| Election Name: | 1849 Maine gubernatorial election |
| Country: | Maine |
| Type: | Presidential |
| Ongoing: | no |
| Previous Election: | 1848 Maine gubernatorial election |
| Previous Year: | 1848 |
| Next Election: | 1850 Maine gubernatorial election |
| Next Year: | 1850 |
| Election Date: | 10 September 1849 |
| Nominee1: | John Hubbard |
| Party1: | Democratic Party (United States) |
| Popular Vote1: | 37,636 |
| Percentage1: | 51.01% |
| Nominee2: | Elijah Hamlin |
| Party2: | Whig Party (United States) |
| Popular Vote2: | 28,056 |
| Percentage2: | 38.03% |
| Nominee3: | George F. Talbot |
| Party3: | Free Soil Party (United States) |
| Popular Vote3: | 7,987 |
| Percentage3: | 10.83% |
| Governor | |
| Before Election: | John W. Dana |
| Before Party: | Democratic Party (United States) |
| After Election: | John Hubbard |
| After Party: | Democratic Party (United States) |
The 1849 Maine gubernatorial election was held on 10 September 1849 in order to elect the Governor of Maine. Democratic nominee and former member of the Maine Senate John Hubbard defeated Whig nominee and former member of the Maine House of Representatives Elijah Hamlin and Free Soil Party nominee George F. Talbot.[1]
On election day, 10 September 1849, Democratic nominee John Hubbard won the election by a margin of 9,580 votes against his foremost opponent Whig nominee Elijah Hamlin, thereby retaining Democratic control over the office of governor. Hubbard was sworn in as the 22nd governor of Maine on 8 May 1850.[2]