* John Parker wrote on Sat, May 31, 2008 at 15:35 -0500:
> > Probability that a "proper" key falls in the space of the
> > "bad debian" keys: 2^15 / 2^2048 = 1 / 2^2033.
> >
> > That's a lot of zeros before the first non-zero digit.
> 
> Put differently, if you were to start generating keys now at a rate
> of, say, 1000/s, how long would you have to wait before you got one of
> the Debian keys?  This is a fun math problem for probability theory
> students.

wow, big numbers, John!

Cool idea to make such a time estimation :)

Maybe we should say `a million keys per second', sounds much more
but just are three digits less in the result :)

Is the calculation that complicated? Aren't the keys independent
of each other, so that each key always have the same probability,
since we are not `searching' but `guessing' when generating?

  (beside, that all those values are so horrible big that
   practically it does not matter of course :-))

With Victor's number of 2013 bits probablility, couldn't we
statistically expect half of that? With a million per second does
this give

(2^2012) / 10^6 / 60 / 60 / 24 / 365.25

years which the 593 digit number

14902094353953870165214353410981143707238235188212334084836694330488\
81602740116106914618746657670317636941551690018457525299578948872878\
36765806488289940028625838604817603080995646449473721456572544453618\
55782431446798772374819591436871325406930507575507226972337350924070\
18286766525605611643878663746554436287030227901811414516143083673080\
28892637223535933402770689260083725677906317276399679998875094201786\
41124284757024653658707346461288521262653417342296719918707161098486\
04762949019240046008945125630714069482285597143371578237868834348990\
3212246280855279993597997641265155474006217516831

of years? Seems there even is a number word[1], so are that
around a hundred quintillion nonagintacentillions? lol

Assuming the age of the universe beeing 13.73 * 10^9 year
(http://en.wikipedia.org/wiki/Age_of_the_universe),

(2^2012) / 10^6 / 60 / 60 / 24 / 365.25 / (13.73 * 10^9)

or `in short':

10853673965006460426230410350314015810078831164029376609495043212300\
66717217855868109700470981551578759607830801178774599635527275216954\
38285365250029089605699809617492791756005569154751435875143877970588\
89863387798105442370589651447102203501041884614353406389175055297938\
95329036071089301998454962670469363646780938020255946479346747030648\
42602066441031269776235024952719392336421207047632687544701452441213\
70083237259304190574440893271149687736819677598176780712823860960295\
73753058280582699205349690918218550242014273228966917871718014820823\
249253188700311725680844693464323049818

universe ages would be needed, slighty more than 10^582, which is
a funny big number... Even when using `googols' (10^100) as
factor it remains terrible...

lol

SCNR.

oki,

Steffen

[1] 
http://en.wikipedia.org/wiki/Names_of_large_numbers#Extensions_of_the_standard_dictionary_numbers
 
About Ingenico Throughout the world businesses rely on Ingenico for secure and 
expedient electronic transaction acceptance. Ingenico products leverage proven 
technology, established standards and unparalleled ergonomics to provide 
optimal reliability, versatility and usability. This comprehensive range of 
products is complemented by a global array of services and partnerships, 
enabling businesses in a number of vertical sectors to accept transactions 
anywhere their business takes them.
www.ingenico.com This message may contain confidential and/or privileged 
information. If you are not the addressee or authorized to receive this for the 
addressee, you must not use, copy, disclose or take any action based on this 
message or any information herein. If you have received this message in error, 
please advise the sender immediately by reply e-mail and delete this message. 
Thank you for your cooperation.
______________________________________________________________________
OpenSSL Project                                 http://www.openssl.org
User Support Mailing List                    openssl-users@openssl.org
Automated List Manager                           [EMAIL PROTECTED]

Reply via email to