Fifty red, fifty blue, two jars, one blind draw

Question

You have 50 red marbles and 50 blue marbles and two empty jars. You may divide the marbles between the jars in any way you like, as long as every marble goes into a jar and neither jar is empty. Afterwards a jar is chosen at random (each with probability 1/2), and then one marble is drawn at random from that jar. How should you distribute the marbles to maximise the probability that the drawn marble is red, and what is that probability?

Accepted Answer

Put a single red marble alone in one jar, and the other 99 marbles (49 red, 50 blue) in the second jar. The chance of red is then 149/198, about 74.7 per cent. The calculation. With probability 1/2 the lone-marble jar is chosen and the marble is certainly red. With probability 1/2 the big jar is chosen and the marble is red with probability 49/99. Total: 1/2 × 1 + 1/2 × 49/99 = 1/2 + 49/198 = 149/198, about 0.747. Why this is best. The overall probability is the average of the two jars' red fractions. To push the average up you want both fractions high, but every red marble you move into one …

Fifty red, fifty blue, two jars, one blind draw

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