What are the possible correct answers for this probability word question?

Question

There is a puzzle I read long ago and whose explanations never left me entirely satisfied.

Recently I've encountered the Boy or Girl paradox, including in particular the line:

The moral of the story is that these probabilities do not just depend on the known information, but on how that information was obtained.

I believe this may have some application to this puzzle, but I'm at a loss to see how.

Also relevant seems to be the aspect of the Monty Hall puzzle that you must be guaranteed to be shown a goat before the game starts for switching doors to have any advantage—but, again, I don't quite see the direct connection to this puzzle.

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Here is the setup:

There are three cards. One card is green on both sides, one is red on both sides, and one is green on one side and red on the other.

You shuffle the cards (with your eyes closed, one would assume) and deal all three out on a table, then cover each card (with flower pots) before opening your eyes.

You uncover a single card. The visible face is green.

What are the odds that the other side of that card is green as well?

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The possible answers seem to be $\frac 1 2$ and $\frac 2 3$, but I'm not sure if there are different possible assumptions/interpretations which could affect the answer.

(The "correct" answer in the book I was reading was $\frac 2 3$.)

1. Under what conditions (meanings, interpretations, scenarios) would $\frac 2 3$ be the correct answer?
2. In what scenarios would $\frac 1 2$ be the correct answer?
3. Are there any other possible correct answers (given other interpretations)?

Answer 1

$\frac{1}{2},$ in fact, is essentially never right and the answer is $\frac{2}{3}.$

Consider the six "sides" to the three cards: R1, G1, R2, R3, G2, and G3. Card 1 has R1/G1, Card 2 R2/R3, and the last card G2/G3. So G1/G2/G3 all show up as green.

But wait. Two of these sides belong to card 3 (which gives green on the other side)... so it's 2 sides in 3, or $\frac{2}{3}.$

The common misconception is $\frac{1}{2}.$ There are no other reasonable answers. The logic behind this is that since there's two cards with green there's a 1/2 chance. People using this logic fail to realize that the probabilities of the cards being chosen are weighted, and card 3 has a $\frac{2}{3}$ chance of being picked in the first place.

Answer 2

The fact that you're looking at a green side tells you a few things:

1. You're NOT holding the red/red card.

2. The green side could be side A of the green/green card, side B of the green/green card, or the green side of the red/green card, but you don't know which one.

Thinking about #2, there are 3 distinct possibilities. In two of them, the green card is green on the other side. In other words, in 2 of 3 possible cases, the other side is green. Thus, the probability of the other side being green is 2/3.

Answer 3

It would be $\frac{1}{2}$ if you knew that the red/green card was shuffled with the green side up.

It is $\frac{2}{3}$ without that knowledge.

Answer 4

As I see it, I think the idea is that the $RG$ card could have been shuffled in two different but equally likely ways: one with the $R$ facing upwards, and one with the $G$ face facing upwards.

I tried using classical probability while being as explicit/obvious as possible in each step.

Let $GG$ be the event 'The revealed card is green-green.'
Let $FG$ be the event 'The revealed face is green.'

Then we wish to calculate the conditional probability

$$\mathbb{P}(GG|FG)=\frac{\mathbb{P}(GG\cap FG)}{\mathbb{P}(FG)}$$

Let $RGG$ be the event 'The red-green card has the green face upwards.'
Let $RGR$ be the event 'The red-green card has the red face upwards.'

Then by the law of total probability:

$$\mathbb{P}(GG\cap FG)=\mathbb{P}(GG\cap FG\cap RGG)+\mathbb{P}(GG\cap FG\cap RGR)$$

Now, we have that

\begin{align} \mathbb{P}(GG\cap FG\cap RGG)&=\underbrace{\mathbb{P}(GG|FG\cap RGG)}_{1/2}\cdot\mathbb{P}(FG\cap RGG)\\ \mathbb{P}(FG\cap RGG)&=\underbrace{\mathbb{P}(FG|RGG)}_{2/3}\cdot\underbrace{\mathbb{P}(RGG)}_{1/2}. \end{align}

Similarly,

\begin{align} \mathbb{P}(GG\cap FG\cap RGR)&=\underbrace{\mathbb{P}(GG|FG\cap RGR)}_{1}\cdot\mathbb{P}(FG\cap RGR)\\ \mathbb{P}(FG\cap RGR)&=\underbrace{\mathbb{P}(FG|RGR)}_{1/3}\cdot\underbrace{\mathbb{P}(RGR)}_{1/2}. \end{align}

Hence $\mathbb{P}(GG\cap FG)=\frac12\cdot(\frac23\cdot\frac12)+1\cdot(\frac13\cdot\frac12)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$.

It remains only to calculate $\mathbb{P}(FG)$. Once again, by the law of total probability:

\begin{align} \mathbb{P}(FG)&=\underbrace{\mathbb{P}(FG|RGG)}_{2/3}\cdot\underbrace{\mathbb{P}(RGG)}_{1/2}+\underbrace{\mathbb{P}(FG|RGR)}_{1/3}\cdot\underbrace{\mathbb{P}(RGR)}_{1/2}\\ &=\frac13+\frac16=\frac12. \end{align}

So that finally,

$$\mathbb{P}(GG|FG)=\frac{\frac13}{\frac12}=\frac23$$