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Math Help - simpe thought experiment that tricks people

  1. #1
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    simpe thought experiment that tricks people[RESOLVED]

    This thought experiment was presented to me the other day.
    Jar A has 100 red marbles and jar B has 100 blue marbles. If you take a handful of marbles out of jar A and put them in jar B and take a handful of marbles out of jar B and put them in jar A, which jar would have more of the non-original coloured marbles. A handful is an arbitrary constant, so both jars end up with 100 marbles but you don't know how many were transferred. So for example would jar A have more blue marbles than jar B has red marbles.

    The correct answer is they always will have the same amount. I can see that this is true, but I don't know why this is true. When I thought about it I thought jar B would have more non-original marbles because when red marbles are put in B, some of those red marbles may be taken back out (when a handful is taken from jar B) and put back to jar A. Why is it wrong to use probability in this situation?

    Another student commented, "I got the answer wrong because I used math". This statement is a falsehood because the wrong mathematical technique was used, right? Or does sometimes math fail us?
    Last edited by superdude; June 22nd 2010 at 11:18 AM. Reason: resolved
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  2. #2
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    I shall do a proof mathematically (well, semi-mathematical) first, then comment on why the given method doesn't work. Also, just to be clear, what the student did was use intuition. If he/she actually worked out the probability he/she should get the correct answer as well (I think the error in the argument of the thought experiment lies in that the extra marbles in Jar B was not taken into account)

    Let handful of marbles = x marbles.
    By 'new marbles' I mean 'non-original marbles'

    When you take out the marbles from Jar A

    Jar A has: 100 - x red marbles
    Jar B has: 100 blue marbles, x red marbles = 100+x marbles.

    Now you take marbles out from Jar B and put in Jar A.
    Say m red marbles and n blue marbles

    Jar A has: (100-x) red marbles + m red marbles +n blue marbles
    So Jar A has: (100-x+m) original marbles and n new marbles

    Jar B has: 100 blue marbles + x red marbles - m red marbles - n blue marbles
    So Jar B has: (100-n) original marbles and (x-m) red marbles

    But since the number of marbles taken out is the same, x = m+n

    Original marbles in A: (100-x+m) = (100-n) original marbles
    New marbles in A: n = (x-m) new marbles

    which corresponds to the expressions for the marbles in B.

    So it's clear that the number of original marbles left in each jar is independent of the proportion of red and blue marbles you take from Jar B.

    Why? Because the more red marbles you take out of Jar B, the less number of new marbles there will be for both Jars. i.e. the more red marbles you take from B, the less number of new marbles are left for B. That also means less blue marbles are transferred to jar A, leading to a decrease of new marbles in that jar as well.
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  3. #3
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    Quote Originally Posted by superdude View Post
    This thought experiment was presented to me the other day.
    Jar A has 100 red marbles and jar B has 100 blue marbles. If you take a handful of marbles out of jar A and put them in jar B and take a handful of marbles out of jar B and put them in jar A, which jar would have more of the non-original coloured marbles. So for example would jar A have more blue marbles than jar B has red marbles.

    The correct answer is they always will have the same amount. I can see that this is true, but I don't know why this is true. When I thought about it I thought jar B would have more non-original marbles because when red marbles are put in B, some of those red marbles may be taken back out (when a handful is taken from jar B) and put back to jar A. Why is it wrong to use probability in this situation?

    Another student commented, "I got the answer wrong because I used math". This statement is a falsehood because the wrong mathematical technique was used, right? Or does sometimes math fail us?
    I remember this question. When I heard it, I think it was in terms of cans of paint. But you didn't state it quite right. Here, we have to mention that after the first transfer of marbles, jar B is mixed so that the distribution is uniform, and also we can't say the amounts will always be equal, rather that on average they will be equal over a long period of time.

    EDIT: Actually, they always will be equal, and the stuff I grayed out above is not true; I just didn't think it through enough. My argument below is overly complex. Sorry about that. Feel free to disregard what follows.

    This is an example of something being counterintuitive. The Monty Hall problem is frequently cited; also take a look at, for example, these paradoxes and Nontransitive dice.

    But this particular problem can be easily dealt with using (correct) mathematics. Suppose a handful means 10 marbles. Then after the first transfer we have

    A: 90 red
    B: 10 red, 100 blue

    Let X be the number of blue marbles in A after the final transfer, and let Y be the number of red marbles in B after the final transfer. We will show that E(X) = E(Y), where E(X) denotes the expected value of X.

    P(X=0) = \displaystyle \frac{\binom{10}{10}\binom{100}{0}}{\binom{110}{10  }}

    P(X=1) = \displaystyle \frac{\binom{10}{9}\binom{100}{1}}{\binom{110}{10}  }

    \dots

    P(X=10) = \displaystyle \frac{\binom{10}{0}\binom{100}{10}}{\binom{110}{10  }}

    E(X) = 1\cdot P(X=1) + \dots + 10\cdot P(X=10) = \frac{100}{11}

    Likewise,

    P(Y=0) = \displaystyle \frac{\binom{10}{10}\binom{100}{0}}{\binom{110}{10  }}

    \dots

    P(Y=10) = \displaystyle \frac{\binom{10}{0}\binom{100}{10}}{\binom{110}{10  }}

    E(Y) = 1\cdot P(Y=1) + \dots + 10\cdot P(Y=10) = \frac{100}{11}

    It can be seen that these expressions are always equal.

    With cans of paint, it's possible to use a different approach, which I could provide if necessary.
    Last edited by undefined; June 19th 2010 at 05:07 PM.
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  4. #4
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    I like gusbob's solution, but here is another approach that may make the result more intuitive.

    Suppose we transfer just one red marble from A to B, so A now has 99 red marbles and B has 100 blue and one red. Now take one marble from B and put it in A. If the marble is red, there are now zero blue marbles in A and zero red marbles in B. If the marble is blue, there is now one blue marble in A and one red marble in B. So this case is "obvious".

    Now suppose we transfer n marbles from A to B and n marbles back from B to A. Break the transfer into n one-marble transfers and apply the one-marble result. If this seems like shakey reasoning (maybe it is), then use induction on n for a formal proof.
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