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Anne and Bernard have between them 100 marbles, Anne the most. Some in each bag of their marbles are red, the rest, though fewer, are blue. If Anne draws at random two marbles from her bag of marbles, the probability that the two marbles are of different colors is preciesely the same as the probability that Bernard draws two marbles at random from his own bag, and they too are of different colors.

How many marbles of each color do Anne and Bernard have?

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  • $\begingroup$ Are the chances nonzero (i.e. both Anne and Bernard have one ore more red marbles and one ore more blue marbles)? $\endgroup$
    – Bubbler
    Dec 7, 2020 at 23:09
  • $\begingroup$ @Bubbler Indeed! $\endgroup$ Dec 7, 2020 at 23:15
  • 2
    $\begingroup$ It feels like you're trying a bit too hard to minimize the number of words :-). $\endgroup$
    – Gareth McCaughan
    Dec 7, 2020 at 23:49
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    $\begingroup$ This is looking more and more like routine mathematical manipulation plus computer search, which it's hard to see as a puzzle in any meaningful sense. @BernardoRecamánSantos, is there something we're all missing? $\endgroup$
    – Gareth McCaughan
    Dec 8, 2020 at 0:14
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    $\begingroup$ @GarethMcCaughan Yes. $\endgroup$ Dec 8, 2020 at 2:07

2 Answers 2

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OK I'll bite.

I find six independent solutions:

Anne red | Anne blue | Bernard red | Bernard blue
_________________________________________________
   28    |     27    |     24      |      21
   30    |     25    |     25      |      20
   34    |     21    |     28      |      17
   36    |     28    |     21      |      15
   39    |     16    |     32      |      13
   44    |     20    |     25      |      11

For each of these solutions, the total number of marbles is 100, Anne has more marbles than Bernard, and each person has fewer blue marbles than red marbles.

Finally, the likelihood of each person drawing two marbles and them being different colours:

The probability here is given by the sum of the probabilities P(red then blue) + P(blue then red):

P(red then blue) = P(red)*P(blue given red)

                 =     Nr           Nb
                   --------- x -------------
                   (Nr + Nb)   (Nr + Nb - 1)

                 =        Nr x Nb
                   -------------------------
                   (Nr + Nb) x (Nr + Nb - 1)

and by symmetry, P(blue then red) is the same calculation, so

P(red then blue) + P(blue then red) =
                          2 x Nr x Nb
                   -------------------------
                   (Nr + Nb) x (Nr + Nb - 1)

Calculating this figure for both Anne and Bernard separately yields the following probabilities (rounded for display) for each of the above solutions:

  Anne   |  Bernard
____________________
0.509091 | 0.509091
0.505051 | 0.505051
0.480808 | 0.480808
0.5      | 0.5
0.420202 | 0.420202
0.436508 | 0.436508

I found these solutions with a simple computer script, but have also confirmed several of them by hand. I also altered my script to operated entirely with integers to avoid potential floating-point problems (though it didn't change the results).

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enter image description here

solving using basics and chances of picking different color depends on the lesser number.

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    $\begingroup$ A picture of hand-written explanation is hard to read and understand. Can you explain in plain text? You can use MathJax to enter math expressions (tutorial is here). $\endgroup$
    – Bubbler
    Dec 8, 2020 at 4:51
  • $\begingroup$ Also, can you elaborate on "chances of picking different color depends on the lesser number" and the equation $\frac{T-x}{T}=\frac{100-T-y}{100-T}$? I don't think it agrees with my (and lxop's) result of $\frac{2x(T-x)}{T(T-1)}=\frac{2y(100-T-y)}{(100-T)(99-T)}$ using your variables. $\endgroup$
    – Bubbler
    Dec 8, 2020 at 7:06

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