The most successful Jeopardy! contestants of all time are finally coming together to answer the question of who would win in a tournament of the show's greatest champions. James Holzhauer, Ken Jennings, and Brad Rutter will compete in a primetime Jeopardy! tournament event that will air on ABC in January.
Host Alex Trebek told USA Today of the news, "When James had his run last year, a lot of people were wondering, well how would he do against Ken Jennings? How would he do against Brad Rutter? [They're] our two most successful players in Jeopardy! history. These three players have won close to $10 million in Jeopardy! prize money, and over 100 games among them, so it was logical."
Jennings, who first competed on the show in 2004, still holds the show's record for most consecutive games won at 74, as well as highest regular season winnings with $2,520,700. But Holzhauer is just behind him on that latter record with $2,462,216. Holzhauer also set the record in April of 2019 for most single-game winnings with $131,127, and he boasts all of the other Top 10 highest single-game winnings. Meanwhile, Rutter has the record for highest all-time winnings (including tournaments) at $4,688,436, well above Jennings' $3,370,700 and Holzhauer's $2,462,216. So beyond the prize money, there are major bragging rights at stake in this tournament.
The tournament will begin with a back-to-back series of games on Tuesday, Jan. 7 at 8/7c on ABC. Whichever contestant emerges from that series with the most combined winnings will win the match. The series will then continue on weeknights (not including Mondays) until one player has won three matches. That winner will then claim the $1 million prize, while the runners-up will earn $250,000 each.
Per USA Today, the ABC event will mark the first time the network has aired Jeopardy! outside of local syndication in 30 years.
Jeopardy!'s Greatest of All Time tournament begins Tuesday, Jan. 7 at 8/7c on ABC.