While US President Barack Obama and Republican US presidential candidate Mitt Romney campaign ahead of Tuesday’s US elections, the NFL’s Washington Redskins could foretell the vote outcome today.
The so-called “Redskins Rule” has been a bellwether for presidential elections since the club moved to the US capital from Boston in 1937.
That fact will draw some extra attention to today’s Redskins home game against the Carolina Panthers. While the Redskins have struggled to a 3-5 record so far, the Panthers are a woeful 1-6.
In 17 of 18 US presidential elections, starting in 1940, the candidate from the party holding the White House won the presidency if the Redskins won their final home game before the election.
If the Redskins lost that contest, the candidate from the party not occupying the White House would claim an election victory.
So Obama backers will hope to see a Redskins triumph, while Romney supporters will be bolstered if Carolina upsets Washington.
The only time the method missed was in 2004, when the Redskins lost 28-14 to Green Bay and Republican former US president George W. Bush beat Democratic US presidential candidate John Kerry, although that anomaly could be explained by Bush’s 2000 election.
Democratic US presidential candidate Al Gore actually won the 2000 popular vote, but Bush won the presidency by collecting the most delegates in the electoral college, in which states are assigned delegates based upon population and the winner of the most delegates takes the election.
Bush won in 2000 only after the US Supreme Court, in a 5-4 decision, halted a controversial recount of votes in Florida, upholding the original findings that handed Bush a narrow victory in the state and national election.
The Redskins Rule returned to order in 2008, when Washington lost 23-6 to Pittsburgh on election eve and the next day, with Bush in the White House, Obama defeated Republican US presidential candidate John McCain to claim the presidency.