I just finished watching the end of the IBM Challenge on Jeopardy where a computer by the name of Watson defeated two of the greatest Jeopardy contestants in history, Ken Jennings and Brad Rutter.
Selfishly people want to know when they can harness the power of such a machine. The answer is 7 years. This computer runs on a whopping 90 servers and is not exactly portable. Jeopardy, in fact, actually had to come to IBM for the contest. According to an online Wall Street Journal article, “The ratio of computer price to performance is now doubling in less than a year, so 90 servers would become the equivalent of one server in about seven years,”
Lucky number seven? This number seemed to stick out during the course of this contest. If you don’t know already Watson would wager odd dollar amounts during Final Jeopardy and the Daily Doubles. These wagers tended to end in the lucky number seven.
Since the number seven is culturally and biblically relevant. This got me wondering so I decided to run some numbers:
There are 673 days and 4 hours until Midnight 12/21/2012. That is 16,156 hours which is divisible by, yup, 7. (16,156/7=2,308).
Watson’s total ended at $77,147, which is divisible by, yup, seven. (77,147/7=11,021) If you take 11,021 and divide that by 24hrs you get 459Days. That’s one year and 94 days.
The number 94 could be relevant. If you take the number 2,308 from above and divide by 24hrs you get 96.16666. Very close to 94 days.
Well 2012 is a leap year so that accounts for an extra day and if we add that to the four hours from the end of the show to midnight we get 28 which is divisible by, yup, 7. Not to mention the additional 24 hour window you get for anything occurring on a certain day this brings us to 96 days.
Giving a variable of +/- 24 hrs, I believe that the beginning of the end will happen 460 and 4 hours from the end of Jeopardy on 05/21/2012, 6 months from 12/21/2012
It’s worth noting that numbers there are thirteen days left in February 2012 from 02/16/10.
