Game of Chance: Exposing Misconceptions

Initially, when the marbles are put into the bag, there are four equally likely possibilities:

A) The first marble dropped in is red , and the second is also red.
B) The first marble dropped in is red , and the second is blue.
C) The first marble dropped in is blue , and the second is red.
D) The first marble dropped in is blue , and the second is also blue.

When the first red marble is picked, it could be any of these possibilities, and all of them are equally likely :

1) The first red marble from A.

2) The second red marble from A.

3) The red marble from B.

4) The red marble from C.

(Bag D is now entirely out of consideration.)

This means in 2 out of the 4 cases, the bag is an all-red bag, and in the other 2 cases, it is a mixed blue-red bag. So, there is a $\frac{1}{2}$ probability of the bag being mixed.

For the next pull to be blue, two things need to happen. One, the bag pick needs be a mixed color bag (a $\frac{1}{2}$ probability), and two, the blue marble instead of red needs to be drawn (another $\frac{1}{2}$ probability). So the overall probability is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of pulling a blue marble.

(The most common mistake people make with this problem is thinking the initial red marble pull only eliminates bag D, while still considering bags A, B, and C to be equally likely. The initial red marble pull does eliminate bag D, but it also tilts the probability towards the bag being A.)