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Jeopardy! ‘Greatest of All Time’ match will crown show's biggest winner

Shawn M. Carter

Tuesday’s “Jeopardy!” episode will be the greatest of all time.

The three biggest winners in the show’s history — Brad Rutter, Ken Jennings and James Holzhauer — will compete for the title of “G.O.A.T” and a grand prize of $1 million.

Rutter, during his time on the show, took home more than $4.6 million in total winnings. Jennings, who holds the longest winning streak ever on “Jeopardy!,” pocketed more than $3.3 million over the course of his 74-day win streak, including a $300,000 second-place prize when he faced off against IBM's Watson computer.

And Holzhauer won more than $2.7 million in his 32 appearances on the show. He also holds the top spot for the highest single-game winnings, raking in $131,000 in April 2019, according to the show’s website.

The special episode, which kicks off at 8 p.m. on ABC, will feature a series of one-hour matches, with each consisting of two complete games. The winners will be decided by total points.

The first-place winner will receive $1 million and the two runners-up will get $250,000 each. Though, all game show winnings are considered by the IRS to be ordinary income and are taxed up to 37 percent along with state taxes.

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Tuesday’s show comes as host Alex Trebek battles pancreatic cancer, which he revealed last March. He told the Associated Press his illness “took a toll” last month during the taping of the “Greatest of All Time” special but was assured by producers no one noticed.

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In an optimistic message to U.S. Rep. John Lewis, who is also facing pancreatic cancer, Trebek said, “We're starting a new year, and let's see if we can't both complete the year as pancreatic cancer survivors.”

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