Las Vegas Freestyle Chess Grand Slam
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format
Levon Aronian, the winner of the Las Vegas Freestyle Chess Grand Slam | |
| Tournament information | |
|---|---|
| Sport | Chess variant (Chess960) |
| Location | Las Vegas, United States |
| Dates | 16 July 2025–20 July 2025 |
Tournament format | Single-elimination tournament with round-robin seeding round |
| Host | Freestyle Chess Operations |
| Venue | Pavillon Chesnaie du Roy |
| Participants | 16 |
| Final positions | |
| Champion | |
| Runner-up | |
| 3rd place | |
The Las Vegas Freestyle Chess Grand Slam was a Chess960 tournament that took place in Las Vegas, United States from 16 July to 20 July 2025. It was the fourth leg of the Freestyle Chess Grand Slam Tour, a series of Chess960 tournaments being held throughout 2025.[1]
While the second leg had 12 participants playing in the main event, the number of participants was increased to 16 for the fourth leg. However, while before only the top eight players after the round-robin stage qualified for the main knockout bracket, now all the remaining eight players would still play for positions 3-16 in a modified double-elimination style format.[2]
Another change to the format was the time control of the knockout stage, which was reduced from 90 minutes per player to 30, with an unchanged increment of 30 seconds. The change was made to have two games played on the same day and shorten the event's length. The round-robin and tiebreak games were unaffected.