Start Your iPods

What the fuck do we start off with this week?

  1. Modest Mouse - Alone Down There. 2
  2. Jimi Hendrix - Wait Until Tomorrow. 3
  3. Flaming Lips - Death Valley 69. 2
  4. Stereolab - Ping Pong. 4
  5. Califone - Lion And Bee. 3
  6. Unrest - When It All Comes Down. 3
  7. Cream - SWLABR. 4
  8. Robyn Hitchcock - Sweet Ghost Of Light. 4
  9. Robyn Hitchcock - Birds In Perspex. 3
  10. The Decembrists - Shanty For The Arethusa. 4.
  11. Bonus: Robyn Hitchcock - Adoration Of The City.

That's a fine mix.

I'd be interested in the calculating the odds of three Hitchcock songs showing up in a list of 11 (out of 6000 songs, roughly 300 of which are RH). But I have no idea how to calculate that... I don't remember enough probability to know how to do it without having to work with numbers like 6000 factorial ... (6000 * 5999 * 5998 * 5997 ... * 1 = 2.689 * 1020065)

2 thoughts on “Start Your iPods”

  1. marklow

    i got home kinda tipsy tonight and felt like sobering up to your probability problem above. No calculator i own can really handle that sorta thing so I may have this all wrong. But isnt it a combinations problem?

    (assuming these are BIG C’s)

    300 C 3
    ——-
    6000 C 11

    or:

    300 x 299 x 298
    ————–
    6000 x ….. 5990

    I think I got, roughly, this:

    4.5
    —-
    5.9 x 10 to the 34th power

    or 1 in 13,110,000,000,000,000,000,000,000,000,000,000

    It may be wrong. But it sure was fun…

  2. cleek

    turns out there are a couple of ways to do this. it’s a “hypergeometric distribution” problem, and there’s a formula for calculating these:

    N = 6000 be the total number of marbles
    n = 10 be the number of marbles drawn (your sample)
    B = 300 be the number of black marbles
    k = 3 be the number you are interested in
    p = (B choose k) * ( (N-B) choose (n-k) ) / ( N choose n)

    that means you have to deal with huge factorials. Excel can handle them, though, and it says answer is 0.010410216

    or, you can treat is as a binomial distribution (ie. with replacement, since N is so large) and get an approximate answer:

    n = # of trials = 10
    x = # of successes among n trials = 3
    p = prob. of success in any one trial = 300/6000 = 0.05
    q = prob. of failure in any one trial = (1 – p) = 0.95
    P(3) = (10!) / ((10 – 3)! * 3!) * (0.05)^3 * (0.95)^(10-3)
    = 0.0104750594

    pretty close.

    of course i didn’t come up with any of that m’self. i had help

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