**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.

### 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(B

_{1}/ B_{1}+ R_{1}) + 0.5(B_{2}/ B_{2}+ R_{2})B

_{1}= blue marbles in jar 1

B_{2}= blue marbles in jar 2

R_{1}= red marbles in jar 1

R_{2}= 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.