On this day, in his major league debut, George Herman “Babe” Ruth pitched seven strong innings to lead the Boston Red Sox over the Cleveland Indians, 4-3. Born February 6, 1895 in Baltimore, Maryland, the young George, known as “Gig” (pronounced jij) to his family, was a magnet for trouble from an early age.
At seven, his truancy from school led his parents to declare him incorrigible, and he was sent to an orphanage, St. Mary’s Industrial School for Boys. Ruth lived there until he was 19, in 1914. That is when he was signed as a pitcher by the Baltimore Orioles. That same summer, Ruth was sold to the Boston Red Sox.
His teammates called him “Babe” for his naivete, but his talent was already maturing. In his debut game against the Indians, the 19-year-old Ruth gave up just five hits over the first six innings. In the seventh inning, the Indians managed two runs on three singles and a sacrifice fly – and Ruth was relieved. His hitting prowess, however, was not on display that first night. He was 0-for-2 at the plate.
A lot of us made outs early in life. Perhaps you have struck out several times. So do what the “Babe” did – just keep swinging. Babe Ruth became the most iconic athlete of his century because on this day, 102 years ago, he refused to give up.