‘Jeopardy! The Greatest of All Time’ tournament to host Jennings, Holzhauer

Alex Trebek speaks during a rehearsal before a taping of Jeopardy! Power Players Week at DAR Constitution Hall on April 21, 2012 in Washington, DC.

Kris Connor | Getty Images

Long-running trivia game show Jeopardy! will air a multi-night tournament event, “Jeopardy! The Greatest of All Time,” in prime time on ABC starting January 7, the show announced on Monday.

The series of matches, hosted by Alex Trebek, will pit against one another the three players who took home the most cash in the game show’s history: Ken Jennings, Brad Rutter and James Holzhauer.

After a series of matches, the first to win three face-offs will bank $1 million. The three players won a combined $10.7 million during their runs on the show.

“We’re excited to bring Jeopardy! to prime time! It’s been a long time in the making – we wanted to create a unique experience sure to wow not just our fans, but all audiences,” said Mike Hopkins, chairman of Sony Pictures Television, in a press release. “We are thrilled to have James, Brad and Ken, three powerhouse players each worthy of the title ‘The Greatest of All Time.'”

Earlier this year, in a 32-episode winning streak, Holzhauer came $58,484 short of surpassing Ken Jennings as the highest-earning contestant in non-tournament winnings. (Jennings took home a total of $3,370,700 across two separate show runs.) 

Rutter holds the title of highest Jeopardy! earner with over $4.6 million in winnings, according to the release.

Holzhauer walked away from winning streak with $2.46 million in total winnings. The 35-year-old trivia whiz got his payback on Friday when he beat Emma Boettcher — the contestant who broke his winning streak — during Jeopardy!’s “Tournament of Champions” to win the first prize of $250,000.

The tournament will run for a maximum of seven shows, between January 7 and January 16, but will end as soon as one of the contestant has tallied up three wins. Each show will air at 8 p.m. ET. 

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