3 biggest ‘Jeopardy!’ winners to face off in ‘Greatest of All Time’ ABC special

Ken Jennings, Brad Rutter and James Holzhauer — the three biggest winners in the history of “Jeopardy!” — will face off in a “primetime throwdown” in January on ABC.

“Jeopardy! The Greatest of All Time” will have the trio competing for a top prize of a million bucks; the first player to win three matches gets the top prize; the two runners up will grab a quarter of a million dollars each.

“Based on their previous performances, these three are already the ‘greatest,’ but you can’t help wondering: who is the best of the best?” longtime host Alex Trebek said.

Jennings holds the record for the longest winning streak in the history of “Jeopardy!” with 74 consecutive games, for a total take of $3,370,700.

Rutter, however, is the highest money winner on “Jeopardy!” — indeed, on any TV game show — winning $4,688,436.

Not to be outdone, Holzhauer holds 15 of the top single-day winnings records on “Jeopardy!,” and recently won the 2019 Tournament of Champions. His earnings total $2,712,216.

“Jeopardy! The Greatest of All Time” kicks off Tuesday, Jan. 7 at 8 p.m. EDT on ABC and will air for three consecutive nights at the same time. However, if no winner has won three matches by then, additional dates have been cleared in the 8 o’clock spot from Friday, Jan. 10 through Thursday, Jan. 16.

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