The Republican Primary Race Is Filling Up

[Hohn]

It is confirmed that Cruz has been mathematically eliminated from the run.

Yes, and math is an important part of this. But not all of this.

Trump must be crazy to want to be President after Obama and get blamed for the mess Obama made.

 

[GodFearinGunTotin]

It is confirmed that Cruz has been mathematically eliminated from the run.

No. He might not be able to get 1237 on the first ballot (which I believe he's more or less acknowledged for a while now) but that does not mean he's been mathematically eliminated...yet.

 

[Alpo]

Yes, and math is an important part of this. But not all of this.

Trump must be crazy to want to be President after Obama and get blamed for the mess Obama made.

 

[Timjoebillybob]

No. He might not be able to get 1237 on the first ballot (which I believe he's more or less acknowledged for a while now) but that does not mean he's been mathematically eliminated...yet.

Actually he can still get 1237 on the first ballot. There are something like 200 unbound delegates, and that doesn't the ones that Rubio that might be released if he officially drops out and not just suspends his campaign. He can't get 1237 bound delegates though.

 

[GodFearinGunTotin]

Actually he can still get 1237 on the first ballot. There are something like 200 unbound delegates, and that doesn't the ones that Rubio that might be released if he officially drops out and not just suspends his campaign. He can't get 1237 bound delegates though.

I'm unclear on how a candidate dropping out affects the delegates versus if they just suspend their campaigns. Doesn't that largely depend on the rules each state's GOP party has established for the delegates? Or are they all the same rule for when the candidate officially drops out?

 

[chipbennett]

No. He might not be able to get 1237 on the first ballot (which I believe he's more or less acknowledged for a while now) but that does not mean he's been mathematically eliminated...yet.

I heard this from some Cruz spokesperson Tuesday night, and it is one of the most absurd things I've ever heard.

The first ballot, and the first ballot only, is what "mathematically eliminated" applies to. It does not apply to any other ballot, because no candidate is "mathematically eliminated" from balloting rounds in which delegates are unbound (which happens, with increasing numbers of delegates, in the second round of balloting).

By this Cruz line of thinking, NO candidate is "mathematically eliminated".

 

[chipbennett]

I'm unclear on how a candidate dropping out affects the delegates versus if they just suspend their campaigns. Doesn't that largely depend on the rules each state's GOP party has established for the delegates? Or are they all the same rule for when the candidate officially drops out?

Suspend campaign: = delegates remain bound
Drop out = delegates unbound

 

[jamil]

I heard this from some Cruz spokesperson Tuesday night, and it is one of the most absurd things I've ever heard.

The first ballot, and the first ballot only, is what "mathematically eliminated" applies to. It does not apply to any other ballot, because no candidate is "mathematically eliminated" from balloting rounds in which delegates are unbound (which happens, with increasing numbers of delegates, in the second round of balloting).

By this Cruz line of thinking, NO candidate is "mathematically eliminated".

I don't see what's so incorrect about that. It is indeed absurd how it works, but it's not factually inaccurate. How else do you suppose it works? Has it occurred to you that Trump, more likely than not, will be mathematically eliminated? In other words it is likely that he won't win enough of the remaining delegates to reach a majority. That means he will need to win enough of the delegates unbound on the second vote. And it is possible for unbound delegates to vote for someone not in the race. So yeah. Mathematical elimination applies to the first vote where delegates are bound by state rules.

 

[ArcadiaGP]

I don't see what's so incorrect about that. It is indeed absurd how it works, but it's not factually inaccurate. How else do you suppose it works? Has it occurred to you that Trump, more likely than not, will be mathematically eliminated? In other words it is likely that he won't win enough of the remaining delegates to reach a majority. That means he will need to win enough of the delegates unbound on the second vote. And it is possible for unbound delegates to vote for someone not in the race. So yeah. Mathematical elimination applies to the first vote where delegates are bound by state rules.

How dare you. How DARE you sir.

"Mathematical elimination" only applies to the rubes, and the riff-raff, and other hoi polloi. Trump should get it if he's close enough.

1237 delegates only matters to the "non-middle finger to establishment" candidates.

 

[GodFearinGunTotin]

I heard this from some Cruz spokesperson Tuesday night, and it is one of the most absurd things I've ever heard.

The first ballot, and the first ballot only, is what "mathematically eliminated" applies to. It does not apply to any other ballot, because no candidate is "mathematically eliminated" from balloting rounds in which delegates are unbound (which happens, with increasing numbers of delegates, in the second round of balloting).

By this Cruz line of thinking, NO candidate is "mathematically eliminated".

Isn't the rule about winning a minimum of 8 states still in play? If so, we're down to 2 people that qualify.

 

[GodFearinGunTotin]

How dare you. How DARE you sir.

"Mathematical elimination" only applies to the rubes, and the riff-raff, and other hoi polloi. Trump should get it if he's close enough.

1237 delegates only matters to the "non-middle finger to establishment" candidates.

Well...if you rely on Fox News, Breitbart, and Alex Jones for your information.

 

[ArcadiaGP]

Well...if you rely on Fox News, Breitbart, and Alex Jones for your information.

Last night, Hannity asked the question... "Should the candidate with the most delegates win the primary?"

 

[jamil]

Isn't the rule about winning a minimum of 8 states still in play? If so, we're down to 2 people that qualify.

I thought that only applied to the first vote as well and that if no one won a majority of delegates on the first vote, then the unbound delegates can vote for whoever. I'd think that if the 8 states rule applied after the first vote Ryan's name would never have come up. And if it did, rather than the dramatic refusal, he'd simply say he's not eligible.

 

[T.Lex]

Last night, Hannity asked the question... "Should the candidate with the most delegates win the primary?"

Depends on the candidate. It isn't a math lesson, its politics.

I thought that only applied to the first vote as well and that if no one won a majority of delegates on the first vote, then the unbound delegates can vote for whoever. I'd think that if the 8 states rule applied after the first vote Ryan's name would never have come up. And if it did, rather than the dramatic refusal, he'd simply say he's not eligible.

As I read it, to even be nominated at the convention to become the party nominee, a candidate has to win 8 states. It is like a pre-condition to even get on the convention ballot. That's why there's such interest in whether to change the rules to allow for more. If it only applied on the first ballot, and no candidate got 1237, there'd be no reason to change that rule.

 

[jamil]

Last night, Hannity asked the question... "Should the candidate with the most delegates win the primary?"

Of course he's priming the Trumpers. But what that would really mean is that he wants a rule change not to requre a majority of delegates. That's not what will happen. And that's not what should happen. I have little doubt that Trump will win on the second vote anyway. But the rules should not strengthen the party's ability to nominate the least liked candidate. A truly representative voting system would not end with the two least liked candidates.

 

[GodFearinGunTotin]

I thought that only applied to the first vote as well and that if no one won a majority of delegates on the first vote, then the unbound delegates can vote for whoever. I'd think that if the 8 states rule applied after the first vote Ryan's name would never have come up. And if it did, rather than the dramatic refusal, he'd simply say he's not eligible.

I think you might be right.

 

[jamil]

.

 

[T.Lex]

I think you might be right.

https://s3.amazonaws.com/prod-static-ngop-pbl/docs/Rules_of_the_Republican+Party_FINAL_S14090314.pdf

Rule #40.
(a) [paraphrasing] Each state can nominate someone for POTUS, unless only one person meets the criteria of (b).

(b) Each candidate for nomination for President of the United States and Vice President of the United States shall demonstrate the support of a majority of the delegates from each of eight (8) or more states, severally, prior to the presentation of the name of that candidate for nomination... to demonstrate the support required of this paragraph a certificate evidencing the affirmative written support of the required number of permanently seated delegates from each of the eight (8) or more states shall have been submitted to the secretary of the convention not later than one (1) hour prior to the placing of the names of candidates for nomination pursuant to this rule and the established order of business.

So, prior to even being nominated by a state, a potential nominee must present evidence of support from 8 states to "put his hat in the ring."

Now, it does look like some additional wiggle room in (a). It doesn't explicitly say that if there are 2 candidates with at least 8 states, what options any given state has.

 

[GodFearinGunTotin]

https://s3.amazonaws.com/prod-static-ngop-pbl/docs/Rules_of_the_Republican+Party_FINAL_S14090314.pdf

Rule #40.
(a) [paraphrasing] Each state can nominate someone for POTUS, unless only one person meets the criteria of (b).



So, prior to even being nominated by a state, a potential nominee must present evidence of support from 8 states to "put his hat in the ring."

Now, it does look like some additional wiggle room in (a). It doesn't explicitly say that if there are 2 candidates with at least 8 states, what options any given state has.

Here's the text. I think your summary text above is conflating the language for if only one candidate for VP is under consideration.

RULE NO. 40
Nominations
(a) In making the nominations for President of the United States and Vice President of the United States and voting
thereon, the roll of the states shall be called separately in each case; provided, however, that if there is only one candidate for
nomination for Vice President of the United States who has demonstrated the support required by paragraph (b) of this rule, a
motion to nominate for such office by acclamation shall be in order and no calling of the roll with respect to such office shall be
required.
(b) Each candidate for nomination for President of the United States and Vice President of the United States shall
demonstrate the support of a majority of the delegates from each of eight (8) or more states, severally, prior to the presentation of
the name of that candidate for nomination. Notwithstanding any other provisions of these rules or any rule of the House of
Representatives, to demonstrate the support required of this paragraph a certificate evidencing the affirmative written support of
the required number of permanently seated delegates from each of the eight (8) or more states shall have been submitted to the
secretary of the convention not later than one (1) hour prior to the placing of the names of candidates for nomination pursuant to
this rule and the established order of business.

(If I'm reading it right, that is. )

My reading is, if you don't have the majority of delegates from 8 states, prior to 1 hour prior time to place the names for nomination, you're out of luck.

 

[chipbennett]

I don't see what's so incorrect about that. It is indeed absurd how it works, but it's not factually inaccurate. How else do you suppose it works? Has it occurred to you that Trump, more likely than not, will be mathematically eliminated? In other words it is likely that he won't win enough of the remaining delegates to reach a majority. That means he will need to win enough of the delegates unbound on the second vote. And it is possible for unbound delegates to vote for someone not in the race. So yeah. Mathematical elimination applies to the first vote where delegates are bound by state rules.

No, it's not. Mathematical elimination only applies to the first ballot.

If Trump also becomes mathematically eliminated, then the result is a contested convention. The root cause of a contested convention is every candidate being "mathematically eliminated" from receiving a majority of delegates, such as to ensure a first-ballot election.

Have I ever held that Trump should be "given" the nomination if he is "close enough", but not at/over 1237?

If Trump doesn't win it on the first ballot, then the convention will be contested, and subsequent ballots will be open season. Whatever happens, will happen. But if Trump enters the convention with hundreds more delegates than anyone else, and with millions more votes than anyone else, it will be a very hard sell for the party to nominate someone else in those subsequent ballots, and the party will have millions of people to answer to, should that happen.

It will be unprecedented, and it will destroy the party.

(But I'll still vote for whomever's name is next to the "R" in the general election.)

How dare you. How DARE you sir.

"Mathematical elimination" only applies to the rubes, and the riff-raff, and other hoi polloi. Trump should get it if he's close enough.

1237 delegates only matters to the "non-middle finger to establishment" candidates.

Straw men are fun to demolish.

Isn't the rule about winning a minimum of 8 states still in play? If so, we're down to 2 people that qualify.

That rule, also, only applies to the first ballot.