Golden State Valkyries forward Janelle Salaun was unanimously selected as the 2026 WNBA Sixth Player of the Year award winner, receiving all 74 first place votes from the voters pool. In order to qualify for the award, players needed to play more games as a reserve than as a starter.
Salaun appeared in 40 contests for the Valkyries in 2026, averaging 12.7 points, 4 rebounds and 1.2 assists per game. She was especially efficient from beyond the arc shooting nearly 40% on the year (39.2%). The 25 year old set career highs in scoring, despite averaging nearly 5 minutes per game less than in her rookie campaign (22.1 minutes per night).
When all was said and done, she set the new WNBA single-season records for points (509), three-pointers made (93) and games with at least 10 points (27) by a reserve.
Salaun proved to be a weapon off the bench for Golden State, able to stretch opposing defences, and provide an offensive spark when her team needed it the most all year long. As a result, the Valkyries posted a 32-12 regular season record, which was good for the 2nd overall seed heading into the WNBA playoffs.
“Whether she starts or she comes off the bench, it doesn’t matter,” coach Natalie Nakase told reporters. “I’m still gonna have a fighter who wants to kill the other opponent, and she’ll exhaust her minutes.”
The French sharpshooter becomes the first Valkyrie player to earn the award, and follows in the footsteps of players who would go on to become stars such as DeWanna Bonner, Kelsey Plum, and Jonquel Jones amongst others. Salaun will receive a $15,000 bonus for winning the award.
Golden State will now turn their attention to their opening game against the Las Vegas Aces which tips off at 4pm ET on Sunday.
Photo: John Mac. This file is licensed under the Creative Commons Attribution-Share Alike 4.0 International license.