NEW YORK (WHTM) – The three highest money winners in “Jeopardy!” history are coming to ABC in a primetime competition hosted by Alex Trebek.
“Jeopardy! The Greatest of All Time” premieres Tuesday, Jan. 7, from 8-9 p.m.
The television event brings together the three highest money winners in the long-running game show’s history: Ken Jennings, Brad Rutter and James Holzhauer.
The fan favorites will compete in a series of matches. The first to win three receives $1 million and the title of “Jeopardy! The Greatest of All Time.”
The two runners up will each receive $250,000.
Jennings became a household name during his record 74-game winning streak, the longest in “Jeopardy!” history. His winnings total $3,370,700.
Rutter is the highest money winner of all time across any television game show, with total “Jeopardy!” winnings of $4,688,436.
Holzhauer holds the record for all 15 of the top single-day winnings records on “Jeopardy!” and just won the 2019 Tournament of Champions. His winnings total $2,712,216.
“Jeopardy! The Greatest of All Time” will air:
- Tuesday, Jan. 7 (8-9 p.m.)
- Wednesday, Jan. 8 (8-9 p.m.)
- Thursday, Jan, 9 (8-9 p.m.)
- *Friday, Jan. 10 (8-9 p.m.)
- *Tuesday, Jan 14 (8-9 p.m.)
- *Wednesday, Jan 15 (8-9 p.m.)
- *Thursday, Jan. 16 (8-9 p.m.)