Sergei Bobrovsky Named Vezina Trophy Finalist

Sunrise, FL – Florida Panthers goaltender Sergei Bobrovsky is a Vezina Trophy finalist for the third time in his career.

Bobrovsky was tied for the most shutouts in the National Hockey League during the regular season with six. He had the third-lowest goals allowed per game among goaltenders with at least 25 games with 2.37. His teammate, Anthony Stolarz allowed 2.03, although he played 27 games to Bobrovsky’s 58.

The Russian netminder became the Florida Panthers franchise second-most winning goalie on October 19th, in a win vs. Toronto. Also, he became the 2nd most games played Russian-born goalie when he made his 698th appearance on April 6th vs Boston. On top of that, he’s one of only four active goaltenders to have made an appearance in 700+ NHL games on April 13th vs. Buffalo.

Sergei Bobrovsky won the Vezina Trophy the previous two times he was a finalist in 2012–13 and 2016–17 with the Columbus Blue Jackets.

Connor Hellebuyck of the Winnipeg Jets and Thatcher Demko of the Vancouver Canucks are the other two finalists.

Hellebuyck (2.39) and Demko (2.45) are right behind Bobrovsky in the goals allowed average leaderboard. Hellebuyck was second in the NHL with 37 wins, just behind Alexandar Georgiev with 38. Bobrovsky was right behind both of them with 36.

The Vezina Trophy is voted on by the NHL’s 32 general managers and does not include postseason performance.