Puzzle | 50 red marbles and 50 blue marbles (original) (raw)

Last Updated : 6 Jun, 2026

There are two boxes, B1 and B2. One of the boxes contains 50 red marbles, and the other contains 50 blue marbles. You are allowed to redistribute the marbles between the two boxes in any way you like, without changing the total number of red and blue marbles (i.e., 50 red and 50 blue marbles in total).

A box is then selected at random, and from the chosen box, one marble is selected at random.

maximize

**Your task is to determine how to redistribute the marbles between the two boxes in order to maximize the probability of selecting a red marble.

Check if you were right -full answer with solution below.

**Solution:

49_red_50_blue

Let J1 and J2 denote the total number of marbles in Box B1 and Box B2, respectively, after redistribution of the marbles and P(R) be the probability of picking a red marble.

**P(R) = P(B1) * P(B1 | J1) + P(B2) * P(B2 | J2)

Let **P(B1) and **P(B2) represent the probabilities of selecting Box B1 and Box B2, respectively. Since the box is selected at random and equally likely,

P(B1)=P(B2)=1/2

**Case 1: Without Redistribution

If no marbles are moved, Box B1 contains 50 red marbles and Box B2 contains 50 blue marbles.

The probability of picking a red marble from B1 is 1, and from B2 is 0. Therefore:

P(R) = ((1 / 2) * 1) + ((1 / 2) * 0) = 0.5

**Case 2: With Redistribution

If we decrease the number of red marbles in B1 and increase the number of red marbles in B2, the probability of getting a red marble will be maximized.

To maximize P(R), place 1 red marble in B1 and move the remaining 49 red marbles to B2. After redistribution:

Therefore:

**P (R) = ((1 / 2) * (1 / 1)) + ((1 / 2) * (49 / 99)) = 0.747474

Hence,

the maximum probability of choosing a red ball is **0.747474