A probability question

I'm with Alan, the probability is 2/3.

There are six possible outcomes. I can draw any one of the three cards and I can then see just one of its sides. Imagine a letter on each side of each card, so, card 1 side A, card 1 side B, card 2 side A etc. If I were to draw card number 1, I could see either A or I could  see B, that's two possibilities, not one. Same for the other two cards.

Of the six possible outcomes, three will be red and three will be white.

Of the three possible reds, two will have red the other side one will have white. (We have two possible ways in which the other side is red, one on which the other side is white.) So, the odds are 2 - 1 on the other side being red, a probability of 2/3.

Alan's first post shows this quite clearly.

You might have drawn 1.  Red/White  2. Red sideA/Red sideB 3. Red sideB/Red sideA  

There are two out of three possibilities that the other side is red, so the probability that the other side is also red is 2/3.

I believe that the answer is a half because the other side has to be either white or red.

the fact that it is red on one side is a given so the all white card has got nothing to do with it.

I'm not great at these kinds of problems, but I always like to give 'em a try!!!.........

Let A be the probabilty of one side showing "red".......and let B be the probabilty that both sides are red.

We already know that the double-sided white card doesn't matter, since we already have a card showing "red."

So we have

(Probability of Two sides Red l One side is Red) = P(A and B) /  P(A)

P(BlA) = [(1/2) * (1/2)] / (1/2)  =    (1/2)

Another way of looking at this is that we either have the double-sided red card, or we don't !!!

That's my two cents worth, anyway !!!

I'm afraid you're wrong guys!  This is a well known problem in recreational maths.  

P(A and B) = Probability of two sides red = 1/3 (there are 3 cards, one of which has two red sides).  

P(A) = Probability of one side red = 1/2 (there are 6 sides, 3 of which are red).

P(A and B)/P(A) = (1/3)/(1/2) = 2/3.

I'm sticking to my answer too.  I believe that is like saying that the chance of a child being a boy is 50%.  If you already have two boys and there is another baby on the way what is the chance of ending up with 3 boys.  it is not 0.5³   It is 0.5

Well the card is chosen.  It is definitly not the fully white one.  that is no longer a possibility so the probability that the other side will be white is 50%. the probablity that it is red is 50%.

I know Alan and Bertie are superior mathematicians but I am STILL sticking with my answer!

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NO I'M NOT - I'm Lady Guinevere - My mind can change as doth the vector of the breeze!

The maker of the glue shall be banished from the land of camelot for it has become Unstuck!

I'm afraid your analogy is misleading you Melody!  There is only one way of "choosing" a boy.  There are two ways of choosing a red side from a card with two red sides.

That wasn't the point Alan.

The point is that the card has already been chosen.  It has 1 red side. therefore

P(2whites sides)=p(2black sides )=p(2purple sides)=0

p(2red sides)=P(one red and one white)=0.5

How can you say differently?????

Please try and explain to me what is wrong with my logic. (or lack thereof)

Okay Alan 

I only just read your answer properly - Now I have something to think about - thank you.

Yea I get it now I think.

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Ok Ok I've changed my mind - I'm wrong - You're right - who is to be Sir Lancelot, head knight of camelot?

Is it Alan? Is it Bertie?  Let the jousting begin!

Melody

Instead of using just the colours red and white, suppose that on one card the colours are light red and dark red, on another they are an ever so slightly red and a dull white and on the third a gleaming white and a creamy white.

You draw a card at random and look at just one side.

In how many different ways can this event happen ?

Answer is six. You could be looking at any of the three different reds or any of the three different whites.

Of those six different outcomes how many would you classify as being red (or reddish) ? Answer is three.

Now look at the other sides of these three cards.

If we see light red, the other side is dark red.

If we see dark red the other side will be light red.

If we see an ever so slightly red, the other side will be a dull white.

That's 2 to 1 on the other side being red, and so a probablitiy of 2/3.

(Sorry, Melody's last answer wasn't there when I started typing, but I'll leave mine anyway.)

And BTW, I doubt that my eyesight is sufficiently good to direct a lance successfully !

And I've never ridden a horse.

Don't worry Alan most likely has the same short falls - Camelot is doomed!

Thanks for the explanation.