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Problem of the day 25.07.09

with 7 comments

In 1896 lord Coin has decided to play a game. From the January 1 till December 31 every day he chooses among two match boxes an arbitrary one and placed a match from it to another box (if the chosen box was not empty). If the chosen box was empty then he placed a match from
the other box to the chosen one. What is the probability that after the December 31 the both boxes will have an equal number of matches if at the beginning each box had a) n = 400 b) n = 200 c) n = 100 matches?


Written by Implex

July 25, 2009 at 3:42 pm

7 Responses

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  1. Complete bouncer …. just waiting for the solution

    nikhil jangi

    July 26, 2009 at 7:22 am

    • just got close to … there are 365 days … arbitary box can be selected with 0.5 probability … different ‘n’ and 365 days will have some effect on probability ..


      nikhil jangi

      July 26, 2009 at 7:26 am

  2. waitin for the soln too .. 1896 is a leap year as its a multiple of 4 . so its 366 days not 365.

    When n is 400 is it 0.5 * 366 for the boxes to be equal ?

    explain pls


    August 8, 2009 at 6:28 am

  3. In the first case you can never get an empty box, while in the second case, if at any moment you exhaust one of the boxes you can never get back to boxes with the same number of matches, so in both cases the probability is equal to the number of sequences of length 366 of 1’s and -1’s summing 0 divided by the total number of such sequences, or binom(366,183) / 2^366, approximately 0.04. Case c must be more complicated.


    September 1, 2009 at 4:12 pm

  4. how about the case in which the same box is chosen in every alternate turn to end up with the same no. of matches (duh…)
    ABABABABABAB… 366 times … so the prob wud be (1/2)^366


    September 11, 2009 at 4:31 pm

  5. case a) 400 matches, 366 days so no way to ever empty a matchbox

    ways in which we can get equal matches = ways of selecting matchbox A 366/2 times = 366 C 183
    total no of ways of selecting matchboxes = 2^366
    therefore prob = 366!/(183!^2)*2^366 (phew!)


    November 10, 2009 at 2:29 pm

  6. case b) 200 matches – caseb.1 no matchbox emptied
    case b.2 matchbox A emptied
    b.1 = same as above
    b.2 = 0 as emptied matchbox cannot be filled up in time

    case c) 100 matches case c.1 same as above
    case c.2 matchbox a emptied…and refilled

    A selected B+200 times
    AAABAA… and A+B = 366, so 2b+200 = 366, b = 83,a = 283,
    ans = (366 C 83)/ 2^366…. multiply by 2 to get probability for b emptied and refilled as well


    November 10, 2009 at 2:36 pm

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