'Jeopardy! The Greatest of All Time' Starts Tuesday
JEOPARDY! is more popular than ever and tonight the syndicated game show goes prime time with the start of a multi-night event on ABC. It’s a multi-night event because whoever the first player to win three matches wins the tournament, and so, ABC has scheduled the specials to run through Thursday the 9th, but could run as long as Thursday, the 16th (skipping next Monday so that the network can air The Bachelor).
Hosted by Alex Trebek (of course), the specials feature three of the highest money winners in the game show’s history:
- Ken Jennings – Won 74 games in a row, the longest in the show’s history. His winnings total $3,370,700.
- James Holzhauer – The Smirkster is the highest money winner of all time across any television game show, with total winnings of $4,688,436.
- Brad Rutter – Is the highest money winner of all time across any television game show, with total winnings of $4,688,436
The JEOPARDY! 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.
“Based on their previous performances, these three are already the ‘greatest,’ but you can’t help wondering: who is the best of the best?” Trebek said.
JEOPARDY! The Greatest of All Time begins tonight at 8 p.m. on ABC.
(Main image: ABC)