Right. Two states do this, Maine and Nebraska. More specifically,
each state produces two electors based on the statewide popular vote
and one for each congressional district. However, this does not affect
the solution below.
The following table shows the voting-age population in each state
and DC as estimated by the US Census Bureau for the date of the 2000
election <http://www.census.gov/Press-Release/www/2000/cb00-125.html>
(in thousands), sorted by this number, and the number of electoral
votes each state had in the 2000 election. The last two columns simply
show the cumulative totals of the other two numbers, reading down.
(Note that the voting-age population is not the exact correct measure
-- we want the number of people actually eligible to vote, which
depends on other things than age -- but it's going to be fairly close.)
California | 24,873 | 54 || 24,873 | 54
Texas | 14,850 | 32 || 39,723 | 86
New York | 13,805 | 33 || 53,528 | 119
Florida | 11,774 | 25 || 65,302 | 144
Pennsylvania | 9,155 | 23 || 74,457 | 167
Illinois | 8,983 | 22 || 83,440 | 189
Ohio | 8,433 | 21 || 91,873 | 210
Michigan | 7,358 | 18 || 99,231 | 228
New Jersey | 6,245 | 15 || 105,476 | 243
Georgia | 5,893 | 13 || 111,369 | 256
North Carolina | 5,797 | 14 || 117,166 | 270
Virginia | 5,263 | 13 || 122,429 | 283
Massachusetts | 4,749 | 12 || 127,178 | 295
Indiana | 4,448 | 12 || 131,626 | 307
Washington | 4,368 | 11 || 135,994 | 318
Tennessee | 4,221 | 11 || 140,215 | 329
Missouri | 4,105 | 11 || 144,320 | 340
Wisconsin | 3,930 | 11 || 148,250 | 351
Maryland | 3,925 | 10 || 152,175 | 361
Arizona | 3,625 | 8 || 155,800 | 369
Minnesota | 3,547 | 10 || 159,347 | 379
Alabama | 3,333 | 9 || 162,680 | 388
Louisiana | 3,255 | 9 || 165,935 | 397
Colorado | 3,067 | 8 || 169,002 | 405
Kentucky | 2,993 | 8 || 171,995 | 413
South Carolina | 2,977 | 8 || 174,972 | 421
Oklahoma | 2,531 | 8 || 177,503 | 429
Oregon | 2,530 | 7 || 180,033 | 436
Connecticut | 2,499 | 8 || 182,532 | 444
Iowa | 2,165 | 7 || 184,697 | 451
Mississippi | 2,047 | 7 || 186,744 | 458
Kansas | 1,983 | 6 || 188,727 | 464
Arkansas | 1,929 | 6 || 190,656 | 470
Utah | 1,465 | 5 || 192,121 | 475
West Virginia | 1,416 | 5 || 193,537 | 480
Nevada | 1,390 | 4 || 194,927 | 484
New Mexico | 1,263 | 5 || 196,190 | 489
Nebraska | 1,234 | 5 || 197,424 | 494
Maine | 968 | 4 || 198,392 | 498
Idaho | 921 | 4 || 199,313 | 502
New Hampshire | 911 | 4 || 200,224 | 506
Hawaii | 909 | 4 || 201,133 | 510
Rhode Island | 753 | 4 || 201,886 | 514
Montana | 668 | 3 || 202,554 | 517
Delaware | 582 | 3 || 203,136 | 520
South Dakota | 543 | 3 || 203,679 | 523
North Dakota | 477 | 3 || 204,156 | 526
Vermont | 460 | 3 || 204,616 | 529
Alaska | 430 | 3 || 205,046 | 532
District of Columbia | 411 | 3 || 205,457 | 535
Wyoming | 358 | 3 || 205,815 | 538
The number of electoral votes required to win in a two-person race is
538/2 + 1 = 270, which is exactly the total of the first 11 states
listed. Their total voting-age population is 117,166,000 out of a
national total of 205,815,000.
So it could happen that in those 11 states there is a turnout of 1 voter
per state, who votes for candidate A, and in the other 39 states and DC,
there is a 100% turnout voting for candidate B. The popular vote is
88,649,000-11 or a bit under 99.999988% for B, but assuming no faithless
electors, A wins the electoral college 270-268.
Since the states listed first have on average fewer electors per unit
population than the others, it's possible that there is another solution
where some of the less populous states vote for A; the popular vote
might then be something like 100,000,000-12 or 110,000,000-13, giving
giving a slightly larger percentage voting for B. Such an alternate
solution might involve a split electoral vote in Maine or Nebraska.
Howver,
I suspect that the question *intended* a uniform turnout through all
the states, to be assumed. In that case, of course, the answer is
very different. Now we want a solution where the less populous states
use their higher ratio of electors to outvote the more populous ones,
so we need a group of more populous states that totals 268 electors,
and these will vote for B instead of A.
For example, take the top 10 on the list, then Massachusetts rather
than North Carolina. For simplicity assume 100% turnout. Then in
these 11 states we could have 111,369,000 + 4,749,000 = 116,118,000
people voting 100% for B, for 268 electoral votes.
In Maine and Nebraska, there is a total of 5 congressional districts.
With the remaining 37 states and DC, this makes a total of 43 contests.
Say the actual total number of voters in each contest is odd (despite
the numbers above being rounded to multiples of 1,000); then it will
suffice to have each race won by 1 vote, or a total of 43 more votes
for A than for B. (The statewide races in Maine and Nebraska will
of course automatically go the same way if every district is won by the
same candidate -- in fact, historically, this has *always* happpened in
those states ever since they adopted the present method.)
So we have 89,697,000 people of voting age in those states and DC;
adding 1 to make the total odd, they would be voting 44,848,522 for A
vs. 44,848,479 for B. A wins all 270 electoral votes and the presidency,
but the total popular vote is 156,217,479-44,848,522 or about 77.694627%
for B.
Since this was a rough "first-fit" solution, it's entirely possible
that a slightly higher percentage can be found by experimenting with
different sets of states. If I was going to experiment further, the
first thing I'd so is re-sort the above table in order of the ratio
of voting-age population to electors.
--
Mark Brader, Toronto | "I conducted a Usenet poll ... on this subject ...
m...@vex.net | Laura is single. By a 2-1 margin." --Ken Perlow
My text in this article is in the public domain.
> Can someone come up with a figure for the theoretical lowest
> proportion of votes cast that would elect a president, assuming no
> "faithless electors"?
>
> I think it was one of the regulars (possibly MSB?) who did it in the
> past. I presume that all that would be required would be sticking new
> population figures in the top and applying the same handle-turning.
Below is a repat of a posting of mine from 2004, which in turn includes
> proportion of votes cast that would elect a president, assuming no
> "faithless electors"?
>
> I think it was one of the regulars (possibly MSB?) who did it in the
> past. I presume that all that would be required would be sticking new
> population figures in the top and applying the same handle-turning.
a repeat of an earlier posting that's the one Phil is thinking of.
I'm not interest in redoing any computations myself, but note that
I did not produce a definitive answer even in 2004.
> As population increases, the minimal proportion should diminish, so
> there should be a more shocking figure than last time.
much of it is concentrated into a few states, because the less
populous states have a higher ratio of electoral votes to voters.
############################## 2004 ##############################
D. Gates writes:
| What's the smallest number of people that could pick the U.S.
| president?
1. This has occurred in 1865, 1881, 1901, 1963 (although some
dispute this), and by different means in 1974.
The next smallest is 5. This occurred in 2000 (although some dispute
this interpretation). A smaller number would be possible in this mode
of picking if there were vacancies on the Supreme Court, or some of
the justices were to recuse themselves.
The next smallest number that has actually occurred is 8, in 1877.
If you meant the smallest number of people to *elect* a president,
the current theoretical minimum is 26, and the practical minimum
is 56. This is based on the 12th Amendment:
# ...The person having the greatest number of votes for President, shall
# be the President, if such number be a majority of the whole number of
# Electors appointed; and if no person have such majority, then from the
# persons having the highest numbers not exceeding three on the list of
# those voted for as President, the House of Representatives shall choose
# immediately, by ballot, the President. But in choosing the President,
# the votes shall be taken by states, the representation from each state
# having one vote; a quorum for this purpose shall consist of a member or
# members from two-thirds of the states, and a majority of all the states
# shall be necessary to a choice...
If at least one member from each of at least 33 states is present in
the House when the question falls to it, and there are 26 states for
which only a single member is present, then those 26 members would
form a majority.
But in practice we can be sure that all members would be present and
voting on such an occasion; and we can assume that a state would cast
a vote only if a majority of its House members supported that candidate.
In that case the practical minimum is 1 vote from the 7 states that
only *have* one representative; 2 votes each from the 10 states that
have 2 or 3; 3 votes each from the 7 states with 4 or 5; and 4 votes
each the 2 states with 6 representatives, making 56 votes from 26 states.
(See the Electoral College table below to see which states these are,
subtracting 2 from the number of EC votes in each case, and ignoring
DC, which is represented on the EC but not in the House.)
If there were more than two candidates (quite likely if the vote was
going to the House in the first place) and a plurality of members from
a particular state, rather than a majority, was considered sufficient
for that state to cast its vote, then 2 votes from one state could be
a plurality, and the above total of 56 would be theoretically reduced
to 45.
| I'm thinking that the answer would lie in finding the least populated
| states, figuring out how many of those states you'd need to get enough
| electoral votes, then adding up the number needed to get 51% in each
| of those states.
Oh, *popular* votes. For popular votes the answer is 0. This has
never happened, but imagine a case where both the President-elect
and the VP-elect die before the Electoral College votes. In practice,
most likely people in their party would select a substitute candidate
and the Electors pledged to the original candidate would vote for
him. Or if no majority in the Electoral College was achieved that
way, the election would fall to the House of Representatives. In
any case, the person elected would not have received any popular
votes for President or VP.
The total of 0 could also theoretically be achieved if there were
270 "faithless electors", i.e. electors who voted for someone other
than the candidate they were pledged to.
Rather than continuing, I'll now quote the related question that
Leroy Quet asked here in March:
> What is the maximum percentage of total votes cast for any
> presidential candidate, in a two-person race, that a candidate can
> receive in the nation-wide POPULAR vote, yet still lose the electoral
> count?
I answered this as follows (having covered the "faithless elector"
case in a separate posting). All text from here on was previously
posted at that time.
*
... I make the answer to be about 99.999988% (using year 2000 data
for the sake of definiteness).
> I wonder if this is really a straight-forward question, since I
> believe that some states do not simply carry out a straight popular
> vote so as to determine who wins those states' electoral votes, but
> instead tally up only the number of counties/districts won by each
> candidate.
Right. Two states do this, Maine and Nebraska. More specifically,
each state produces two electors based on the statewide popular vote
and one for each congressional district. However, this does not affect
the solution below.
The following table shows the voting-age population in each state
and DC as estimated by the US Census Bureau for the date of the 2000
election <http://www.census.gov/Press-Release/www/2000/cb00-125.html>
(in thousands), sorted by this number, and the number of electoral
votes each state had in the 2000 election. The last two columns simply
show the cumulative totals of the other two numbers, reading down.
(Note that the voting-age population is not the exact correct measure
-- we want the number of people actually eligible to vote, which
depends on other things than age -- but it's going to be fairly close.)
California | 24,873 | 54 || 24,873 | 54
Texas | 14,850 | 32 || 39,723 | 86
New York | 13,805 | 33 || 53,528 | 119
Florida | 11,774 | 25 || 65,302 | 144
Pennsylvania | 9,155 | 23 || 74,457 | 167
Illinois | 8,983 | 22 || 83,440 | 189
Ohio | 8,433 | 21 || 91,873 | 210
Michigan | 7,358 | 18 || 99,231 | 228
New Jersey | 6,245 | 15 || 105,476 | 243
Georgia | 5,893 | 13 || 111,369 | 256
North Carolina | 5,797 | 14 || 117,166 | 270
Virginia | 5,263 | 13 || 122,429 | 283
Massachusetts | 4,749 | 12 || 127,178 | 295
Indiana | 4,448 | 12 || 131,626 | 307
Washington | 4,368 | 11 || 135,994 | 318
Tennessee | 4,221 | 11 || 140,215 | 329
Missouri | 4,105 | 11 || 144,320 | 340
Wisconsin | 3,930 | 11 || 148,250 | 351
Maryland | 3,925 | 10 || 152,175 | 361
Arizona | 3,625 | 8 || 155,800 | 369
Minnesota | 3,547 | 10 || 159,347 | 379
Alabama | 3,333 | 9 || 162,680 | 388
Louisiana | 3,255 | 9 || 165,935 | 397
Colorado | 3,067 | 8 || 169,002 | 405
Kentucky | 2,993 | 8 || 171,995 | 413
South Carolina | 2,977 | 8 || 174,972 | 421
Oklahoma | 2,531 | 8 || 177,503 | 429
Oregon | 2,530 | 7 || 180,033 | 436
Connecticut | 2,499 | 8 || 182,532 | 444
Iowa | 2,165 | 7 || 184,697 | 451
Mississippi | 2,047 | 7 || 186,744 | 458
Kansas | 1,983 | 6 || 188,727 | 464
Arkansas | 1,929 | 6 || 190,656 | 470
Utah | 1,465 | 5 || 192,121 | 475
West Virginia | 1,416 | 5 || 193,537 | 480
Nevada | 1,390 | 4 || 194,927 | 484
New Mexico | 1,263 | 5 || 196,190 | 489
Nebraska | 1,234 | 5 || 197,424 | 494
Maine | 968 | 4 || 198,392 | 498
Idaho | 921 | 4 || 199,313 | 502
New Hampshire | 911 | 4 || 200,224 | 506
Hawaii | 909 | 4 || 201,133 | 510
Rhode Island | 753 | 4 || 201,886 | 514
Montana | 668 | 3 || 202,554 | 517
Delaware | 582 | 3 || 203,136 | 520
South Dakota | 543 | 3 || 203,679 | 523
North Dakota | 477 | 3 || 204,156 | 526
Vermont | 460 | 3 || 204,616 | 529
Alaska | 430 | 3 || 205,046 | 532
District of Columbia | 411 | 3 || 205,457 | 535
Wyoming | 358 | 3 || 205,815 | 538
The number of electoral votes required to win in a two-person race is
538/2 + 1 = 270, which is exactly the total of the first 11 states
listed. Their total voting-age population is 117,166,000 out of a
national total of 205,815,000.
So it could happen that in those 11 states there is a turnout of 1 voter
per state, who votes for candidate A, and in the other 39 states and DC,
there is a 100% turnout voting for candidate B. The popular vote is
88,649,000-11 or a bit under 99.999988% for B, but assuming no faithless
electors, A wins the electoral college 270-268.
Since the states listed first have on average fewer electors per unit
population than the others, it's possible that there is another solution
where some of the less populous states vote for A; the popular vote
might then be something like 100,000,000-12 or 110,000,000-13, giving
giving a slightly larger percentage voting for B. Such an alternate
solution might involve a split electoral vote in Maine or Nebraska.
Howver,
I suspect that the question *intended* a uniform turnout through all
the states, to be assumed. In that case, of course, the answer is
very different. Now we want a solution where the less populous states
use their higher ratio of electors to outvote the more populous ones,
so we need a group of more populous states that totals 268 electors,
and these will vote for B instead of A.
For example, take the top 10 on the list, then Massachusetts rather
than North Carolina. For simplicity assume 100% turnout. Then in
these 11 states we could have 111,369,000 + 4,749,000 = 116,118,000
people voting 100% for B, for 268 electoral votes.
In Maine and Nebraska, there is a total of 5 congressional districts.
With the remaining 37 states and DC, this makes a total of 43 contests.
Say the actual total number of voters in each contest is odd (despite
the numbers above being rounded to multiples of 1,000); then it will
suffice to have each race won by 1 vote, or a total of 43 more votes
for A than for B. (The statewide races in Maine and Nebraska will
of course automatically go the same way if every district is won by the
same candidate -- in fact, historically, this has *always* happpened in
those states ever since they adopted the present method.)
So we have 89,697,000 people of voting age in those states and DC;
adding 1 to make the total odd, they would be voting 44,848,522 for A
vs. 44,848,479 for B. A wins all 270 electoral votes and the presidency,
but the total popular vote is 156,217,479-44,848,522 or about 77.694627%
for B.
Since this was a rough "first-fit" solution, it's entirely possible
that a slightly higher percentage can be found by experimenting with
different sets of states. If I was going to experiment further, the
first thing I'd so is re-sort the above table in order of the ratio
of voting-age population to electors.
--
Mark Brader, Toronto | "I conducted a Usenet poll ... on this subject ...
m...@vex.net | Laura is single. By a 2-1 margin." --Ken Perlow
My text in this article is in the public domain.
Phil Carmody
Sep 13, 2012, 7:26:42 PM9/13/12
to
m...@vex.net (Mark Brader) writes:
> Phil Carmody:
> > Can someone come up with a figure for the theoretical lowest
> > proportion of votes cast that would elect a president, assuming no
> > "faithless electors"?
> >
> > I think it was one of the regulars (possibly MSB?) who did it in the
> > past. I presume that all that would be required would be sticking new
> > population figures in the top and applying the same handle-turning.
>
> Below is a repat of a posting of mine from 2004, which in turn includes
> a repeat of an earlier posting that's the one Phil is thinking of.
> I'm not interest in redoing any computations myself, but note that
> I did not produce a definitive answer even in 2004.
Thanks fine, Mark, I can't believe it was 2004! It's begging for a
> Phil Carmody:
> > Can someone come up with a figure for the theoretical lowest
> > proportion of votes cast that would elect a president, assuming no
> > "faithless electors"?
> >
> > I think it was one of the regulars (possibly MSB?) who did it in the
> > past. I presume that all that would be required would be sticking new
> > population figures in the top and applying the same handle-turning.
>
> Below is a repat of a posting of mine from 2004, which in turn includes
> a repeat of an earlier posting that's the one Phil is thinking of.
> I'm not interest in redoing any computations myself, but note that
> I did not produce a definitive answer even in 2004.
perl script to take care of the details, including not necessarily
greedily settling for the first fit.
> > As population increases, the minimal proportion should diminish, so
> > there should be a more shocking figure than last time.
>
> The critical number is not the country's total population but how
> much of it is concentrated into a few states, because the less
> populous states have a higher ratio of electoral votes to voters.
bloat merrily in order to decrease the ratio.
Cheers,
Don Del Grande
Sep 13, 2012, 8:59:38 PM9/13/12
to
Mark Brader wrote:
>Phil Carmody:
>> Can someone come up with a figure for the theoretical lowest
>> proportion of votes cast that would elect a president, assuming no
>> "faithless electors"?
>
>Phil Carmody:
>> Can someone come up with a figure for the theoretical lowest
>> proportion of votes cast that would elect a president, assuming no
>> "faithless electors"?
>
>Oh, *popular* votes. For popular votes the answer is 0. This has
>never happened, but imagine a case where both the President-elect
>and the VP-elect die before the Electoral College votes. In practice,
>most likely people in their party would select a substitute candidate
>and the Electors pledged to the original candidate would vote for
>him. Or if no majority in the Electoral College was achieved that
>way, the election would fall to the House of Representatives. In
>any case, the person elected would not have received any popular
>votes for President or VP.
>
>The total of 0 could also theoretically be achieved if there were
>270 "faithless electors", i.e. electors who voted for someone other
>than the candidate they were pledged to.
Also, there is nothing in the Constitution of the USA that says that
>never happened, but imagine a case where both the President-elect
>and the VP-elect die before the Electoral College votes. In practice,
>most likely people in their party would select a substitute candidate
>and the Electors pledged to the original candidate would vote for
>him. Or if no majority in the Electoral College was achieved that
>way, the election would fall to the House of Representatives. In
>any case, the person elected would not have received any popular
>votes for President or VP.
>
>The total of 0 could also theoretically be achieved if there were
>270 "faithless electors", i.e. electors who voted for someone other
>than the candidate they were pledged to.
electors have to be chosen based on popular votes - or any other vote,
for that matter. "Each state shall appoint, in such manner as the
Legislature thereof may direct, a number of electors, equal to the
whole number of Senators and Representatives to which the State may be
entitled in the Congress." (Article II, Section 1; the 23rd Amendment
gives three to the District of Columbia) In theory, each state can
choose its electors based on random draw, rock-paper-scissors, or best
two out of three falls.
-- Don
Eric Sosman
Sep 13, 2012, 9:23:55 PM9/13/12
to
On 9/13/2012 8:59 PM, Don Del Grande wrote:
>[...]
--
Eric Sosman
eso...@ieee-dot-org.invalid
"The speed at which the system fails is usually not important."
>[...]
> Also, there is nothing in the Constitution of the USA that says that
> electors have to be chosen based on popular votes - or any other vote,
> for that matter. "Each state shall appoint, in such manner as the
> Legislature thereof may direct, a number of electors, equal to the
> whole number of Senators and Representatives to which the State may be
> entitled in the Congress." (Article II, Section 1; the 23rd Amendment
> gives three to the District of Columbia) In theory, each state can
> choose its electors based on random draw, rock-paper-scissors, or best
> two out of three falls.
Me! Me! I'm the first to mention Jesse Ventura!
> electors have to be chosen based on popular votes - or any other vote,
> for that matter. "Each state shall appoint, in such manner as the
> Legislature thereof may direct, a number of electors, equal to the
> whole number of Senators and Representatives to which the State may be
> entitled in the Congress." (Article II, Section 1; the 23rd Amendment
> gives three to the District of Columbia) In theory, each state can
> choose its electors based on random draw, rock-paper-scissors, or best
> two out of three falls.
--
Eric Sosman
eso...@ieee-dot-org.invalid
"The speed at which the system fails is usually not important."
Mark Brader
Sep 13, 2012, 11:48:55 PM9/13/12
to
Mark Brader:
Indeed, in the 1789 election some of them weren't.
However, if we're assuming no changes to existing laws, then the popular
vote is applicable.
--
Mark Brader, Toronto | Actor sent to jail for not finishing sentence
m...@vex.net | --Knoxville, TN, News-Sentinel, 1989-01-21
>> The total of 0 could also theoretically be achieved if there were
>> 270 "faithless electors", i.e. electors who voted for someone other
>> than the candidate they were pledged to.
Don Del Grande:
>> 270 "faithless electors", i.e. electors who voted for someone other
>> than the candidate they were pledged to.
> Also, there is nothing in the Constitution of the USA that says that
> electors have to be chosen based on popular votes - or any other vote,
> for that matter...
> electors have to be chosen based on popular votes - or any other vote,
Indeed, in the 1789 election some of them weren't.
However, if we're assuming no changes to existing laws, then the popular
vote is applicable.
--
Mark Brader, Toronto | Actor sent to jail for not finishing sentence
m...@vex.net | --Knoxville, TN, News-Sentinel, 1989-01-21
Harry Vaderchi
Sep 17, 2012, 4:34:00 AM9/17/12
to
On Thu, 13 Sep 2012 19:34:13 +0100, Mark Brader <m...@vex.net> wrote:
> Phil Carmody:
>> Can someone come up with a figure for the theoretical lowest
>> proportion of votes cast that would elect a president, assuming no
>> "faithless electors"?
>>
>> I think it was one of the regulars (possibly MSB?) who did it in the
>> past.
[snip detailed analysis]
> Phil Carmody:
>> Can someone come up with a figure for the theoretical lowest
>> proportion of votes cast that would elect a president, assuming no
>> "faithless electors"?
>>
>> I think it was one of the regulars (possibly MSB?) who did it in the
>> past.
American democracy leading the world!
--
[dash dash space newline 4line sig]
Albi CNU
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