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'Jeopardy! The Greatest of All Time' premiering in January

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Ken Jennings, Brad Rutter and James Holzhauer will compete for the title{ } "Jeopardy! The Greatest of All Time" and $1 million. (CNN Newsource)

The three top winners in the history of "Jeopardy!" will go head to head during a prime-time event on ABC in January.

Ken Jennings, Brad Rutter and James Holzhauer will compete for the title as the best of the best on the popular game show.

Jennings became a household name during his record 74-game winning streak -- the longest in the game show's history.

Rutter holds the title for the most money won by a contestant -- across any television game show. He managed to rake in $4,688,436 in prize money on "Jeopardy!" He has never lost to a human opponent.

Holzhauer won the Tournament of Champions and his winnings more than $2.7 million. He also holds the record for all 15 of the top Single-Game Winnings records on "Jeopardy!"

The first person to win three matches of will receive the title of "Jeopardy! The Greatest of All Time" and $1 million.

The runner ups will each receive $250,000.

Alex Trebek will host the special "Jeopardy! The Greatest of All Time" will air between Jan. 7 - Jan. 16 at 8:00 p.m.

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