Bucks’ Giannis Antetokounmpo wins NBA Defensive Player of the Year
It is award season in the NBA. Elite players across the league being rewarded for their efforts during the 2019-20 NBA regular season. One player that is in the running for multiple awards, Giannis Antetokounmpo won one of potentially two awards Tuesday night.
According to ESPN, Giannis was awarded the NBA Defensive Player award after having another incredible year for the Bucks during what could be a potential championship run for the team and superstar.
Milwaukee Bucks star Giannis Antetokounmpo was named the 2019-20 NBA Defensive Player of the Year on Tuesday.
Antetokounmpo won the award in a landslide, receiving 75 first-place votes from a panel of 100 sportswriters and broadcasters and earning 432 points total.
Antetokounmpo was thought to have some stiff competition in the form of Anthony Davis of the Los Angeles Lakers. However, it wasn’t as close as many anticipated. Tge Greek superstar, or ‘Greek Freak’ ran away with the award with Davis only securing 14 first-place votes.
Bucks head coach Mike Budenhoizer was able to hand the trophy to his superstar as Milwaukee awaits their second-round playoff matchup, saying the following via NBC Sports.
“Congrats to Giannis on being named Defensive Player of the Year,” Budenholzer said in announcing the award on TNT. “His commitment to defending, his commitment to winning is beyond incredible. He impacts the game with his blocked shots, his rebounding, his ability to guard all five positions, his chase-down blocks, his challenge to do everything on defense.”
Giannis is in select company now becoming one of only five players to win NBA MVP and NBA Defensive Player of the Year during their career. Antetokounmpo joins Kevin Garnett, Hakeem Olajuwon, Michael Jordan, and David Robinson.