A box contains 3 cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random , and the color on one side is observed. If this side is green, what is the probability that the other side is also green?
My attempt:
3 cards
R R
G G
R G
3/6=1/2 chance of getting green
P(GG|G)=$\frac{1/2)(1/2)}{1/2}$
$=1/2$
Does this look ok?
Same question I jsut found: Conditional probability question with cards where the other side color is to be guessed
Now I am torn between 1/2 and 2/3
Best Answer
No, this scenario is counter intuitive.
The model is that a card is picked at random, and then a side is displayed at random.
There are six equally probable results of this -- one for each face. Three of these result in displaying a green side and only one of those will have red on the other side.
The probability that the other side is also green given that a green side is displayed is: $2/3$.
Remark The information that "a green side is displayed" is not the same as knowing "the card has a green side", or even that "the card isn't double red."
By listing all the outcomes of the card drawn and side displayed (first or second).
$$\{\overbrace{\underbrace{GG:1, GG:2, GR:1}_\text{you see a green side}, GR:2}^\text{the card isn't double red}, RR:1, RR:2\}$$
The likelihood of drawing each card is equal, but if a card with two green sides is drawn it is twice as likely to display a green side than if the card with one green side were drawn. Therefore, if a green side is displayed the card is twice as likely to be a double green card than a single green card.
Let $C$ be the card drawn, taking values $\{0, 1, 2\}$ for the number of green sides. (So $C=0$ is the red-red card, et cetera).
Let $D$ be the colour displayed, taking values $\{0, 1\}$ for red or green respectively.
So we want to find: $\mathsf P(C=2 \mid D=1)$
$$\begin{align} & = \cfrac{\mathsf P(C=2)\cdot\mathsf P(D=1\mid C=2)}{\mathsf P(C=2)\cdot\mathsf P(D=1\mid C=2)+\mathsf P(C=1)\cdot\mathsf P(D=1\mid C=1)+\mathsf P(C=0)\cdot\mathsf P(D=1\mid C=0)} \\[1ex] & = \cfrac{\cfrac 1 3 \cdot\cfrac{2}{2}}{\cfrac 1 3 \cdot\cfrac{2}{2}+\cfrac 1 3 \cdot\cfrac{1}{2}+\cfrac 1 3 \cdot\cfrac{0}{2}} \\[1ex] & = \dfrac 2 3 \end{align}$$