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Thread: The Maths Thread

  1. #571
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    Quote Originally Posted by Days of Grace View Post
    Apologies if this if I have asked this before, but...

    How do I get microsoft excel to give a higher value to a bowling average of 20 than a bowling average of 30?

    I want a linear progression, so a bowling average of 40 is worth 80 points, an average of 20 is worth 160 points, and an average of 10 would be worth 320 points, and so on.

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    LOL, you're right.
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    Quote Originally Posted by Days of Grace View Post
    How do I enter that into the formula input?
    Average Points
    20 160
    40 80
    10 320

    Suppose, averages are in column A and Points are in column B, then in cell B2 write this:
    =3200/A2

    Copy B2 and paste in B3, B4 etc...
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  4. #574
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    Hulululu has 3/4th chance of dying and 1/4th chance of breaking into 2 Hulululus. If broken into 2, each of these 2 Hulululus have the same chances - 3/4th chance of dying and 1/4th chance of breaking into 2 further Hulululus....and so on.

    What's the probability that the Hulululu species will never be extinct?


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    I'm massively sleep deprived but: 0. Monkeys, Shakespeare, etc.

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  6. #576
    Global Moderator Spark's Avatar
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    Yeah the sum you'd set up would converge to 1, wouldn't it?

    **** I hate probability. Prefer something nice and clean like finding two irrationals a and b such that a^b is rational.
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    Quote Originally Posted by silentstriker View Post
    I'm massively sleep deprived but: 0. Monkeys, Shakespeare, etc.

    Right?
    Right. Yes, that's one good way to think about it, if you leave the small mathematics out...

    Quote Originally Posted by Spark View Post
    Yeah the sum you'd set up would converge to 1, wouldn't it?
    Yes. There's a much easier approach than setting up the sum and breaking into geometric progressions etc though..

  8. #578
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    My immediate reaction was to keep it simple and go with tree diagrams, but the splitting in two part threw me off.

    Immediately I thought it was (1/4)^n with n becoming very large, but obviously that doesn't work.

    Secondarily I thought in terms of series; the probability being quartered each step as there are two, but I doubt that holds up mathematically. That would have been 1/4 * 1/16 * 1/64 ...

    So whatever the formula is [I got P(n)=1/(4^n), but I may be very wrong], as n becomes very large you approach zero, so it is almost certain they will become extinct at some point.
    Last edited by Dan; 27-12-2012 at 06:34 AM.

  9. #579
    International Captain weldone's Avatar
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    Repeating patterns, mate.

    OK. Assume 'if we start with a bacteria, the probability of the species finally going extinct is p'. Then,

    p = 3/4 + 1/4*p*p
    => (p-2)^2 = 1
    => p = 1 [Rejecting the other solution as that can't be a probability number]

    So, the probability of of the Hulululu species never going extinct = 1-p = 0
    Last edited by weldone; 27-12-2012 at 07:00 AM.

  10. #580
    International Captain weldone's Avatar
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    In fact, for any x where 0<x<1, if you replace 3/4 by x and 1/4 by (1-x) in the original problem, then also p will come out as 1 from the equation I think. I'm also sleep-deprived, so not putting an effort to solve that generic one lol.

  11. #581
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    Quote Originally Posted by Spark View Post
    Prefer something nice and clean like finding two irrationals a and b such that a^b is rational.
    I'll try that tommorrow maybe (assuming it doesn't need your continuous geometric algebra or stuff sounding similar to that ). Too sleepy now...

  12. #582
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    Alternatively you can look at the expected population at every year.

    E(0) = 1
    E(1) = 0.5
    E(2) = 0.25
    E(3) = 0.125

    Et voila...
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    Also love the fact that he wasn't completely correct.

  15. #585
    Global Moderator Spark's Avatar
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    Quote Originally Posted by weldone View Post
    I'll try that tommorrow maybe (assuming it doesn't need your continuous geometric algebra or stuff sounding similar to that ). Too sleepy now...
    Two line proof. Very, very sneaky proof too. The specific problem, I should say, is proving that there exists two irrationals a, b such that a^b is rational.
    Last edited by Spark; 27-12-2012 at 09:09 PM.

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