Report: Redd staying with Milwaukee

The Milwaukee Journal Sentinel reported Thursday that the former All-Star has agreed to a six-year deal worth between $90 million and $96 million. According to the report, the Cleveland Cavaliers had also offered a lucrative contract to Redd, believed to be in the neighborhood of $70 million for five years.

Cleveland is Redd’s home-state team, but even the opportunity to play with All-Star LeBron James was not enough to lure the Ohio State product away from the Bucks.

"It was a difficult decision for Mike," Redd’s agent Kevin Poston told the Journal Sentinel, "but in his heart he felt the Milwaukee Bucks stepped up. Cleveland was a great situation, intriguing. But Milwaukee is Michael’s town."

Redd, a 2004 All-Star, played in 75 games last season and averaged 23.0 points, 4.2 rebounds and 2.3 assists. He shot 44.1 percent from the field and converted 35.5 percent of his attempts from beyond the arc.

Originally selected by Milwaukee with the 43rd overall pick of the 2000 NBA Draft, Redd has registered 17.7 points, 4.2 boards and 1.8 assists in 312 regular-season contests. He has participated in 11 playoff games, averaging 13.5 points in postseason action.

On June 28, Milwaukee chose Utah center Andrew Bogut with the first overall pick in the draft. The addition of Bogut had to help Redd in his decision to stay with the only team that he has played for during his career.

Redd, who will turn 26 on August 24, will try and form one of the top inside- outside duos in the NBA with his new teammate.

The contract will not be official until July 22nd, the first day free agents can officially sign.