Top Jeopardy! winners Ken Jennings, James Holzhauer, and Brad Rutter to face off in primetime event

Jeopardy.
(Image credit: Screenshot/Twitter/ABC)

Who is the greatest Jeopardy! player of all time?

That question will be answered this January as the show's top three contestants — James Holzhauer, Ken Jennings, and Brad Rutter — will face off in a special Greatest of All Time tournament, as ABC announced Monday. The event will be airing on prime time.

Rutter holds the record for all-time Jeopardy! winnings including tournaments with $4.6 million, followed by Jennings with $3.3 million and Holzhauer with $2.4 million. Jennings, however, holds the record for most consecutive games won at 74, as well as highest winnings during regular season play. During Holzhauer's unbelievable run on the show earlier this year, he repeatedly shattered the record for single-game winnings, reaching a high of $131,127. Jennings teased in June he would "love to" face off against Holzhauer, adding the idea would be "irresistible," although he noted, "it's been 15 years since I was in Jeopardy! shape."

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In this tournament, the first player to win three matches will take home $1 million and be declared the show's all-time greatest player, per The Hollywood Reporter. "Based on their previous performances, these three are already the 'greatest,' but you can't help wondering: who is the best of the best?" host Alex Trebek said.

This, USA Today reports, will be the first time Jeopardy! has aired on a major network on prime time since 1990. The event will last at least three days and is set to kick off on Jan. 7.

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Brendan Morrow

Brendan worked as a culture writer at The Week from 2018 to 2023, covering the entertainment industry, including film reviews, television recaps, awards season, the box office, major movie franchises and Hollywood gossip. He has written about film and television for outlets including Bloody Disgusting, Showbiz Cheat Sheet, Heavy and The Celebrity Cafe.