Question: You have 50 blue marbles, 50 red marbles, and two jars to put them in. A marble will be selected at random from a jar selected at random. How do you divide the marbles among the jars as to maximize the probability of choosing a blue marble. You must use all the marbles.
As always, try solving the problem yourself or keep reading for the solution.
We know that the probability of picking either jar will always be 50%. The probability of picking a blue marble in either jar will depend on the number of blue marbles divided by the total number of marbles in that jar. Using this, we can construct a formula for the total probability of picking a blue marble – 50% times the probability of picking a blue in jar 1 plus 50% times the probability of picking blue in jar 2.
P(B) = 0.5(B1 / B1 + R1) + 0.5(B2 / B2 + R2)
B1 = blue marbles in jar 1
B2 = blue marbles in jar 2
R1 = red marbles in jar 1
R2 = red marbles in jar 2
What happens if we just split the marbles evenly, 25 of each in both jars:
P(B) = 0.5(25 / 50) + 0.5(25 / 50)
P(B) = 0.5
We would do no better than random. What if we keep the red split evenly and put all the blue marbles on one side:
P(B) = 0.5(50 / 75) + 0.5(0 / 25)
P(B) = 0.33
We end up doing worse. Dropping the probability of one jar to zero hurt us more than putting all the blue on one side helped. If going to 0% blue hurt, going to 100% blue in one jar should help. More specifically, we want to raise the probability of drawing blue in one jar to 100% while minimizing the loss of blue in the other jar. The solution is just to put one blue marble and no red marbles in one of the jars:
P(B) = 0.5(1 / 1) + 0.5(49 / 99)
P(B) = 0.74
It turns out that 74% is the best we can do.