Wednesday, June 09, 2010

Hide and Seek Counting

Five children are hiding in two closets. What is the least number of kids in the closet with the most kids?

Just to make it a little more interesting, what's the probability that there are the least number of kids in the closet with the most kids?


  1. Three. If five kids hide in two closets, the most even split would be 3 in one, 2 in the other.

    I pretty much failed the probability section of Math in high school...I got like a 40% on that test. So all I can tell you about the second part is P(me getting the right answer)= not good.

  2. Kira has answered the first part.

    Assuming that a closet always has at least one kid in it.
    For the second part there are 4 combinations.
    Then the probability is 50%.

  3. The splits can be:

    5,0 4,1 3,2 2,3 1,4 0,5

    between the two closets.

    So there are 2 out of 6 permutations where the split is 3 children in one closet and two in another, so the probability is 1/3 or 0.33333... (or 33%)

    In reality, as a man with 2 small kids, I can assure you that Heisenberg's Uncertainty Principle kicks in here, and many of the children could be in at least two places at once, one will hurt themselves and cry, and a further two could disappear altogether. At any time, three of their friends from up the road will appear. At no point will you be able to determine both the number AND the position or momentum of the children, so normal probability calculations break down.

    Hope that clears it up!

  4. Your answer made me laugh Kira and reallyfatbloke. Considering the day, I really needed it. Thanks!

    I have to say between the three of you, you've covered all the bases. I'll leave it at that.


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