President-elect Joe Biden will finish the 2020 vote count with a larger share of Alaska’s presidential vote than any Democrat since Lyndon Johnson won the state in 1964.
Despite that, Biden will still lose the state by almost 10% to President Donald Trump, who earned a larger share of the state’s votes this year than he did in 2016.
By Tuesday night, 99.3% of Alaska’s estimated 360,000 ballots were counted, and turnout had already set a record for most ballots cast.
In the presidential race, Trump has received about 53% of Alaska’s votes, compared to under 43% for Biden.
Four years ago, Trump earned 51.3% of the state’s vote and Democratic candidate Hillary Clinton had 36.6%.
Many of the remaining uncounted votes are from Democratic-leaning rural Alaska, but even before they are added to the tally, Biden has already had a notable performance. In 1992, Bill Clinton came within 9 points of Alaska winner George H.W. Bush, but that was with third-party candidate Ross Perot earning 28% of the vote. Bill Clinton had only 30% of the state’s votes.
As of Tuesday night, Biden had 42.79% of Alaska’s votes, higher than any Democratic presidential candidate since Lyndon Johnson won 66% of the state’s presidential votes in 1964. Until this year, the No. 2 candidate had been Hubert Humphrey in 1968.
Preliminary results indicate Biden won a majority of the votes cast in Anchorage, a switch from 2016.
Biden performed that feat in part by earning more votes than many local Democratic candidates. Through Tuesday in House District 27, for example, Biden had about 200 votes more than Democratic state House candidate Liz Snyder. In nearby House District 28, Biden had 6,225 votes, about 530 more than the Democratic-endorsed independent, Suzanne LaFrance, who lost her race.
With Anchorage closely divided, Trump’s victory came courtesy of massive support in the Kenai Peninsula and the Matanuska-Susitna Borough. In House District 10, which includes parts of Wasilla, Trump had more than 8,000 votes; Biden had fewer than 3,000.