Eddie Robinson, Last Living Member of 1948 Cleveland Indians World Series Team, Will Be at Game 6 Tonight
The 95-year-old is hoping to see the team's first World Series win in 68 years.
— -- Eddie Robinson, the last living member of the Cleveland Indians' 1948 World Series championship team, will be at Progressive Field tonight as today's Indians attempt to win their first World Series title in 68 years.
Robinson, 95, was the starting first baseman on the ’48 team. He drove in the decisive run in Cleveland's victory over the Boston Braves in Game 6 of the '48 Fall Classic, according to MLB Network.
"That was such an exciting time," Robinson told MLB Network of the '48 season. "All we wanted to do was win the American League pennant. Once we did that, we were a little bit free and easier in the World Series."
The retired baseball player and his wife flew from their home in Fort Worth, Texas, on Monday to Cleveland, MLB Network reported. They will be hosted in a suite during Game 6 of the World Series tonight.
The Indians currently lead the Chicago Cubs, 3-2, in the best-of-seven set.
Related Topics
Related Stories
ABC News Live