We are ready for knowledge and this trio of intelligent gentlemen! Vulture can confirm that very smart players James Holzhauer, Ken Jennings, and Brad Rutter will face off against one another in the first-ever Jeopardy! “Greatest of All Time” tournament, which is set to air on ABC beginning Tuesday, January 7. The decision to create such a tournament — besides the guaranteed smash ratings — is to celebrate the show’s 30th anniversary on the network, and it’s going to have a pretty unique structure. Follow along with us: It’ll consist of a set of two back-to-back games; the player with the most combined winnings from these two games wins the “match.” The play will continue on successive nights “until one of them has won three matches and takes home a $1 million prize.” (Runners-up get $250,000.) Crunching the numbers, that means the tournament could end in as quickly as three days or last as long as seven.
Holzhauer, Jennings, and Rutter are the three most celebrated contestants in the history of Jeopardy! Rutter, who has never lost a match to a human opponent (congrats, Watson), is the show’s all-time champion with $4,688,440 in total winnings, with Jennings in second with $3,370,700 and Holzhauer in third with $2,464,216. Due to Rutter’s competing during the show’s five-game-maximum era, however, Jennings holds the record for the most consecutive games won (72), with Holzhauer in second (32). Holzhauer also holds the top 15 spots for his single-game winnings.
We’ll take “Oh God, This Is Extremely Good Programming” for $800, Alex.