Tuesday , Dec , 06 , 2005 C.Y. Ellis

Kobe, Lakers stop off in Milwaukee

(Sports Network) – The Milwaukee Bucks go for their fourth straight win
tonight, as they welcome the Los Angeles Lakers to the Bradley Center.

This is the first of two meetings between the clubs this season. The Lakers
are scheduled to host the Bucks on March 24th at the Staples Center.

The Lakers have won eight straight and 17 of the last 19 meetings in this
series. The last time the Bucks beat the Lakers was on March 21, 2001 in front
of their home crowd.

LA has won four in a row and eight of its last nine at Milwaukee.

The Bucks closeout a brief two-game homestand. On Saturday, Michael Redd
netted a game-high 30 points and Milwaukee pulled away in the fourth quarter
to earn a 104-84 victory over the Orlando Magic at the Bradley Center.

Dan Gadzuric scored 14 of his 16 points in the final period, while fellow
reserve Toni Kukoc dropped in 14 points on 6-of-6 shooting in the win over
Orlando for the Bucks, who made 56.6 percent of their field-goal attempts en
route to their third consecutive triumph. Maurice Williams added 12 points and
11 assists in the victory.

Milwaukee is 5-2 at the Bradley Center this season. The Bucks, who have won
two in a row at home, are scheduled to visit Allen Iverson and the
Philadelphia 76ers on Wednesday at the Wachovia Center.

The Lakers begin a six-game road trip. On Sunday, Kobe Bryant hit two free
throws with seven seconds remaining and finished with 29 points overall to
lift Los Angeles over the Charlotte Bobcats, 99-98, at the Staples Center.

Center Chris Mihm scored a season-high 21 points and collected nine rebounds
in the victory over Charlotte for the Lakers, who have won two of their last
three games overall. Sasha Vujacic added 12 points off the bench in the win.

LA is 3-4 on the road this season. The Lakers will also visit Toronto,
Chicago, Minnesota, Dallas and Memphis on the current swing.