They don’t make any sense to me.
Nate Cohn, the Times poll maven, writes:
Despite low national approval ratings and the specter of impeachment, President Trump remains highly competitive in the battleground states likeliest to decide his re-election, according to a set of new surveys from The New York Times Upshot and Siena College.
…
The Times/Siena results and other data suggest that the president’s advantage in the Electoral College relative to the nation as a whole remains intact or has even grown since 2016, raising the possibility that the Republicans could — for the third time in the past six elections — win the presidency while losing the popular vote.
But but but but…
I looked at the Iowa poll information as analyzed by 538.com.
The poll requires respondents to rank-choice rate their Democratic candidates. The results — 14 pages in that above link for Iowa — are really interesting.
But Trump is the sole Republican in question. So presumably a large percentage of Republican voters who are polled are selecting Trump, while the 439 Democratic voters polled spread their votes over 19 candidates plus an option for “other.”
So Trump represents 100 percent of his voters while the Democratic candidates represent a percentage of 20 possibilities. Can the polling — rank-choice voting notwithstanding — accurately pinpoint what happens to those ranked votes after the primary selects that one candidate? I don’t know how polls can evaluate what happens after a primary, when voters turn their support to the nominee, usually enthusiastically.
How does this suggest that Trump can win again in the Electoral College?
When the Democrats choose their candidate, then rational state polls comparing Trump to the Democratic candidate can have some meaning. Until then, I’m not getting depressed — which a number of Times’ commentators said they were.
Cheer up, folks. None of this stuff means anything until after the primary. Except that so many people seem to be going crazy.
Note: If anyone can contradict me by telling me how polls can indeed measure one against many with accuracy, please let me know.