The winners of the 3-point shootout will square off between Stephen Curry and Sabrina Ionescu

NBA All-Star Weekend is set to feature a new addition, courtesy of Sabrina Ionescu and Stephen Curry. The stars from the Golden State Warriors and New York Liberty will compete in a 3-point shootout in Indianapolis next month, as reported by Shams Charania of The Athletic. Curry, a two-time NBA 3-point contest winner (2015, 2021), will face off against Ionescu, who set a WNBA record with 37 points in the final of her 3-point contest victory last year.

Details about the specific rules of the contest, including which 3-point line the players will use, are yet to be clarified. The potential for this matchup was hinted at by Curry earlier on Thursday in an apparently spontaneous segment with Warriors rookie Brandin Podziemski, which was coincidentally captured by TNT’s cameras.

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Curry and Ionescu have been hinting at this for some time. Following Ionescu’s impressive performance at the WNBA shoot-out, she reached out to Curry on Twitter, suggesting a shootout. Days later, Curry responded, affirming his interest in the contest.

In previous NBA All-Star Weekends, WNBA stars have typically participated in events like the celebrity game and the shooting stars competition, with no history of involvement in the 3-point contest. However, Ionescu is breaking that trend, providing an opportunity to showcase women’s basketball on one of the NBA’s grandest stages.

The NBA All-Star Weekend is set to take place from Feb. 16-18 in Indianapolis. The announcement of the game’s starters occurred earlier on Thursday, revealing that Curry fell short of securing his ninth career start. Alongside the Curry-Ionescu contest, another notable change in the weekend’s format is the return to the traditional East-West format for the game, as opposed to selecting teams.