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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

Community
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user181304
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3 Answers3

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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: $\mathsf P(C=2 \mid D=1)$

$$\begin{align} & = \cfrac{\mathsf P(C=2)\cdot\mathsf P(D=1\mid C=2)}{\mathsf P(C=2)\cdot\mathsf P(D=1\mid C=2)+\mathsf P(C=1)\cdot\mathsf P(D=1\mid C=1)+\mathsf P(C=0)\cdot\mathsf P(D=1\mid C=0)} \\[1ex] & = \cfrac{\cfrac 1 3 \cdot\cfrac{2}{2}}{\cfrac 1 3 \cdot\cfrac{2}{2}+\cfrac 1 3 \cdot\cfrac{1}{2}+\cfrac 1 3 \cdot\cfrac{0}{2}} \\[1ex] & = \dfrac 2 3 \end{align}$$

Graham Kemp
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  • I don't get it...if only 2 cards have a green side, and you know that one card is only red, won't the probability just be 1/2 (either you have the RG or GG card)? – user181304 Feb 17 '15 at 03:10
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    @user181304 The probability of drawing each card is equal, but you will see a green side of a double green card twice as often as you will see a green side of the two-colour card. – Graham Kemp Feb 17 '15 at 03:45
  • When you see a green side then it's more likely that you have the double-green card, since this has twice as many green sides as the red-green card. – Graham Kemp Feb 17 '15 at 03:55
  • @GrahamKemp Interesting interpretation of the question and on re-reading yours is probably correct. I interpreted it that you were deliberately shown the green side i.e. a Monty Hall problem. – Dale M Feb 17 '15 at 22:02
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You are seeing one of three green sides. Two of the three green sides have green opposite. Therefore your chance of green opposite is $\frac23$.

paw88789
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Everything fine.

$$Pr (GG \mid G)=\frac{Pr GG}{Pr G}=Pr G=1/2.$$

  • But someone answered differently and now I am not so sure? – user181304 Feb 17 '15 at 03:13
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    @user181304 Because it is: $$\begin{align}P(GG \mid G) & = \frac{P(G\mid GG)P(GG)}{P(G\mid GG)P(GG)+P(G\mid RG)P(RG)} \ & = \frac{1}{1+\frac 1 2} \ & = \frac 2 3\end{align}$$ – Graham Kemp Feb 17 '15 at 04:03
  • @GrahamKemp where does bottom part of your equation come from? according to Bayes shouldn't it just be P(GG) ? – dark_ruby Nov 22 '22 at 21:25
  • No, @dark_ruby it is $\mathsf P(G)$. $~G$ is the event that you see green. $~GG$ is the event that both sides are green. $~RG$ is the event that the sides are red and green. $~$You see green if at least one side is green and you see one of those. $$\begin{align}\mathsf P(G) &=\mathsf P(GG)~\mathsf P(G\mid GG)+\mathsf P(RG)~\mathsf P(G\mid RG) \[1ex]&=\tfrac 13\cdotp(1+\tfrac 12)\end{align}$$ – Graham Kemp Nov 22 '22 at 22:46