After reportedly signing a two-year, $8 million deal with the Milwaukee Bucks, let’s hand out our grade on the acquisition of Bobby Portis.
The Milwaukee Bucks kicked off an eventful Saturday by bringing in a big man off the free agent market.
First reported by The Athletic’s Shams Charania, the Bucks are signing Bobby Portis on a two year deal worth around $8 million, which fits right into the Bucks’ bi-annual exception. The deal also features a player option in the second year of his deal.
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Now that he’s Milwaukee-bound, this will serve as Portis’ fourth stop in the NBA since being drafted with the 23rd overall pick in the 2015 NBA Draft after having previously been with the Chicago Bulls, Washington Wizards and New York Knicks.
Portis brings plenty to the table as a basketball player, and this signing has the potential to come up big down the stretch. With that out of the way, let’s take a look at Bobby Portis and what this signing means for the Milwaukee Bucks.
The value Bobby Portis’ deal with the Milwaukee Bucks
At $8 million over two years, this is a solid deal from the Bucks’ side of things. Portis originally was under contract with the Knicks earlier in the week, but they smartly declined his $15 million team option to offload some of the talent in their overloaded frontcourt.
That does not mean Portis was not of value to the Knicks, as he put together a decent 2019-20 season with the franchise, even under a rebuilding situation.
In 66 games, five of them coming in the starting lineup, the big man averaged 10.1 points on .450/.358/.763 shooting splits, 5.1 rebounds, and 1.6 assists across 21.1 minutes per game. If Portis brings that same type of production to the Bucks on a far cheaper deal, that is a solid acquisition to fortify their frontcourt behind Giannis Antetokounmpo and Brook Lopez.
The fit is made especially compelling, given the team’s overwhelming need for talent in the frontcourt, and if everything breaks right, Portis could be up for the job.