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Puzzle – Circus

January 31, 2007 am31 8:20 am

A group of children totalling less than a hundred go to the circus and sit in one long row of seats. There are five brothers there, and a girl at each end of the row, and the probability of that happening is 1/3. How many boys and how many girls are there?


goes to Bertie Taylor, a regular at compuserve’s SCIMAT forum puzzles section. The forum is gone, and I last saw Bertie’s signature a decade ago, but he provided some gems.


Standing challenge for any problem on this website: Take a kiddie problem, already solved, and make it challenging for strong high school students or even math teachers. Or, take a problem for advanced high school students, and twist it to make it a good challenge for middle school kids. In any event, this is a standing challenge:

Take a good problem, and rewrite it into a good problem for someone else.

10 Comments leave one →
  1. Clueless permalink
    January 31, 2007 pm31 9:36 pm 9:36 pm

    Neat problem. It is interesting how the numerical clues help make the solution unique. (In the absence of these, it looks like there are infinite solutions, but not all would fit in a single pragmatic circus tent).

  2. rdt permalink
    January 31, 2007 pm31 11:49 pm 11:49 pm

    Can you clarify what “there are five brothers there” means and what the “that” in the “probability of that happening” is 1/3 refers to? Just the probability of there being a girl at each end of the row?

    The probability of there being 5 brothers from the same family in a group of less than 100 kids seems pretty low, and relies on information about family size that’s not included in the problem, so I assume the 5 brothers is there to place a minimum on the number of boys?

  3. January 31, 2007 pm31 11:59 pm 11:59 pm

    There are five brothers = amongst all the children in the row, there is a group of five male siblings (doesn’t say where they are sitting) And yes, the five is there to establish a minimum number of boys.

    that = girls are sitting on both ends.

    Fwiw, I agree that the wording is not the best. Ideas for a sharp rewording?

  4. rdt permalink
    February 1, 2007 pm28 9:53 pm 9:53 pm

    I brute-forced it with a little fortran routine: there are 36 kids, 21 girls, 15 boys.

    As far as wording goes, how about…
    A group of children — less than 100 total, and including 5 brothers — go to the circus and sit in one long row of seats. When they are seated, there is girl at each end of the row, and the probability of that happening was 1/3. How many boys and how many girls are there?

  5. February 1, 2007 pm28 10:26 pm 10:26 pm


    and thanks.

    Follow up, same idea, a little easier, but can help generate more:

    I picked up part of a deck of cards, shuffled what I had, and then turned over the top three. All were aces! The probability of that happening was just 1 in 204. How many cards had I picked up?

  6. JBL permalink
    February 4, 2007 am28 2:16 am 2:16 am

    Given b boys and g girls, the probability of a girl sitting at each end is g/(b + g) * (g – 1)/(b + g – 1) = 1/3. So
    3g^2 – 3g = b^2 + 2bg + g^2 – b – g. Bringing everything over to the left side,
    2g^2 – (2b + 2)g – b^2 + b = 0. This is a quadratic equation in g (or in b), with roots g = (2b + 2 +- sqrt(4b^2 + 8b + 4 + 8b^2 – 8b))/4 which simplify to (b + 1 +- sqrt(3b^2 + 1))/2. Now, we just need to find the integers larger than 5 such that 3b^2 + 1 is a perfect square; this is some sort of Pell-type equation, which means there is a method for generating solutions, but I don’t happen to remember what the method is. (There’s some sort of nice recursion for them, I think.) On the other hand, we can check by hand or computer that the smallest three answers are b = 3, 15, 56; 3 is too small, and 56 is too large, so 15 it is, as rdt wrote. I also like rdt’s re-wording. I’ll think about the aces question when I get the chance.

  7. February 5, 2007 pm28 3:57 pm 3:57 pm

    “There’s some sort of nice recursion for them, I think.”

    Indeed so. Saying that 3b^2 + 1 = a^2 means that a/b is a very good approximation of sqrt(3). In fact a/b must be one of the convergents of the continued fraction of sqrt(3). It is pretty quick to calculate that the continued fraction expansion of sqrt(3) is [1,1,2,1,2,1,2,…], which means that sqrt(3) = 1 + 1/( 1 + 1/( 2 + 1/( 1 + 1/( 2 + 1/ …. (Yes the formatting is awful.) Once you know this expansion, the “magic box” algorithm calculates the possible values of a and b. The whole thing is explained at this page:

  8. JBL permalink
    February 6, 2007 am28 6:16 am 6:16 am

    Jonathan, I think there’s something wrong with the aces question: if you select n (greater than or equal to 3) cards from a normal deck, shuffle them, and turn over the top 3, the probability of getting 3 aces is 1/5525 (= 4*3*2/(52*51*50)) regardless of what n is.

  9. February 6, 2007 am28 6:25 am 6:25 am


    consider that I had picked up 4 cards, and three were aces. Given that, the probability that the top three of those four would be aces would be (3/4)(2/3)(1/2) = 1/4

    (which it better be, since that should match the probability that the bottom card was the non-ace).

    The wording should indicate that we want to identify the mix of aces and non-aces such that P(AAA) = 1/204. Suggestions for clarifying the language?

  10. JBL permalink
    February 7, 2007 am28 3:55 am 3:55 am

    Ah, I see! Well, we must have either 3 or 4 aces, to be sure. If we have 3 aces and n cards, the probability that the top 3 are aces is 3*2*1 / (n(n – 1)(n – 2)) = 1/204, which gives us the equation 1224 = n(n – 1)(n – 2). Now, there are all sorts of pieces of technology I could apply here, including the rational root theorem and Cardano’s formula for the roots of a cubic. My prefered method is to note that the right-hand side is increasing in n, and when n = 11 it’s too small and when n = 12 it’s too big. So this case is a bust.

    If we have 4 aces, everything works out the same except that on the left side of that equation we get 4896 instead of 1224. Any of the same tools would work here; one of the ways to “see” the root is to note that the right side is just a tad smaller than (n – 1)^3, while the left side is just a tad smaller than 17^3; and indeed n = 18 works.

    I think replacing the final sentence with, “How many cards had I picked up and how many of them were aces?” drives home the point that the number of aces is also considered fixed. There are lots of other possible re-wordings, but the ones I thought up were all too clunky.

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