Here at Milwaukee Record, we employ a sentient machine known as the Great Job, Milwaukee! Bot to keep track of the city’s many appearances on dopey online lists, as well as any time a national publication deigns to acknowledges our Midwest existence. Back in April, for instance, we were told we were ugly as dirt. Today, the Bot brings us the annual (monthly?) report that yes, Milwaukee is one of the drunkest cities in America. Again. beep-boop-beeeeeep…
In a new “Drunkest (and Driest) Cities in America” list from 24/7 Wall Street, Milwaukee stumbles in at #20 out of 20 (phew!), right behind Ames, Iowa. “To identify the U.S. cities with the highest and lowest excessive drinking rates,” the site explains, “24/7 Wall St. reviewed the percentage of adults 18 and older who report binge or heavy drinking within a 30 day period across 381 metro areas.” Here’s what it has to say about Milwaukee:
> Pct. adults drinking to excess: 22.5%
> Pct. driving deaths involving alcohol: 36.4%
> Est. number of bars: 494
> Median household income: $56,247
There are 3.1 bars in the Milwaukee metro area for every 10,000 residents, well more than double the nationwide concentration of 1.3 bars per 10,000 people. More venues to consume alcohol can lead to more widespread consumption. Some 22.5% of metro area adults drink excessively, well above the 18.0% share of American adults.
(Not) surprisingly, nine other Wisconsin cities pop up on the shitfaced list: Sheboygan (#15), Fond du Lac (#12), La Crosse (#10), Wausau (#9), Oshkosh (#6), Madison (#4), Appleton (#3), Eau Claire (#2), and Green Bay (#1). So yeah, Wisconsin makes up 50% of the list.
Once again, great job (?), Milwaukee, and great job (?), Wisconsin. beep-boop-boooooop…