Thursday, February 07, 2008
Romney's Departure may Cause Key 2 to Stand
I said yesterday that it is especially important that Romney stay in the race. Now he's done it. He quit. That may cause Lichtman Key 2 to stand. The key reads: "There is no serious contest for the incumbent-party nomination.", and Lichtman defines "serious" as that the nominee gets 2/3 of the delegates at the convention. As of today the delegate count is
McCain 714
Romney 286
Huckabee 181
Paul 16
McCain has more than Romney and Huckabee combined but not more than 2/3 that. He has 60% of the delegates. If things were to continue like this, McCain would win with less than 2/3 of the delegates, and Key 2 would fall. However, with Romney quitting, his delegates will split somehow between the candidates. If they split equally, the result is:
McCain 857
Huckabee 412
Paul 16
Assuming Paul does not get any Romney delegates. This gives McCain 72% of the delegates, more than enough to clinch Key 2. At what percentage of the Romney vote going to Huckabee would be needed to topple the key? Go to Hilbert's Hotel, my mathematical blog, to find out. It's a nice algebra problem, and it shows that algebra can be useful in real life. The result is 72%, or 202 delegates. In other words, 202 of Romney's 286 delegates would have to go to Huckabee to topple Key 2. Any less, and Key 2 is likely to stand.
I am not sure where his support is going to go, and I certainly hope he doesn't support McCain! Support Huckabee, Romney. If 72% of Romney's delegates go to Romney, it would indicate that the UltraConservatives of the Republican Party object so much to McCain that they are willing to make a knockdown dragout ding-dong fight out of it, and that indicates Key 2 falling.
But if any sizable proportion of the Romney vote goes to McCain, Key 2 would stand, vindicating my claim earlier that eventually one candidate would run away with it. Sure enough it seems to have. Now let's hope that the other keys down mean an Obama victory in November.
McCain 714
Romney 286
Huckabee 181
Paul 16
McCain has more than Romney and Huckabee combined but not more than 2/3 that. He has 60% of the delegates. If things were to continue like this, McCain would win with less than 2/3 of the delegates, and Key 2 would fall. However, with Romney quitting, his delegates will split somehow between the candidates. If they split equally, the result is:
McCain 857
Huckabee 412
Paul 16
Assuming Paul does not get any Romney delegates. This gives McCain 72% of the delegates, more than enough to clinch Key 2. At what percentage of the Romney vote going to Huckabee would be needed to topple the key? Go to Hilbert's Hotel, my mathematical blog, to find out. It's a nice algebra problem, and it shows that algebra can be useful in real life. The result is 72%, or 202 delegates. In other words, 202 of Romney's 286 delegates would have to go to Huckabee to topple Key 2. Any less, and Key 2 is likely to stand.
I am not sure where his support is going to go, and I certainly hope he doesn't support McCain! Support Huckabee, Romney. If 72% of Romney's delegates go to Romney, it would indicate that the UltraConservatives of the Republican Party object so much to McCain that they are willing to make a knockdown dragout ding-dong fight out of it, and that indicates Key 2 falling.
But if any sizable proportion of the Romney vote goes to McCain, Key 2 would stand, vindicating my claim earlier that eventually one candidate would run away with it. Sure enough it seems to have. Now let's hope that the other keys down mean an Obama victory in November.
