I experienced a double hard drive failure that crippled both primary and secondary laptops, leaving me scrambling to score a replacement off Craigslist using my old Bronze Powerbook. The one with the cracked display and the keys for most of the swear words missing and a modem that makes phone noises expecting Juno to pick up. I had to get Donny to drive me around until we found someone broadcasting an unprotected version of 802.11 still on speaking terms with MacOS, which meant I had to listen to him whine about the latest Game of Thrones still not being on Netflix. Tragic I suppose, but not nearly as tragic as LabKitty.com going postless, which troubled him, I thought, insufficiently.
Still, the alternative was elbowing through the hobos on the machines over at the public library, the keyboard equivalent of a theater floor. So I am now intimate with the trials of John Snowden, and Bionca of Tusk, the conniving Little Feet, Theon Lovejoy and Deloris Dragonborn and all the rest. Can't wait to see who's going to win the battle of five armies.
Go house Snuffelpuff.
The good news is I found a nice man named Jeeter looking to unload a dead 13" MacBook for negotiated cheap. All of this over the phone, I would note, because nobody makes a browser running under 8.5 that can get into gmail anymore. Not really what I would call advanced functionality. Yeah, looking at you Mozilla. And Classilla. And Opera, And iCab. Not only did you people make me desperate enough to try Internet Explorer, you made me talk on the phone. To another human. You and I are enemies until we are both dead.
I asked Jeeter if the hard drive was an ATA/66 and the line went quiet with just the sound of breathing. After long minutes, Jeeter said all he knowed, man, was his old lady was using it over at the beauty college before she fell off the back of his chopper with it and now it don't turn on. This was cause for some concern. Although I have never fallen off a chopper, I once fell over in a recumbent when my feet got stuck in the toe clips and I chipped a tooth. "Ah, she fine," Jeeter chimed in unprovoked, "ceptin for a little road rash 'cross her Ronnie James Dio tat."
Jeeter said he'd meet us over at Carmikes cause they's going to the 8 PM of Paul Blart. I don't know what that is, but we went there and found Jeeter standing at the snack bar working a Big Drink in one hand and holding the MacBook by its power cord in the other. His black t-shirt was emblazoned with a Harley Davidson logo and "If you can read this my old lady fell off" printed on the back, which, as we know, she had, at least once. I considered telling him I owned a similar shirt emblazoned with the Linux penguin and "live to code," thinking it might build solidarity and maybe knock a couple points off the asking price. But in the end I demurred. I pressed a wad of bills into a hand the size of a catcher's mitt and Jeeter held out the MacBook like a severed head. Inside, presumably, was a ripe 60 gig SeaGate ready for a new life.
I didn't get to meet Jeeter's old lady (she was parking the minivan) but I'm willing to live with that disappointment as long as there's nothing on her hard drive that brings the federales to my door. Just to be safe I had Donny run it past the big degausser they got at the Votech.
But LabKitty is back, for now at least. If I remember, we had just finished up a math contest, so I suppose I should post the solution.
To refresh your memory, here was the question:
Suppose the average inconsiderate movie-going rat bastard receives 10 text messages per day, each of which upon arriving on their iThing (which they carry with them at all times) plays a sample from Shrieking Teenage Harpy at maximum volume. What is the probability of enjoying a screening of Anna Kareninna (starring dreamy hunk Jude Law!) uninterrupted by this ringtone if a crowd of three attends the show? Assume the assumptions of the Poisson distribution are applicable (wink, wink).
Solution
There is a none-too-subtle hint that the Poisson distribution is applicable. The Poisson says: P(X=k) = lambda^k ⋅ exp(-lambda) / k!, where X is the random variable being described, k is any integer, and lambda is an average arrival rate for the process of interest. Let X be the number of interruptions during the movie. We seek P(X=0). We must figure out lambda, and we turn to the Internet for help.
IMDB informs me the running time of Anna Karenina (the Jude Law version) is 129 minutes. Ten text messages per day is equivalent to 10 * (129/1440) = 0.90 messages per 129 minutes. Multiplying this by three people attending the show gives a mean rate of 1.8 interruptions per viewing. (I would have also accepted multiplying this by two, assuming you yourself have manners. Oddly, all of the solutions submitted assumed three.) Applying the Poisson equation shown above, we find P(X=0) = exp(-1.8) which is about 0.17.
So roughly one chance in five viewings. Not too terrible, I guess, but only because the theater was mostly empty. (Sorry, Mr. Law. I know you poured your heart into the project, but remakes of 19th-century Russian costume dramas just don't put butts in the seats anymore. Next time, maybe try to work zombies into the script.)
Footnote: Has there ever been an important text message? Something that can't wait? Do movie goers get last minute questions from some twitchy intern in re a critical clarification on the end-to-end anastomosis they're about to dig into? I'll bet a beer the answer is no. A good beer too. You know, something you would take the trouble to put in a glass. But I digress.
Our winner was LabKitty reader P, who is basking in Amazon gift coupon glory as we speak while the rest y'all play out the gift-couponless comings and goings of your little lives.
What will the next LabKitty contest be? And when? Stay tuned.
Nice info (y)
ReplyDeleteThank for sharing :))