Ken Jennings, Brad Rutter will play Jeopardy against IBM-Watson (computer) on February 14. TV Jeopardy to Feature IBM's Watson vs Ken Jennings & Brad Rutter on Feb 14
January 16, 2011
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I don't know what you'll be doing on February 14, 2011 at 7:00 pm.
But I'll either be glued to my TV or my DVR will be taping.
I know . . . I know . . . It's Valentine's Day.
But there's actually something "bigger" scheduled that night.
A computer named "IBM-Watson" is going to play Jeopardy on TV that night.
Competing against Watson will be two of the most celebrated players ever to appear on Jeopardy -- Ken Jennings & Brad Rutter.
Ken Jennings broke the Jeopardy! record for the most consecutive games played by winning 74 games in a row during the 2004-2005 season, resulting in winnings of more than $2.5 million.
Brad Rutter won the highest cumulative amount ever by a single Jeopardy! player, earning $3,255,102. The total amount is a combination of Rutter's original appearance in 2002, plus three Tournament wins: the "Tournament of Champions" and the "Million Dollar Masters Tournament" in 2002 and the "Ultimate Tournament of Champions" in 2005.
The grand prize for this competition will be $1 million with second place earning $300,000 and third place $200,000. Rutter and Jennings will donate 50 percent of their winnings to charity and IBM will donate 100 percent of its winnings to charity.
Watch the three youtube videos below . . . and then tell me you're not going to watch.
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O. J. Cunningham |
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