Thursday, August 07, 2008

San Francisco: Less Drunk Than Milwaukee!

Forbes magazine carved out a lucrative niche for itself with its world's-richest-people lists, so they decided to start making lists about everything. It's getting a little ridiculous: the 10 most luxurious port-a-potties, the 100 smelliest CEOs, the five best places to find a cheap Ukrainian prostitute...

Okay, I made those up, but this one is real: the 15 drunkest cities. They tried to find the U.S. cities that consume the most alcohol, and lo, San Francisco is right up there at No. 3.

Here's the ranking:
1. Austin, Texas
2. Milwaukee
3. San Francisco
4. Providence, Rhode Island
5. Chicago
8. (tie) Cleveland
8. (tie) Seattle
8. (tie) St. Louis
9. Boston
10. Cincinnati
11. Pittsburgh
12. Virginia Beach, Virginia
13. Portland, Oregon
14. Jacksonville, Florida
15. Detroit

Apparently 59.4 percent of San Franciscans reported having at least one drink in the past month.

Wait, that means 40.6 percent of people haven't had a single drink in a month. That actually seems really low. I mean, aside from Kelly and the guys chain-smoking in front of the detox clinic around the corner from us, where are all these teetotalers?

Maybe the next list will be "the most stoned cities." If so — sorry Austin — but I think we'll have that one in the bag. Especially if we can include Berkeley in the metro area (and maybe Humboldt County).