Jeopardy! ‘Greatest of All Time’ match will crown show's biggest winner
The winner will take home $1M
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.”