There have been plenty of memorable Jeopardy! winners over the years, but three stand above everyone else as the best to ever play the TV game show — 2019 Tournament of Champions winner James Holzhauer, who holds the record for single-game winnings; Ken Jennings, whose 74-game winning streak is the longest in the show’s history; and Brad Rutter, who holds the record for most money won by a contestant — and they’re finally going to face off to determine who’s the greatest.
The three contestants will compete in a primetime special called Jeopardy! The Greatest of All Time hosted by Alex Trebek that will air Jan. 7 through Jan. 16 on ABC at 8 p.m. EST. The first competitor to win three matches will be crowned the ultimate Jeopardy! champion and win $1 million. The other two will each take home $250,000.
Currently, Rutter’s winnings on the show stand at $4,688,436, while Jennings has won $3,370,700 and Holzhauer’s total $2,712,216.
“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.
“I am always so blown away by the incredibly talented and legendary Alex Trebek, who has entertained, rallied and championed the masses for generations — and the world class Jeopardy! team who truly are ‘the greatest of all time,’” Karey Burke, president of ABC Entertainment, said. “This timeless and extraordinary format is the gift that keeps on giving and winning the hearts of America every week. We can’t wait to deliver this epic and fiercely competitive showdown — with these unprecedented contestants — to ABC viewers and loyal fans everywhere.”
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