Do you feel lucky?

raffle ticket of the beast

If so, you might want to go buy a raffle ticket at Clusterflock, a scintillating group blog where I have occasionally been known to talk some trash. What’s being raffled?

iPhones, baby, iPhones.

You know you want one.

All the money received will be disbursed in the form of iPhones—with recipients to be chosen at random by celebrity random-name-chooser Jason Kottke—or will go to charity. So the more $20 tickets get bought, the more $499 phones get given away. Your chance of winning is therefore pretty close to 1 in 25, unless I’m really bad at math. (Which I am, by the way, so work it out for yourself; you may use a separate sheet of paper; I’ll wait. Correct me in the comments if I’m being totally stupid.) The no doubt exorbitant monthly service plan will, of course, be solely your problem.

The deadline for buying raffle tickets is noon CST (that’s Dallas time, y’all?) on June 26. Winners will be announced on June 27.

Photo: raffle ticket of the beast by squacco; some rights reserved.

4 Responses

  1. Schizohedron
    Schizohedron June 12, 2007 at 9:51 am |

    The play has a slightly positive expectation for a single raffle ticket in a 25-ticket pool, assuming no sales tax (that is, it only actually costs $499). A $20 raffle ticket has an expected value of $0.76; 24 times you would lose $19.20, and 1 time, you would win a prize worth $499.

    If Clusterflock picks up the tab for sales tax, your EV goes up. If this were in NJ, the iPhone would cost $533.93 with 7% sales tax. Your EV would increase to $2.1572 for a single $20 bet. The same math could be used to calculate how much you save on shipping.

    All this is for 25 total tickets with each person only buying one. If only 25 are in the drawing for a single phone, and you buy two, your EV goes up to $3.12. On the other hand, unless my math is fouled up, once 50 tickets are out there, your EV goes negative for a single ticket purchase on a $499 possible win, and it’s a losing play. There may be a number of tix you can buy that would bring +EV to a 50-ticket pool with two draws, but that’s a touch outside my math ability.

    Now, the fact that you NEED multiples large enough to purchase the phone for the chance your money will actually count as in the pot for the phone (and therefore be an influence on the odds of winning), with the remainder of the cash going to charity if they don’t clear enough for the next phone, makes this a John Nash–level problem that I — a low-level poker degenerate and English major — cannot possibly fathom. However, it does demonstrate that I find statistics more palatable than looking for a job. :)

  2. India Amos
    India Amos June 12, 2007 at 10:12 am |

    Whoa—see, now, I didn’t get one bit of that. What’s EV even stand for? I never got anywhere near statistics in school, and I was far too protective of my hoarded allowance to ever learn to play poker from my parents.

  3. Schizohedron
    Schizohedron June 12, 2007 at 12:15 pm |

    EV = expected value, or how much you stand to win or lose, over time, in that particular wager or game situation. If it’s positive, as it was in the first paragraph of my math rant, then over time that play should win you money, and you should make that bet every time those conditions are present. That upper-level poker players can calculate these sort of algebraic problems in their heads, over a few seconds, is what keeps me in the ultra-cheap seats when I go to the casino!!

  4. Sillium
    Sillium June 15, 2007 at 5:21 pm |

    I don’t get it. My very simple view would be: In one of 25 times, I would win $479 ($499 – $20 raffle ticket). In all the other 24 times, I’d lose $20. So if I played long enough, I’d lose $1 every 25 turns. This makes a negative EV. Where’s the error in my calculation?

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