As the Republican presidential campaign moves into January, 2012, with the first round of primaries and the Iowa caucuses, the Romney campaign is emphasizing its contention that Mitt is the most electable Republican. They are basing this assertion on polls which show him running better against President Barack Obama nationwide than his chief rival for the GOP nomination, Newt Gingrich.
Polls are a very thin reed, however, upon which to base an argument for any prospective candidate’s electability. In presidential election years, the electorate is often fickle, and that appears to be the case in this campaign.
As an example, the Rasmussen Polls taken in November indicated that Newt was more electable than Mitt. These polls reported that Newt led Obama by two points nationwide, 45-43, while Obama led Mitt by a score of 44-38. Polls taken this month indicate that Mitt is substantially more electable than Newt.
The best way to measure electability is to do an Electoral College assessment, which I have done. In making this assessment, one must first determine the Electoral College base of each candidate – i.e., those states which are highly likely to be won by each candidate, respectively.
Obama and the generic GOP candidate each have a base of 22 states, with the incumbent president also carrying the District of Columbia. Yet Obama’s base gives him a total of 268 electoral votes – only two votes shy of the 270 total which wins the election. By contrast, the generic GOP candidate’s base only gives him or her a total of 180 electoral votes.
The reason for this discrepancy is that Obama’s 22 state Electoral College base includes states with large numbers of electoral votes, such as California (55), Illinois (20), Pennsylvania (20), and New York (29). The generic GOP candidate’s base only includes one state with a large electoral vote total, to wit, Texas (38).
There are six battleground states: New Hampshire (4), Ohio (18), Indiana (11), Virginia (13), Florida (29), and North Carolina (15), for a total of 90 electoral votes. In order for the GOP candidate to win, he or she must carry all six of these battleground states. All Obama needs to do to win reelection is to win one of these states.
It is quite possible and, in fact, likely for the GOP nominee to win five of these states: New Hampshire, Ohio, Indiana, Virginia, and Florida. The problem for the Republican candidate is North Carolina. Due to a number of demographic and economic factors, it appears increasingly likely that Obama will carry the Tar Heel state. This would give him 283 electoral votes and thus a second term as president.
Thus, for a GOP candidate to defeat Obama, he or she will have to win one or more of the states that would normally be a base state for Obama. The only realistic possibility of this would be Michigan (16), and the only Republican candidate capable of defeating Obama in Michigan is Mitt Romney. A victory in Michigan, together with all the 22 GOP base states and Florida, Ohio, Indiana, New Hampshire, and Virginia would give Mitt Romney a total of 271 electoral votes and thus, the election as president.
I cannot foresee any circumstances under which Newt Gingrich could win any of the 22 Obama base states, including Michigan. Nor do I think he could deny Obama a victory in North Carolina.
It is ironic that it will be virtually impossible for Mitt Romney to win the electoral votes of Massachusetts, a state where he was a highly successful governor. Yet there have been various polls in recent months which show Romney leading Obama in Michigan, Mitt’s native state, in spite of all the assistance the President gave General Motors.
The Romney name is magic in Michigan, due to the enduring popularity of Mitt’s father, the late George Romney, a highly popular governor of the Wolverine State from 1963 to 1969. Prior to that, he was the Chief Executive Officer of American Motors Corporation, the producer of the Rambler automobile.
Michigan voters have warm historical memories of George Romney. If Mitt Romney is elected president, at his inaugural on January 20, 2013, he may owe his greatest debt of gratitude for his election victory to his late father.
Alan J. Steinberg served as Regional Administrator of Region 2 EPA during the administration of former President George W. Bush. Region 2 EPA consists of the states of New York and New Jersey, the Commonwealth of Puerto Rico, the U.S. Virgin Islands, and eight federally recognized Indian nations. Under former New Jersey Governor Christie Whitman, he served as Executive Director of the New Jersey Meadowlands Commission. He currently serves on the political science faculty of Monmouth University.
