View Full Version : Brokered Convention Question

01-30-2008, 09:46 AM
How many delegates would a candidate need for there not to be a brokered convention? Anybody know?

01-30-2008, 09:50 AM
If one candidate gets 1191 or more then they win...otherwise it's brokered

Bradley in DC
01-30-2008, 09:57 AM

01-30-2008, 10:04 AM
I see McCain, Romney, Huck as the three front runners now.

Huck will likely drop out within the week, based on the debate tonight.

There are 2245 delegates remaining.

If Ron Paul, Mitt Romney, and John McCain split them evenly... that ends up with the following:

Mitt Romney: 748+29 = 776
John McCain: 748+89 = 837
Ron Paul: 748+8 = 756

And that is if we split ALL equally.

If we get 20% of the delegates, and the other two split the remaining 80%..


Mitt Romney: 898 + 29 = 927
John McCain: 898 + 89 = 987
Ron Paul: 448+8 = 456

If we get 10% , and the remaining get 45% each..

Mitt Romney: 1010 + 29 = 1039
John McCain: 1010 + 89 = 1099
Ron Paul: 224 + 8 = 232

The point is, the best we can hope for out of a brokered convention is to be enough to push one of the two candidates over the top. Then we can swing being a VP out of this. That's IF they don't agree to do it themselves.

We need to WIN this. We need to not worry about or focus on these "games" to win a brokered convention, because we needed four candidates splitting votes to do that. Now that Ghoul is out, our chances are even lower. We needed Ghoul to stay in through Feb 5 and steal some delegates there.

01-30-2008, 10:05 AM
If one candidate gets 1191 or more then they win...otherwise it's brokered

Cool! Thanks :cool: