The Tampa Bay Buccaneers are sitting at a cool 4-1 in their first five games. It’s a team effort on all sides to make that happen, but one of the biggest reasons that the Bucs are on top and tied for the NFL’s best record is because of their rookie wide receiver.
The Buccaneers drafted wideout Emeka Egbuka in the first round of the 2025 NFL Draft with the No. 19 pick, and many were confused at the time — the Bucs had needs on defense, after all, and it seemed like a luxury pick for a team that already had a loaded wide receiver room. As it turns out, Egbuka is exceeding expectations, and he’s the favorite to win an award that only one Buccaneers player has ever won.
Tampa Bay Buccaneers wide receiver Emeka Egbuka (2) receives a pass against the New York Jets. / Nathan Ray Seebeck-Imagn Images
It should come as no surprise to anyone who has seen him, but Egbuka is currently odds-on for the NFL’s Offensive Rookie of the Year award, per Fanduel Sportsbook. He’s sitting at +115 for the award, beating out Las Vegas Raiders running back Ashton Jeanty (+380) and New York Giants quarterback Jaxson Dart (+650).
The hype is warranted. Egbuka has put up 445 yards on 25 passes, and he has five touchdowns in five games — that’s tied for second in the NFL. He’s not just one of the NFL’s best rookies, he’s one of the NFL’s best wide receivers, and he continues to show out week after week when he’s needed the most.
The only other Buccaneers player to win Offensive Rookie of the Year was running back Warrick Dunn, who did it in 1997 after rushing for 978 yards and four touchdowns to go with 462 receiving yards and three receiving touchdowns. Egbuka would be in great company there were he to win the award, and he’s been doing everything he needs to in search of that goal through the team’s first five games.
Egbuka will have a further chance to show what he can do when the Buccaneers return home to face off against the San Francisco 49ers in Week 6.
