Friday, April 5, 2013

A Post-Easter Puzzler

Recently, a good friend gifted me a bag of Easter SweeTarts, with ducks and bunnies in place of the usual round candies.  They come in five flavors, which are officially blue raspberry (blue), cherry (pink), grape (purple), orange (orange) and green apple (green).  (This according to Wikipedia, at least.)

I don't like the blue ones, so because I'm that way, I culled those out and they're still sitting in front of me in a bowl.  The rest I ate at my leisurethough rather less leisurely than is quite becoming, I'm afraid.

At some point, near the end of the bag, I wondered what the probability was, if I pulled out two candies at random, that they'd have the same flavor.  I looked at the candies that were in there, and it was a simple matter to figure the answer out.  It wasahh, but then I'd be giving the puzzle away. 

I then turned to look away and puckishly reached into the bag and pulled out a pair of green apples.  (Those are my second favorite, after the artificial grape ones, which, as you'll know if you eat them, taste nothing at all like actual grapes.)  I wondered if that affected the probability at allfiguring that if anything, it would have reduced the oddsbut after looking in and doing some quick figuring, I found that the odds were exactly the same as before.

"Hunh!" I thought to myself in surprise, and (if you know me at all, you know what happened next) I wondered what the odds were of that.  As it so happens, that's not a question you can answer rigorously at all, without knowing the priors.  But perhaps you can rigorously answer

Q1: What were the odds of drawing a matching pair of candies?

I reached in again, and this time pulled out a cherry and a grape; I did it again, and pulled out a matching pair of grapes.  So this time, I thought, for sure there must be a change in the odds of drawing a match, but when I looked in again, the odds were once more exactly the same as the first time.  What's more, I reached in again, pulled out a pair of green apples, looked in one last time, and the odds were again exactly the same as before!

Q2: Knowing there were no blue raspberry candies at all, how many of each of the other flavors were there, before I pulled out the first pair of green apples?

3 comments:

  1. My answers to Q1 and Q2 don't fit in this window, so I'll post them one at a time.

    I'd rather add than subtract, so I'll start at the end and work backward. I'll suppose that after drawing the second pair of green apples, you were left with A green apples, B grapes, C cherries, and D oranges.
    Let N = A + B + C + D.

    Let p be the chance of drawing a matching pair at this point. Since this was the same probability of a matching pair at the very start of the exercise, when you had at least three different flavors in the bag (at least four green apples, three grapes, and a cherry), we know p < 1.
    This rules out the case where A = N. Therefore A < N.
    But since you actually did draw matching pairs, we know that p > 0.
    This implies that N > 1.

    Then the probability that the next two candies drawn will match is

    p = (A(A - 1) + B(B - 1) +
    C(C - 1) + D(D - 1))/(N(N - 1)). [1]

    Just before you drew the last two green apples, you had the same probability of drawing a matching pair,

    p = ((A + 2)(A + 1) + B(B - 1) +
    C(C - 1) + D(D - 1))/((N + 2)(N + 1)). [2]

    Let
    q = A(A - 1) + B(B - 1) + C(C - 1) + D(D - 1)
    and
    r = N(N - 1).

    Then [1] implies that p = q/r. But also

    (A + 2)(A + 1) + B(B - 1) +
    C(C - 1) + D(D - 1) = q + 4A + 2
    and
    (N + 2)(N + 1) = r + 4N + 2.

    Therefore [2] implies that
    p = (q + 4A + 2)/(r + 4N + 2),
    and so
    q/r = (q + 4A + 2)/(r + 4N + 2). [3]

    Cross-multiply the terms of [3]:

    q(r + 4N + 2) = (q + 4A + 2)r. [4]

    Cancel the qr term on each side of [4] and divide by 2:

    q(2N + 1) = (2A + 1)r. [5]

    That is,
    q(2N + 1) = (2A + 1)N(N - 1)
    = (2AN + N - 2A - 1)N
    = ((2N + 1)(A - 1) + 3(N - A))N
    and
    (q - (A - 1)N)(2N + 1) = 3(N - A)N. [6]

    But 2N + 1 and N have no common prime factors. It follows from [6] that all the prime factors of 2N + 1 (including repetitions) are factors of 3(N - A), so

    3(N - A) = m(2N + 1) [7]

    for some integer m. Since A < N, it follows that m > 0.
    Moreover, [7] implies

    3N - 2mN = 3A + m > 0,
    so
    3N > 2mN,
    so
    3 > 2m > 0.

    Therefore m = 1, so

    3(N - A) = 2N + 1 [8]
    and
    N = 3A + 1. [9]

    But from [5] we can also derive

    p = q/r = (2A + 1)/(2N + 1) [10]

    and using [9] to substitute for N in [10], we get

    p = (2A + 1)/(2(3A + 1) + 1)
    = (2A + 1)/(6A + 3)
    = 1/3,

    which is the answer to question Q1.

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  2. I have a possible answer for the second part, but no proof that it is unique.
    I observed that [9] greatly reduces the number of possible cases to be examined for any particular N (in fact it rules out many values of N immediately). I observed that [6] and [8] imply that

    q − (A − 1)N = N,
    so
    q = AN. [12]

    q = A(A − 1) + B(B − 1) + C(C − 1) + D(D − 1)
    = A² − A + B² − B + C² − C + D² − D
    = A² + B² + C² + D² − N.
    So
    A² + B² + C² + D² − N = AN,
    and so
    B² + C² + D²
    = AN − A² + N
    = A(3A + 1) − A² + 3A + 1
    = 2A² + 4A + 1. [13]

    But meanwhile,

    A + B + C + D = N = 3A + 1,
    so
    B + C + D = 2A + 1, [14]

    with the constraint that none of B, C, or D can be negative.

    This even further limits the number of possibilities. For example, if A = 1,
    then B + C + D = 3 and B² + C² + D² = 7.
    But this has no solution, so A cannot be 1.
    If A = 2, then B + C + D = 5
    and B² + C² + D² = 17.
    This has solutions where B, C, and D are any permutation of 0, 1, and 4, but no other solutions.

    I then applied a spreadsheet and brute force. For each value of A, I examined each ordered set of three integers whose sum was 2A + 1, and subtracted 2A² + 4A + 1 from the sum of the their squares. If the result was zero, [13] was satisfied.

    I also observed that if A were the number of green apples after drawing the first pair of green apples, then A, B, C, and D would have to satisfy [13] and [14] for the same reasons they have to satisfy [13] and [14] when A is the number of green apples after drawing the second pair. So I was looking for two values of A, one greater than the other by 2, each of which allowed solutions for [13] and [14].

    For example, the spreadsheet found a solution for [13] and [14] when A = 2, but it showed there is no solution for [13] and [14] when A = 4, so this could not be the solution to the puzzle.

    The least pair of values that worked were 6 and 8. If A = 6, the spreadsheet showed (by exhaustive search) that B, C, and D must be a permutation of 0, 4, and 9; and indeed, when I tried all six permutations, I found that for B=9, C=0, D=4, the probability of drawing a matching pair from a bag of A+4, B+3, C+1, and D candies (i.e., before drawing the first pair) was 1/3, and the probability of drawing a matching pair from a bag of A+2, B+3, C+1, and D candies also was 1/3.

    It is therefore possible that you started with 10 green apples, 12 grapes, 1 cherry, and 4 oranges, a total of 27 candies.

    You could not have started with fewer candies. But I have not yet determined that you could not have started with more candies.

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  3. Somehow I got the impression initially that there was supposed to be a simpler (and more complete) solution than I found. I wonder what I missed.

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