Originally Posted by

**Soroban** Hello, holly123!

This is a classic trick question.

Here is an elementary explanation.

The "obvious" answer is $\displaystyle \frac{1}{2}$

. . (The other side is either Red or Green.)

However, this is wrong . . .

Let's label the six faces of the cards:

. . $\displaystyle \begin{array}{ccc}

\text{Card 1} & \text{Card 2} & \text{Card 3} \\

\boxed{R_1,R_2} & \boxed{G_1,G_2} & \boxed{R_3,G_3} \end{array}$

We draw one card and see a Green face.

. . (Obviously, we did *not* draw Card 1.)

There are *three* possible scenarios:

(1) You drew Card 2.

. . .You are looking at $\displaystyle G_1$; the other side is $\displaystyle G_2.$

(2) You drew Card 2.

. . .You are looking at $\displaystyle G_2$; the other side is $\displaystyle G_1.$

(3) You drew Card 3.

. . .You are looking at $\displaystyle G_3$; the other side is $\displaystyle R_3.$

In *two* out of the *three* cases, the other side is Green.

Therefore: .$\displaystyle P(\text{other side is Green}) \;=\;\dfrac{2}{3}$