Hancock, Maryland AA Meetings

  • Please note that AA meeting locations and times tend to change often and quickly, so always check before assuming these times are accurate.
  • Do you have new information about an AA meeting? Please let us know!

Hancock is a town in Washington County that is located at the narrowest part of the state of Maryland. Despite a population of just a couple thousand, it has a history of alcohol addiction. The Western Maryland community has Alcoholics Anonymous programs in place to offer help to its residents who are battling alcohol addiction. The disease has become a major public health issue in America. One of the primary reasons for the epidemic is the wide accessibility of alcohol. Moreover, not only is alcohol easily accessible throughout America and most other countries in the world, it is also legal. Alcohol consumption has become so socially accepted that it is considered unusual if an individual refuses to drink in a social setting. Alcoholics Anonymous (AA) programs can help people to learn how to manage these situations and permanently stay away from alcohol. Through its one-hour-long counseling sessions called AA meetings, AA provides opportunities for people with an alcohol dependency to receive guidance from senior members, or ”sponsors,” who have successfully overcome alcohol addiction after going through the AA program. If you live in Hancock with an alcohol dependence, you can get a “sponsor” by attending AA meetings in Hancock. The locations to these meetings along with AA meetings Maryland from other areas in the state are listed on alcoholicsanonymous.com.

Where do calls go?

Calls to numbers on a specific treatment center listing will be routed to that treatment center. Additional calls will also be forwarded and returned by one of our treatment partners below.

Calls to any general helpline (non-facility specific 1-8XX numbers) for your visit will be answered by ARK Behavioral Health, a paid advertiser on AlcoholicsAnonymous.com.

All calls are private and confidential.

Who Answers?