Help with a math problem

Discussion in 'The Pub' started by ow my eyes, Mar 30, 2015.

  1. ow my eyes

    ow my eyes Member

    Messages:
    1,892
    Joined:
    Sep 1, 2009
    Location:
    th gr8 NW
    It seems like I should be able to do this, but I cant seem to create the formula correctly.

    Q: An event has occurred 61.7% of the time. ( rounded only to the .1%)
    It took the event to happen exactly 576 more times out of the next 882 "attempts" for the rate to change to 61.8%
    How many times has the event (and attempts) occured before?

    x/y = 61.7%
    x/y + 576/882 = 61.8%

    I dont know X or Y, can we solve for them?
     
  2. lefort_1

    lefort_1 Nuzzled Firmly Betwixt Gold Supporting Member

    Messages:
    11,601
    Joined:
    Aug 4, 2012
    Location:
    Oregon
    EDIT, my first AND second answer were bunged up....re-thinking:



    AGAIN
     
    Last edited: Mar 30, 2015
  3. jhc

    jhc Senior Member

    Messages:
    12,541
    Joined:
    Dec 12, 2005
    Location:
    Seattle
    previously was 6170 events in 10,000 attempts

    (probably not the right answer but that's the ballpark)
     
  4. Rockyrollercat

    Rockyrollercat Member

    Messages:
    2,037
    Joined:
    Apr 16, 2010
    Location:
    South Jerrrsey not Noith Jiosey
    Occurred before the change or before now after the change?
     
  5. Mark McPheeters

    Mark McPheeters Member

    Messages:
    908
    Joined:
    Jan 31, 2005
    Location:
    mid-Missouri
  6. Fantom1

    Fantom1 Member

    Messages:
    1,997
    Joined:
    Feb 6, 2004
    Are there only two events?
     
  7. Kitten Cannon

    Kitten Cannon Member

    Messages:
    4,703
    Joined:
    Sep 27, 2010
    Location:
    I'M ON AN INTERNET!
    It's basically a system of equations, which I haven't worked out in forever (literally not since high school). Your two equations were close, but the second one has a grouping error, which I've fixed below. If you isolate x as a factor of y in one equation, you can substitute it into the second equation, and solve for y. Then go back and solve for x using that value, and check your work:

    x/y = 61.7% --> x = .617y
    (x + 576) / (y + 882) = 61.8% --> (1000/618) (x + 576) = y + 882


    1000/618 (.617y + 576) = y + 882
    .998y + 932 = y + 882
    50 = .002y

    y = 25000

    And since x = .617y, x = 15425

    Checking the work:
    15425 / 25000 = 61.7%
    (15425 + 576) / (25000 + 882) = 16001/25882 = 61.8%

    My rounding might be a little off here because I think I used some rounded numbers in multiplications midway through the process, but it more or less bears out to be true.
     
  8. John Coloccia

    John Coloccia Cold Supporting Member

    Messages:
    9,469
    Joined:
    Dec 31, 2007
    Location:
    Connecticut, outside of Hartford
    x/y = .617
    (x+576)/(y+882) = .618

    2 equations, 2 unknowns

    x = .617y (eq1)

    sub x into equation 2

    (.617y + 576)/(y+882) = .618

    solve for Y
    .617y+576 = .618y + 545.076
    .001y = 30.824

    y = 30824
    That's the number of attempts before the additional attempts. 30824*.617 = 19018 occurrences.

    And we can check that it all makes sense

    (19018+576)/(30824+882) = 19594/31706 = .618
     
  9. John Coloccia

    John Coloccia Cold Supporting Member

    Messages:
    9,469
    Joined:
    Dec 31, 2007
    Location:
    Connecticut, outside of Hartford
    Well, we can't both be right. LOL.
     
  10. lefort_1

    lefort_1 Nuzzled Firmly Betwixt Gold Supporting Member

    Messages:
    11,601
    Joined:
    Aug 4, 2012
    Location:
    Oregon
    ^^^ YEAH, WHAT HE SAID !!!
     
  11. Serious Poo

    Serious Poo Armchair Rocket Scientist Graffiti Existentialist Gold Supporting Member

    Messages:
    5,322
    Joined:
    Jun 9, 2005
    Location:
    SF Bay Area
    29070

    Using iterations in Excel. No idea of the actual formula that would render the number, though.
    Started with 617/1000 = 61.7% (treated as exactly 61.700000% per OP)
    Added the new elements = 576/885 = 65.0847%
    Then worked up multiples of 617 and 1000 respectively until the combined totals of the numerators divided by the combined total of the denominators equaled exactly 61.8000000%.

    17936.19 / 29070 + 576 / 885 = 18512.19 / 29955 = 0.61800000000

    Sloppy, I know.
     
    Last edited: Mar 30, 2015
  12. jhc

    jhc Senior Member

    Messages:
    12,541
    Joined:
    Dec 12, 2005
    Location:
    Seattle
    This works, but so does my example and Mark McPheeters'. There are many combos where the initial % is 61.7 and then increases to 61.8, but that's not what the OP is specifically asking.

    I'm pretty sure Mark is right, but don't know how he got it
     
  13. Kitten Cannon

    Kitten Cannon Member

    Messages:
    4,703
    Joined:
    Sep 27, 2010
    Location:
    I'M ON AN INTERNET!
    I'm not about to pull out an actual calculator, but I'm guessing that since our process was the same, your unrounded numbers probably yielded more accurate results.

    The variance is due to the fact that a LOT of combinations would round to the right answer. John rounded to a decimal at intermediate steps. I rounded to whole numbers, because I am lazy.
     
  14. Mark McPheeters

    Mark McPheeters Member

    Messages:
    908
    Joined:
    Jan 31, 2005
    Location:
    mid-Missouri
    You have to establish a required degree of precision, methinks. Otherwise, you can get multiple answers. The rounding error manifests itself in different ways depending on your approach to solving the equation.....
     
  15. John Coloccia

    John Coloccia Cold Supporting Member

    Messages:
    9,469
    Joined:
    Dec 31, 2007
    Location:
    Connecticut, outside of Hartford
    He specified .1%. That's why I went out as far as I did. You need to have 3 significant digits to get .1% accuracy.
     
  16. Kitten Cannon

    Kitten Cannon Member

    Messages:
    4,703
    Joined:
    Sep 27, 2010
    Location:
    I'M ON AN INTERNET!
    Yes, funny that this comes up, because it's the crux of a lot of my troubles in my day job. We occasionally have to build reports or load data into data warehouse schemas for BI consumption that involves doing compound calculations, and if you don't round everything the right way at the right time, numbers can go pretty sideways. Which is frustrating, because we're usually rebuilding some arcane calc that a former employee with NO math/stats background put together in Excel many moons ago, that became the gold standard and that nobody can mess with because That's How It Was Reported Since The Beginning. So, this usually means replicating the bad math with rounding to whole numbers every step, or whatever, making our calculations A) much harder to write and B) less accurate.

    AKA, this is why we can't have nice things.
     
  17. John Coloccia

    John Coloccia Cold Supporting Member

    Messages:
    9,469
    Joined:
    Dec 31, 2007
    Location:
    Connecticut, outside of Hartford
    What's funny is I think most of the answers DO come within .1%, depending how to read the question. :D
     
  18. scotth

    scotth Member

    Messages:
    1,154
    Joined:
    Mar 13, 2012
    Location:
    Morden, MB
    I got that you started with 18,510 out of 30,0000.

    No formula. I got it by trial and error using fractions & ratios. Took about 20 minutes. Lol
     
  19. Kitten Cannon

    Kitten Cannon Member

    Messages:
    4,703
    Joined:
    Sep 27, 2010
    Location:
    I'M ON AN INTERNET!
    If someone was feeling really nerdy, they could calculate the upper and lower bounds of both x and y, for all responses that would still round to .1%.

    Spoiler alert: that someone will not be me.
     
  20. John Coloccia

    John Coloccia Cold Supporting Member

    Messages:
    9,469
    Joined:
    Dec 31, 2007
    Location:
    Connecticut, outside of Hartford
    That's a problem for a mathematician. Any real engineer knows that no one is ever going to actually check anyway.
     

Share This Page