‘Jeopardy! The Greatest of All Time’ premiering in January

The three top winners in “Jeopardy!” history are set to compete against each other during a prime time event on ABC in January.

It was announced Monday that “Jeopardy” alums Ken Jennings, Brad Rutter and James Holzhauer will vie for the title as top champ of the champs on the popular game show.

They will compete in a series of matches. The first competitor to win three matches will receive $1 million and the title of “Jeopardy! The Greatest of All Time.”

The two other contestants will each receive $250,000.

Alex Trebek will host the special.

“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 in a statement.

Jennings became a household name during his record 74-game winning streak — the longest in Jeopardy! history — and his winnings totaled $3,370,700.

Rutter holds the title for the most money won by a contestant — across any television game show — raking in $4,688,436 in “Jeopardy!” prize money. He has never lost “Jeopardy!” to a human opponent.

Holzhauer is the record holder for all 15 of the top Single-Game Winnings records on the show. He won the 2019 Tournament of Champions and his innings total $2,712,216.

“Jeopardy! The Greatest of All Time” will air between Jan. 7 – Jan. 16 at 8 p.m. EST.

“Jeopardy! The Greatest of All Time” is being produced by Sony Pictures Television with Harry Friedman serving as executive producer.

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