Fantasy Take: Bobrovsky Taking His Talents To Florida

Ian Gooding

2019-07-01


The Florida Panthers have signed goalie Sergei Bobrovsky to a seven-year contact worth $70 million total.

The Panthers get: A two-time Vezina Trophy winner who will immediately upgrade the Panthers’ goaltending. Although his 2018-19 ratios (2.58 GAA, .913 SV%) weren’t among the elite, Bob is still widely considered a top-10 (maybe even top-5) fantasy goalie and the prize goalie of this season’s free agent crop. He is also well-known as a better second-half goalie, and his 2018-19 pre/post-All Star Game splits will attest to that (2.91 GAA, .904 SV% pre-ASG/2.20 GAA, .924 SV% post-ASG).

The contract was expected to be huge, and it didn’t disappoint. Only Carey Price ($10.5 million) earns more than Bobrovsky ($10 million) on his new deal. For this contract to pay off, Bobrovsky will need to be better in the playoffs (2.80 GAA and .908 SV% over the last three seasons). The Panthers will also need to make it to the playoffs first, which will be no easy task in an Atlantic Division that already regularly sends Tampa Bay, Boston, and Toronto to the playoffs. Like many of these contracts, there’s a high likelihood it will not age well (Bobrovsky will turn 31 just before the season starts).

Adding Bobrovsky will upgrade the Panthers’ goaltending from the injured and now retired Roberto Luongo and the inconsistent James Reimer. But how will the Panthers be for Bob’s numbers? The Blue Jackets and Panthers gave up a similar number of shots (29-30) and shot attempts (about 3700) last season, so this move shouldn’t have much of an effect on Bob’s numbers overall. In fact, the move could be even better for his numbers with defensively responsible Joel Quenneville now behind the bench. Given the investment, you should pencil Bob in for at least 60 starts (barring injury), a number he has reached in each of the previous three seasons. So you can probably draft him where you would have had he stayed in Columbus, at the very least.  

The defensemen should also benefit from Bobrovsky.