1998
Born October 7, 1998
Harvey, IL, USA
Bob Seymour Notable
b. October 7, 1998 · Tampa Bay Rays 2025
Bob Seymour has played his whole career for the Tampa Bay Rays as a first baseman, since 2025.
Across 26 games he batted .205, with 16 hits and 9 runs. He finished with 1 home run, 1 double and 1 stolen base.
He played 25 games at first base, with 156 putouts and 1 error. He turned 19 double plays.
He was born in Harvey, IL, USA on October 7, 1998. He batted left-handed and threw right-handed. He has not retired; our record has no final game for him.
Batting, season by season
| Year | Team | G | AB | R | H | 2B | 3B | HR | RBI | BB | SO | SB | AVG |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 2025 | TBR | 26 | 78 | 9 | 16 | 1 | 1 | 1 | 5 | 4 | 32 | 1 | .205 |
| Career | — | 26 | 78 | 9 | 16 | 1 | 1 | 1 | 5 | 4 | 32 | 1 | .205 |