Ken, Brad or James? Jeopardy tournament to determine 'greatest of all time'

1st player to win 3 matches will land $1M, runners-up split $500K

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Caption: Jeopardy, hosted by Alex Trebek, will hold a tournament to determine its 'greatest player of all time,' pitting the three best-known champions against one another in January. (Ben Hider/Getty Images)

The three top money winners in Jeopardy history will vie for a share of $1.5 million US in January.
ABC and the quiz show's producer said Monday that Ken Jennings, Brad Rutter and James Holzhauer will compete in prime-time episodes on the network.
The first contestant to win three matches will receive $1 million. Each runner-up will take home $250,000.
Alex Trebek will host the contest, titled Jeopardy: The Greatest of All Time and debuting 8 p.m. ET on Jan. 7.
Rutter is the top all-time money winner with $4.7 million, followed by Jennings with $3.4 million and Holzhauer with $2.7 million.

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Jennings holds the record for longest winning streak (74 games), while Holzhauer holds the top 15 of Jeopardy's single-game winnings records.
In a statement, Trebek said the three have proved they qualify as the "greatest," and now will compete to be "the best of the best."