JEOPARDY! THE GREATEST OF ALL TIME Airs Jan. 7

By: Dec. 23, 2019
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JEOPARDY! THE GREATEST OF ALL TIME Airs Jan. 7

On the heels of the iconic Tournament of Champions, "JEOPARDY!" is coming to ABC in a multiple consecutive night event with "JEOPARDY! The Greatest of All Time," premiering TUESDAY, JAN. 7 (8:00-9:00 p.m. EST), on ABC.

Hosted by Alex Trebek, "JEOPARDY! The Greatest of All Time" is produced by Sony Pictures Television. Harry Friedman will executive produce.

This epic television event brings together the three highest money winners in the long-running game show's history: Ken Jennings, Brad Rutter and James Holzhauer. The "JEOPARDY!" fan favorites will compete in a series of matches; the first to win three receives $1 million and the title of "JEOPARDY! The Greatest of All Time." The two runners up will each receive $250,000.

"Based on their previous performances, these three are already the 'greatest,' but you can't help wondering: who is the best of the best?" Trebek said.



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