m1b@rayssd.UUCP (M. Joseph Barone) (10/09/85)
In 'Errand of Mercy', where Kirk and Spock try to help
the Organians from Klingon domination, Spock computes their odds
of success to be some_large_number.7 to 1. I want to know where
the .7 came from. Any number divided by 2 will have either an
integral quotient or a quotient with a remainder of one (0.5).
Did Spock put weights to each Klingon of the garrison in deter-
mining that ratio? What is the answer to this burning question?
In that same scene, a good line from Spock is, "I endeavor
to be accurate."
Joe Barone, {allegra, decvax!brunix, linus, ccice5}!rayssd!m1b
Raytheon Co, Submarine Signal Div., Box 330, Portsmouth, RI 02871ccs020@ucdavis.UUCP (Kevin Chu) (10/10/85)
> [...] Any number divided by 2 will have either an > integral quotient or a quotient with a remainder of one (0.5). > [...] > > Joe Barone, {allegra, decvax!brunix, linus, ccice5}!rayssd!m1b > Raytheon Co, Submarine Signal Div., Box 330, Portsmouth, RI 02871 1.4 / 2 = 0.7 I guess 1.4 isn't any number. :-) -- --Kevin Chu ..ucbvax!ucdavis!vega!ccs020 /ex
brown@utflis.UUCP (Susan Brown) (10/11/85)
In article <1137@rayssd.UUCP> m1b@rayssd.UUCP (M. Joseph Barone) writes: > In 'Errand of Mercy', where Kirk and Spock try to help >the Organians from Klingon domination, Spock computes their odds >of success to be some_large_number.7 to 1. I want to know where >the .7 came from. Any number divided by 2 will have either an >integral quotient or a quotient with a remainder of one (0.5). >Did Spock put weights to each Klingon of the garrison in deter- >mining that ratio? What is the answer to this burning question? The most satisfying answer to this that *I* have ever read is the one Jean Lorrah (author of The Vulcan Academy Murders) has Amanda, Spock's mother, suggest in one of her earlier fanzine-published stories, namely that "I think they make those numbers up at least half the time." Of course she had to live with two Vulcans around the house, so it might be a necessary human defense mechanism.:-) sb
erosenth@aecom.UUCP (Elazar Rosenthal) (10/15/85)
How about good ol' human arithmitic remember how kirk said that the computers audio sensers could aplify sound by a factor of 1 to the forth power? Where I come from 1^4=1.
m1b@rayssd.UUCP (M. Joseph Barone) (10/15/85)
> > [...] Any number divided by 2 will have either an > > integral quotient or a quotient with a remainder of one (0.5). > > [...] > > 1.4 / 2 = 0.7 > > I guess 1.4 isn't any number. :-) So sue me! Have you seen 1.4 Klingons lately? :-) Joe Barone, {allegra, decvax!brunix, linus, ccice5}!rayssd!m1b Raytheon Co, Submarine Signal Div., Box 330, Portsmouth, RI 02871
johnw@astroatc.UUCP (10/16/85)
In article <298@utflis.UUCP> brown@utflis.UUCP (Susan Brown) writes: >In article <1137@rayssd.UUCP> m1b@rayssd.UUCP (M. Joseph Barone) writes: >> In 'Errand of Mercy', where Kirk and Spock try to help >>the Organians from Klingon domination, Spock computes their odds >>of success to be some_large_number.7 to 1. I want to know where >>the .7 came from. Any number divided by 2 will have either an >>integral quotient or a quotient with a remainder of one (0.5). >>Did Spock put weights to each Klingon of the garrison in deter- >>mining that ratio? What is the answer to this burning question? > Why should spock divide by 2?? (There's him+Spock, but there are also 400+ people somwhere on the enterprise, and an undetermined? number or organians, who might be perswaded to lend a hand too eh? * \ == ) * / To err is Human, to really screw up world news requires the net. John W {...allegra,ihnp4,siesmo....} !uwvax!astroatc!johnw
johnw@astroatc.UUCP (10/17/85)
In article <1149@rayssd.UUCP> m1b@rayssd.UUCP (M. Joseph Barone) writes: >> 1.4 / 2 = 0.7 >> >> I guess 1.4 isn't any number. :-) > > So sue me! Have you seen 1.4 Klingons lately? :-) > Maybe there was a crippled Klingon that Spock counted as equivilant to 4/10ths of a normal Klingon. Maybe Kilgons, being stronger than humans count at 1.4 humand equvilants each.
slerner@sesame.UUCP (Simcha-Yitzchak Lerner) (10/18/85)
> >>Did Spock put weights to each Klingon of the garrison in deter- > >>mining that ratio? What is the answer to this burning question? > > > Why should spock divide by 2?? (There's him+Spock, but there are > also 400+ people somwhere on the enterprise, and an undetermined? > number or organians, who might be perswaded to lend a hand too eh? Also, given their track record, shouldn't Kirk and Spock receive a greater weighting factor than 2? 'nuf said! -- Opinions expressed are public domain, and do not belong to Lotus Development Corp. ---------------------------------------------------------------- Simcha-Yitzchak Lerner {genrad|ihnp4|ima}!wjh12!talcott!sesame!slerner {cbosgd|harvard}!talcott!sesame!slerner talcott!sesame!slerner@harvard.ARPA
phillips@cisden.UUCP (Tom Phillips) (10/21/85)
>>In article <1137@rayssd.UUCP> m1b@rayssd.UUCP (M. Joseph Barone) writes: >>> In 'Errand of Mercy', where Kirk and Spock try to help >>>the Organians from Klingon domination, Spock computes their odds >>>of success to be some_large_number.7 to 1. I want to know where >>>the .7 came from. Any number divided by 2 will have either an >>>integral quotient or a quotient with a remainder of one (0.5). Who says that the odds against success are strictly (# of enemies)/(# of friendlies)? Perhaps there were modifications for armament, organization, desperation, historical precedents? In any case, "(some large number).7 to 1" is totally bogus. Too much uncertainty in the calculations. Perhaps the writers left out a :-)? Tommy Phillips cisden!phillips
m1b@rayssd.UUCP (M. Joseph Barone) (10/22/85)
johnw@astroatc.UUCP (John F. Wardale) writes: > Why should spock divide by 2?? (There's him+Spock, but there are > also 400+ people somwhere on the enterprise, and an undetermined? > number or organians, who might be perswaded to lend a hand too eh? The Enterprise was getting reinforcements and the Organians were wimps (as far as Spock knew), with a combat limit approaching 0! johnw@astroatc.UUCP (John F. Wardale) also writes: > Maybe there was a crippled Klingon that Spock counted as equivilant > to 4/10ths of a normal Klingon. Maybe Kilgons, being stronger > than humans count at 1.4 humand equvilants each. Klingons don't have cripples or doctors. Sickly specimens are destroyed! Tell Kirk they're stronger than him! Joe Barone, {allegra, decvax!brunix, linus, ccice5}!rayssd!m1b Raytheon Co, Submarine Signal Div., Box 330, Portsmouth, RI 02871
cipher@mmm.UUCP (Andre Guirard) (11/07/85)
In article <298@utflis.UUCP> brown@utflis.UUCP (Susan Brown) writes: >> ... Spock computes their odds >>of success to be some_large_number.7 to 1. I want to know where >>the .7 came from. > >Jean Lorrah (author of The Vulcan Academy Murders) has Amanda, Spock's >mother, suggest in one of her earlier fanzine-published stories >that "I think they make those numbers up at least half the time." I think the real answer is that the authors of the Star Trek scripts don't know enough about statistics to realize that it's usually not possible to compute the odds exactly. In "The Trouble with Tribbles", Spock computes the number of tribbles in the grain storage compartment (it's some large number ending with three, I think) and then tells us how he computed it... obviously it is not possible to compute the exact number, but he pretends to do so. I would think a "real" Vulcan would have said something like, "Somewhere between 12,340,000 and 12,500,000".
mcewan@uiucdcs.CS.UIUC.EDU (11/12/85)
> In "The Trouble with Tribbles", > Spock computes the number of tribbles in the grain storage compartment > (it's some large number ending with three, I think) and then tells us > how he computed it... obviously it is not possible to compute the exact > number, but he pretends to do so. I would think a "real" Vulcan would > have said something like, "Somewhere between 12,340,000 and 12,500,000". Actually, it was possible to compute it exactly, given the assumptions that Spock made. He assumed that each tribble had exactly x babies every y hours (I don't remember what x and y were). I think I even checked the arithmetic once and found that Spock did give the right number. Scott McEwan {ihnp4,pur-ee}!uiucdcs!mcewan "A flash in front of my eyes ... I blink ... open my eyes to ... discover I am a dog in a pickup truck full of garbage ... no one but me sees the lid blow off the can ... it's 14 miles to the dump ... this is ... at last ... heaven."
ins_akaa@jhunix.UUCP (Kenneth Adam Arromdee) (11/14/85)
In article <288@mmm.UUCP> cipher@mmm.UUCP (Andre Guirard) writes: >In article <298@utflis.UUCP> brown@utflis.UUCP (Susan Brown) writes: >>> ... Spock computes their odds >>>of success to be some_large_number.7 to 1. I want to know where >>>the .7 came from. >I think the real answer is that the authors of the Star Trek scripts >don't know enough about statistics to realize that it's usually not >possible to compute the odds exactly. In "The Trouble with Tribbles", >Spock computes the number of tribbles in the grain storage compartment >(it's some large number ending with three, I think) and then tells us >how he computed it... obviously it is not possible to compute the exact >number, but he pretends to do so. I would think a "real" Vulcan would >have said something like, "Somewhere between 12,340,000 and 12,500,000". It doesn't end in 3 and in fact is a power of 11. Spock claims that it allows for the volume of the storage compartment, etc... but it doesn't. (Interestingly, the James Blish novelization includes a different number, which presumably does allow for it.) -- ------------------------------------------------------------------- If you know the alphabet up to 'k', you can teach it up to 'k'. Kenneth Arromdee BITNET: G46I4701 at JHUVM and INS_AKAA at JHUVMS CSNET: ins_akaa@jhunix.CSNET ARPA: ins_akaa%jhunix@hopkins.ARPA UUCP: ...{decvax,ihnp4,allegra}!seismo!umcp-cs!aplvax!aplcen!jhunix!ins_akaa ...allegra!hopkins!jhunix!ins_akaa
cipher@mmm.UUCP (Andre Guirard) (11/28/85)
In article <1187@jhunix.UUCP> ins_akaa@jhunix.ARPA (Kenneth Adam Arromdee) writes: >In article <288@mmm.UUCP> cipher@mmm.UUCP (Andre Guirard) writes: >> In "The Trouble with Tribbles", >>Spock computes the number of tribbles in the grain storage compartment >>(it's some large number ending with three, I think) and then tells us >>how he computed it... >It doesn't end in 3 and in fact is a power of 11. Spock claims that it >allows for the volume of the storage compartment, etc... but it doesn't. It could allow for the volume of the storage compartment and the other factors he mentions if in fact the storage compartment was larger than necessary for the computed number of tribbles, etc, etc. The real problem is that he computes an exact number from an "average" litter size and an "assumed" starting time and initial number of tribbles. -- /''`\ Andre Guirard ([]-[]) De Tuss from de Tonn \ o / ihnp4!mmm!cipher `-'