Steve O’Neill Net Worth 2018: What is this baseball player worth?
Steve O’Neill was a pro baseball player who played Catcher / Manager . O’Neill was born on July 6, 1891, in Scranton, Pennsylvania. O’Neill died on January 26, 1962, in Cleveland, Ohio. This page will take a closer look at Steve O’Neill’s net worth.
Steve O’Neill Career, Earnings
O’Neill batted Right and threw Right. O’Neill debuted in the MLB on September 18, 1911 for the Cleveland Naps. In all, O’Neill played for the Cleveland Indians, Boston Red Sox, New York Yankees, St. Louis Browns, Cleveland Indians, Detroit Tigers, Boston Red Sox, and Philadelphia Phillies. O’Neill’s career ended with the St. Louis Browns in 1928.
Some of O’Neill’s most prominent statistics in the MLB included a Batting average stat of .263, a Home runs stat of 13, and a Run batted in stat of 537. Some of the highlights of Steve O’Neill’s career included: 2× World Series champion, and Cleveland Indians Hall of Fame.
Steve O’Neill Net Worth 2018
Baseball player earnings can range widely. In professional baseball, the median pay is near $3 million per year. Top pro baseball players can get $25 million or more per year, and lower paid players make $1 million or less. In the minor leagues, most contracts are worth less than $10,000 a year.
Steve O’Neill net worth: baseball salary distribution
So what was baseball player Steve O’Neill’s net worth at the time of death? Our estimate for Steve O’Neill’s net worth at death is:
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