Gentlemen, start your buzzers.
If you've ever wondered how some of best and brightest in "Jeopardy!" history would do if they faced off against each other, you're about to find out.
"Jeopardy! The Greatest of All Time," is a prime time event that will premiere on ABC on Jan. 7. It will bring together the long running series three highest money winners - Ken Jennings, Brad Rutter and James Holzhauer.
The trio will compete in a series of matches over multiple nights, with the first to win three receiving $1 million dollars and the runners up earning $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?” said "Jeopardy!" host Alex Trebek.
Ken Jennings became famous during his record 74-game winning streak. His winnings total $3,370,700. James Holzhauer was the 2019 Tournament of Champions winner. He's won $2,712,216 as a "Jeopardy!" contestant.
Brad Rutter is the highest money winner of all time on any TV game show. His winnings total $4,688,436.
“JEOPARDY! The Greatest of All Time” will air as follows:
• TUESDAY, JAN. 7 (8-9 p.m. EST)
• WEDNESDAY, JAN. 8 (8-9 p.m. EST)
• THURSDAY, JAN. 9 (8-9 p.m. EST)
• *FRIDAY, JAN. 10 (8-9 p.m. EST)
• *TUESDAY, JAN. 14 (8-9 p.m. EST)
• *WEDNESDAY, JAN. 15 (8-9 p.m. EST)
• *THURSDAY, JAN. 16 (8-9 p.m. EST)