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This question already has an answer here:

You have three cards in a hat. One is white on both sides, one is red on both sides, the third is white on one side and red on the other. You pull a card out of the hat and lay it down. The face is white. What are the chances that the other side of the card is also white?

Edit: to those who have marked this as a duplicate, I invite you to explain how it can be so when the answer in the "duplicate" is different from this one.

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marked as duplicate by KSmarts, Rand al'Thor, Gamow, mdc32, Alexis Feb 19 '15 at 13:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ knee jerk feeling is 1/2 but there has to be a trick. yeah it could be 2/3. $\endgroup$ – user3453281 Feb 17 '15 at 21:18
  • $\begingroup$ Nope, not the same. Solution is entirely different. $\endgroup$ – BobRodes Feb 17 '15 at 21:22
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    $\begingroup$ This is a retelling of the classic Bertrand's box paradox. en.wikipedia.org/wiki/Bertrand%27s_box_paradox $\endgroup$ – Engineer Toast Feb 18 '15 at 11:10
  • $\begingroup$ Precisely. It's also referred to as the "three-card swindle" because people use the fact of others not realizing the real probabilities to win bets. $\endgroup$ – BobRodes Feb 19 '15 at 3:34
  • $\begingroup$ The other question asks why it is 1/2 in one case and 1/3 in the other case. This is a duplicate of the 1/3 case. $\endgroup$ – f'' May 29 '16 at 19:02
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I think the correct answer is

2/3

We use the law of conditional probability.

Let $WW$ be the event that we get the all-white card, $RW$ the mixed card and $RR$ the all-red card. Let $W$ be the event that we get a white face up when we pull a card from the bag.

The law of conditional probability gives us $$P(WW \mid W) = \frac{P(WW \cap W)}{P(W)} = \frac{1/3}{1/2} = \frac{2}{3}$$ as $P(WW \cap W)$ simply reduces to the probability that we drew the all-white card, and half the faces are white, so $P(W)$ is a half.

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  • $\begingroup$ The mathematical solution. $\endgroup$ – BobRodes Feb 17 '15 at 22:19
  • $\begingroup$ Sample Space would be: {W1W2, W2W1, WR}, where first letter is faced up side and second letter is faced down side. Right? $\endgroup$ – Bhaskar Feb 19 '15 at 13:38
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The answer is obviously 2/3...

Note that the set of cards you describe is symmetrical with respect to swapping white and red surfaces, so you could have asked exactly the same sequence of questions with the words white and red swapped at each occurrence - i.e. "...the face is red. What are the chances that the other side of the card is also red?..." - and would get the same answer.

Using this symmetry the question is equivalent to asking - "What is the probability that the other side of the card is the same color as the exposed side?".

As there are three cards, and two of them are the same on both sides, there is a 2/3 probability that you will have chosen one of these two cards.

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  • $\begingroup$ The logical solution. $\endgroup$ – BobRodes Feb 17 '15 at 22:29
  • $\begingroup$ A great intuitive explanation! $\endgroup$ – xnor Feb 18 '15 at 0:16
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I think the correct answer is

2/3

let's imagine that we do this game 600 times. when we draw a red card on our first try, the game stops. this will happen about 300 times. for the other 300 (white) draws, we flip the card. one might think that the chances of revealing a red card is 50% then, but that would be wrong. we would get about 100 reds and 200 whites after a card flip. the trick is that even if drawing the W/R card is exactly as likely as drawing the W/W card, for half of the W/R draws, the game will stop before the card gets to be flipped.

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  • $\begingroup$ The brute force solution. $\endgroup$ – BobRodes Feb 17 '15 at 22:29
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You have three cards in a hat. One is white on both sides, one is red on both sides, the third is white on one side and red on the other. You pull a card out of the hat and lay it down. The face is white. What are the chances that the other side of the card is also white?

There are 3 cards with 2 sides each:

  • C1: W1 and W2
  • C2: R1 and R2
  • C3: W and R

When you draw a card and place it with one face up, you actually have 6 possible results:

  • C1 W1
  • C1 W2
  • C2 R1
  • C2 R2
  • C3 W
  • C3 R

The question then removes all the possibilities where a red face is showing, so the only possible results are the following faces:

  • C1 W1
  • C1 W2
  • C3 W

Then the question asks what's on the other side of the card, and these are the possible other sides of the card:

  • C1 W2
  • C1 W1
  • C3 R

So the chance that the other side of the card is white, based on the fact that the face up card is white, is 2/3.

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My gut-reaction to this was 1/2, and after reading everyone's answers I had a heck of a time wrapping my head around the correct answer. Here's the really simple, non-technical way I finally convinced myself:

Let's simplify the original experiment and imagine there are only two cards in the hat: a W-W card and a W-R card. Now let's reach in and draw one out, and imagine we see a white face. What are the chances that the other face is also white? Well, you night say, there are only two possible outcomes: I drew the all white card or I drew the white and red card. The answer must be 50%.

In fact, there are three possible outcomes: (1) You may have drawn red and white card, (2) you may have drawn the all white card, or (3) you may have drawn the all white card. No, that's not a typo. Yes, outcomes (2) and (3) are the same. Or are they? The all white card may be drawn in two different ways: either one white side up, or the other white side up. The unfortunate fact that both ways look the same is inconsequential.

In other words, looking at my white card, the possible outcomes are:

  1. I draw W, the flip side is R.
  2. I draw W1, the flip side is W2.
  3. I draw W2, the flip side is W1.

I don't know whether I'm seeing W, W1 or W2, but the chance the flip side is also white (that is, either W2 or W1) is 2/3.

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Another interpretation for those having a hard time to wrap their hand around 2/3.

Just label all card faces: A,B,C,D,E,F

The all white card gets A on one side B on the other. The white red card gets C on the white side and D on the red. The all red card gets E and F.

Our question now becomes: We pick a card and the upper Face shows a letter of (A,B,C) what is the chance we picked the A/B card?

Three possible faces and two of them belong to AB so chance is 2/3

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Answer

2/3

Reasoning

In randomly picking a card and randomly picking a side from that card, you have effectively picked a random card-side from all the card-sides in the hat. There are 3 white card-sides in total. 2 of which have another white side on the back. Given that you've picked a white card-side, it follows that there's a 2/3 chance you've picked one of the white card-sides with another white side on the back.

This reasoning extends to any number of cards with any number of colours:

Given that a randomly picked side of a randomly picked card is colour A, what is the probability that the opposite side is colour B?

Count the total number of card-sides of colour A. Call this $N_A$. Then count the number of those card-sides with colour B on the back. Call this $N_{A,B}$. Then the probability of the opposite side being colour B is $N_{A,B}/N_A$.

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I think the chances are 1/2, as you described your card.

Card 1: Red - Red  
Card 2: White - White  
Card 3: White - Red

And pulled card has face White. By this we are clear that its not the Card 1. So it can be either Card 2 or Card 3. As first face is White, Second face may be White or Red so the probability will 1/2.

Card 1: Red - Red  //This case can't be possible as we got first face is white
Card 2: White - White  //This is the first case (Second face may be white)
Card 3: White - Red    //This is the Second case (Second face may be Red)
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  • $\begingroup$ You're missing that there are actually two possible cases in Card 2: you could draw one side of it and you could draw the other side of it. If you think about it, you'll realize that if you draw white, you are twice as likely to have drawn one side of the white-white card as you are to have drawn the white side of the white-red card. Therefore, the other side of the card is twice as likely to be white. It's very easy to miss; it took me two or three hours to realize what I was doing wrong. $\endgroup$ – BobRodes Feb 19 '15 at 3:43
  • $\begingroup$ Another way of phrasing the question might be convincing. You have three cards: red-red, white-red, and white-blue. You pull a card and turn it up. The color is either white or blue. What are the chances that the other side is either white or blue? This is exactly analogous to the original question if you think about it. $\endgroup$ – BobRodes Feb 19 '15 at 3:57
  • $\begingroup$ @BobRodes, what are talking about, man. If you are getting face 1 white then other face will be either face 2 white (obviously not face 1 white) or it may be red . or if you are getting face 2 white other will be either face 1 white or red. $\endgroup$ – Himanshu Feb 19 '15 at 5:25
  • $\begingroup$ Let's use face 1 as one side of WW, face 2 as the other, face 3 as one side of WR. You draw a card, and it comes up white, so you turned up either face 1, face 2, or face 3. (Agreed?) Now, if you turn over face 1, you get face 2. (Agreed?) If you turn over face 2, you get face 1. (Agreed?) If you turn over face 3, you get red. Two white possibilities and one red possibility, or 2 out of 3 are white. Does that help? :) $\endgroup$ – BobRodes Feb 20 '15 at 18:17
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For those who find the fact that the answer is not 1/2 hard to swallow, some intuitive explanations have already been proposed. Here's another.

When you see a white face, it's more likely you're looking at the white-white card than the white-red card, because that's where most of the white faces are.

To scale up the numbers a little, imagine you have two six-sided dice, only on one of them you've replaced all the numbers with sixes. I choose one of these dice randomly and roll it and get a six. Which die do you think it is?

Or let's make it more extreme. I have two lottery machines, only one of them always churns out the balls 1 to 6 in order. I pick one of the machines at random, play a lottery game, and win. Which machine did I pick?

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