‘Jeopardy’ announces ‘greatest of all time’ primetime specials
By Matt Collins Article may include affiliate links
“Jeopardy!” will air a primetime series to find the “greatest” player of all time.
- The matches will put Ken Jennings, Brad Rutter and James Holzhauer — who, between themselves, hold a myriad of “Jeopardy!” records — against each other.
- The series will take on a different format than most “Jeopardy!” tournaments — with the first person to win three total games to win $1 million.
- The two non-winners will receive $250,000 each.
- The matches are scheduled to start airing Tuesday, Jan. 7, 2019 on ABC from 8 to 9 p.m. eastern.
- It’s important to note these games will not necessarily air on the same station that carries the syndicated, daily version of “Jeopardy!”
- Instead, viewers should tune in to the channel they normally receive ABC primetime programming such as “Dancing with the Stars” and “How to Get Away with Murder” to catch the “Greatest” games.
- Games will run through at least Jan. 9, with Jan. 10, 14, 15 and 16 slated as “if needed” slots.
- It’s not immediately clear how the show will fill the hour, though, based on the schedule laid out, it appears there won’t be more than one game per night (if there were, it could conceivably end Jan. 8).