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3 biggest ‘Jeopardy!’ winners to face off in ‘Greatest of All Time’ ABC special

The three biggest winners in “Jeopardy!” history, Ken Jennings, Brad Rutter and James Holzhauer will face off in primetime on ABC beginning in the new year.

“Jeopardy! The Greatest of All Time” will have the trio competing for a top prize of $1 million. The first player to win three matches gets the top prize; the two runners up will grab $250,000 each.

Ken Jennings holds the record for the longest winning streak in “Jeopardy!” history with 74 straight wins.

Rutter won more money than anyone else in the show’s history winning $4,688,436.

Most recently, Holzhauer holds 15 of the top single-day winning records. He just won the 2019 Tournament of Champions.

“Based on their previous performances, these three are already the ‘greatest,’ but you can’t help wondering: who is the best of the best?” host Alex Trebek said.

The first episode of “Jeopardy! The Greatest of All Time” will air on Tuesday, January 7, 2020 at 7 p.m. on WXOW.

The program will air on WXOW as follows: All of the shows will be from 7-8 p.m. CT

Tuesday, Jan. 7
Wednesday, Jan 8
Thursday, Jan 9
Friday, Jan 10*
Tuesday, Jan 14*
Wednesday, Jan 15*
Thursday, Jan 16*
* If Necessary

Kevin Millard

Kevin Millard-Social Media Digital Content Manager for WXOW.

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