1. #1

    Combinatorics Question of the Day (2/15)

    A raid leader has to setup this week's schedule. Her team always raids two times a week at the same time, but the team is flexible and can raid any of the 7 days during the lockout. However, she knows from the past that it is always a disaster if she schedules two consecutive days.

    A.) So how many different schedules can she create that do not include two consecutive raid nights during the lockout?

    B.) If she is equally likely to choose any of the possible schedules (from A), what are the chances her raid team would be raiding Tue & Thur this week?

  2. #2
    Mechagnome
    Join Date
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    Moscow, Russia
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    This is easier to write down than to calculate =))
    "do not include two consecutive raid nights" is a bit hard to formulate.
    14 schedules. 7.14%.
    1234567
    !_****_
    _!_****
    __!_***
    ___!_**
    ____!_*
    1 = first raiding day.
    * = possible 2nd raiding day.

    EDIT : Yeah, I was an idiot, counted wrong.
    Last edited by Thiron; 2013-02-15 at 02:39 PM.
    Old Gods made me do it.

  3. #3
    Missing one? *_____!_

    How would you generalize this to work with larger numbers? Suppose lockouts lasted 30 days instead of 7, how many ways could you select 5 raid nights among any of those 30 days without two consecutive?

  4. #4
    Mechagnome
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    Quote Originally Posted by Laurabelle View Post
    Missing one? *_____!_

    How would you generalize this to work with larger numbers? Suppose lockouts lasted 30 days instead of 7, how many ways could you select 5 raid nights among any of those 30 days without two consecutive?
    I marked "!" as earliest raiding day in week, it can't be after "*". *_____!_ is !_____*_ from first line.
    I know how to name it, but I don't remember the actual formula for those things. Gotta think a bit.

    ---------- Post added 2013-02-15 at 06:58 PM ----------

    Yeah, I was overthinking it.
    It's C(N,K) - N;
    C(7,2) - 7 in first case, C(30,5)-30 in second case.
    My math terms might be a bit different, since some stuff does depend on country, but C(n,k) is number of combinations of n items in k places. C(n,k) = n! / (k! * (n-k!))
    Old Gods made me do it.

  5. #5
    Hmm, I still think we're missing one. Perhaps it is the Monday + Tuesday pair. You'd think it's forbidden, but it's not. You raid the first night on Tuesday:

    !

    Then, you skip all the other nights until Monday (day before next lockout):

    !_____

    Then you raid on that Monday (day before the next lockout)
    !_____*

    You'd think this would be forbidden (consecutive) but they don't actually touch during that week... right?

    So that would make 15?


    To add: they do not touch because X!______*X these X mark the start or end of some other week, so they don't actually wrap around and touch each other.
    Last edited by Laurabelle; 2013-02-15 at 03:24 PM.

  6. #6
    Mechagnome
    Join Date
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    503
    Well, if you consider non-constant chedule and only think of current week, you do get +1.
    But then we also don't know about previous week, what if we raided on Monday on previous week? We'd get -4 then.

    Ooh, I see. So we do get 15 different schedules, but the chance is not 1/15, since we also need co calculate chances for previous week.
    Old Gods made me do it.

  7. #7
    When I was writing the first question, I was thinking that you have 5 days you do not raid during the 7. So let's use X's or bottles, or some object for each of the 5 non-raid days and space them out:

    X X X X X

    Then we have gaps between each non-raid day where we could raid:
    | X | X | X | X | X |

    We count the start and end gap since those butt up against some other weeks:

    end of some other week | X | X | X | X | X | start of some other week

    When I was writing the question, I didn't consider what the raid leader chose to do in those other weeks. So, if we ignore the other weeks.

    Then we are simply choosing 2 of the gaps. In other words, 6 choose 2 or C(6,2) which is 15

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